Unit 1: Transmission Genetics

DSC-14 · TRANSMISSION AND MOLECULAR GENETICS Unit 1: Transmission Genetics Heredity and variation, Mendel and his extensions, chromosome theory, probability, gene interactions and polygenic traits. Every topic ends with recap points and exam questions with answers. Course: B.Sc. (Hons) Biological Sciences, Sem V Unit 4 notes: separate file CONTENTS 1. Principles of heredity 2. Genetic variation 3. Molecular basis of genetic information 4. Mendel's work on transmission of traits 5. Principles of inheritance and chromosome theory 6. Laws of probability 7. Incomplete dominance 8. Codominance 9. Multiple alleles 10. Lethal alleles 11. Epistasis and modified ratios 12. Pleiotropy 13. Penetrance and expressivity 14. Norm of reaction 15. Phenocopy 16. Polygenic inheritance 17. Continuous vs discontinuous variation, and heritability DSC-14 Unit 1 Notes | page 1 of 85 How to use these notes: green boxes are key ideas, red boxes are exam traps, gold boxes are exam questions. Answers are printed under each question, so cover them and try the question on paper first. Diagrams are drawn for these notes so they load anywhere and work in dark mode. UNIT 1 Transmission Genetics Principles of heredity to polygenic inheritance and heritability. DSC-14 Unit 1 Notes | page 2 of 85 1.1 Principles of heredity Genetics is the study of heredity (how traits pass from parents to offspring) and variation (why offspring differ from parents and from each other). This first topic sets the vocabulary and history that every later topic uses. Branches of genetics Transmission genetics: how traits are passed from one generation to the next, how genes are arranged on chromosomes, and gene mapping. This is the branch this whole unit belongs to. Molecular genetics: how genetic information is encoded, replicated and expressed (replication, transcription, translation). It focuses on gene structure, organisation and function. Population genetics: the study of the group of genes found in a population, and so fundamentally the study of evolution. Model organisms Six organisms have been studied so intensively that most of modern genetics comes from them. They share short life cycles, many offspring, cheap upkeep and simple genetics, which is what makes crosses and statistics practical. Organism Common name Why genetics likes it Drosophila Fruit fly 10 day life cycle, giant salivary gland chromosomes, easy to score melanogaster Escherichia coli mutants (Morgan) Gut bacterium Divides every 20 minutes, haploid, ideal for mutants, conjugation and gene expression Caenorhabditis Nematode Transparent, exactly 959 somatic cells in the adult hermaphrodite, elegans worm cell lineage fully mapped Arabidopsis thaliana Thale cress Small plant, tiny genome, self fertilises, seeds in about 6 weeks Mus musculus House mouse A mammal with genetics close to humans, knockout mice are routine Saccharomyces cerevisiae Baker's yeast Single celled eukaryote, can be haploid or diploid, easy to grow and manipulate DSC-14 Unit 1 Notes | page 3 of 85 Ideas about heredity before Mendel Before 1866 people had guesses rather than data. It matters to know each idea and, more importantly, why it fails, because exam questions ask you to differentiate them. Idea Who What it claimed Why it fails Pangenesis Hippocrates; Every part of the body sheds Galton transfused blood between Darwin revived it as tiny particles (gemmules) rabbit varieties (1871) and the gemmules (1868) that travel to the offspring showed no mixing, so no reproductive organs and are gemmules travel in blood. Also passed to the embryo at cannot explain why a lost limb conception does not pass on Leeuwenhoek and A tiny, complete miniature Would need infinite nesting of Hartsoeker (1600s) adult (a homunculus) sits homunculi, and cannot explain inside the sperm (spermists) resemblance to both parents. or egg (ovists) and just Development is epigenetic (new enlarges during structures form step by step) Preformationism development Inheritance of Lamarck Traits acquired during a Weismann cut the tails of mice for acquired lifetime become part of 22 generations and every characteristics hereditary information and newborn still had a normal tail. are passed on Only germ line changes are inherited Blending Kolreuter, Galton; Offspring are a blend or Variation would vanish within a inheritance used against mixture of parental traits, few generations. And the Darwin by and the hereditary material recessive trait reappears Fleeming Jenkin itself blends and cannot be unchanged in F2, which blending separated again cannot explain (Mendel) Exam trap: Pangenesis needs a material (gemmules) coming from body parts to the gonads. Inheritance of acquired characteristics is about traits gained during life. Weismann's mouse tail experiment disproves the second, and Galton's rabbit transfusion tests the first. Do not mix the two. DSC-14 Unit 1 Notes | page 4 of 85 Timeline you should know Year Who Contribution c. 520 Alcmaeon of Croton Dissected animals, proposed that the brain is the seat of perception and BC the origin of semen 1833 Robert Brown Described the cell nucleus 1838 to Schleiden and Schwann Cell theory 1859 Charles Darwin Evolution by natural selection, On the Origin of Species 1866 Gregor Mendel Principles of heredity from pea plant experiments 1879 Walther Flemming Described mitosis and chromosome behaviour; by 1885 the nucleus 39 was accepted as the seat of heredity 1885 August Weismann Germ plasm theory: reproductive cells carry a complete set of genetic information passed to egg and sperm 1900 de Vries, Correns, Independently rediscovered Mendel's results Tschermak 1902 Sutton and Boveri Chromosome theory of inheritance: genes are located on chromosomes 1902 Archibald Garrod Inborn errors of metabolism (alkaptonuria): first link between a gene and an enzyme 1905 to Bateson; Johannsen 09 1910 Bateson coined "genetics"; Johannsen coined gene, genotype and phenotype T. H. Morgan Discovered the first mutant (white eye) in Drosophila; showed a gene is on the X chromosome 1928 Griffith Transformation in Streptococcus pneumoniae 1930s Fisher, Haldane, Wright Foundation of population genetics 1940s Bacteria and viruses Rapid reproduction and simple genetic systems allowed detailed study enter genetics of gene organisation and structure Watson, Crick, Wilkins, 3D structure of DNA 1953 Franklin 1966 to Molecular genetics 73 advances Led to the first recombinant DNA experiments DSC-14 Unit 1 Notes | page 5 of 85 Year Who Contribution 1977 Gilbert and Sanger Methods for DNA sequencing 1983 Kary Mullis PCR technique 1990 First gene therapy on a human genetic disease; Human Genome Project launched Key terms (the working vocabulary) Term Meaning Heredity Transmission of traits from parents to offspring Variation Differences in characteristics among individuals of a population Gene Functional unit of heredity, a DNA sequence controlling a trait or product Allele Alternative form of a gene Locus (plural loci) Specific position of a gene on a chromosome Genome Complete genetic material of an organism Chromosome / DNA and protein structure carrying genes; a chromatid is one of the two identical chromatid copies of a replicated chromosome Homologous Pair of chromosomes with the same genes at corresponding loci, one from each chromosomes parent Genotype / phenotype Genetic constitution of an organism / its observable characteristics Homozygous / Two identical alleles (AA or aa) / two different alleles (Aa) heterozygous Hemizygous Only one copy of a gene present, such as X linked genes in a male Dominant / recessive Allele expressed in the heterozygote / allele expressed only when no dominant allele is present Wild type The most common naturally occurring form of a trait or allele Mutant Organism or cell carrying a mutation Hybrid Offspring of genetically different parents DSC-14 Unit 1 Notes | page 6 of 85 Term Meaning True breeding (pure A line that produces only its own type when self crossed, meaning it is homozygous line) Test cross Crossing an individual of unknown genotype with a homozygous recessive individual Back cross Crossing an F1 hybrid with either of its parents Reciprocal cross Two crosses where the sexes of the parents for each trait are swapped (A female x B male, then B female x A male) Segregation Separation of the two alleles of a gene during gamete formation Independent assortment Alleles of different genes assort independently during gamete formation when the genes are unlinked Linkage Tendency of genes located close together on the same chromosome to be inherited together Non-disjunction Failure of chromosomes to separate properly during cell division Chromosomes, chromatids and DNA molecules through the cell cycle This is a favourite trap. Count chromosomes by centromeres, and count DNA molecules by chromatids. Take a diploid cell with 2n = 12. Stage Chromosomes DNA Reason molecules G1 12 12 Each chromosome is a single chromatid After S phase 12 24 DNA replicated, each chromosome now has two sister (G2) Metaphase of chromatids joined at one centromere 12 24 Still 12 centromeres 24 24 Centromeres split, each chromatid is now counted as mitosis Anaphase of mitosis Each daughter a chromosome 12 12 Back to the starting state 6 12 Homologues separated, chromatids still paired cell (G1) Meiosis I end (each cell) DSC-14 Unit 1 Notes | page 7 of 85 Stage Chromosomes DNA Reason molecules Gametes 6 6 Haploid, one chromatid per chromosome QUICK RECAP Genetics has three branches: transmission, molecular and population; this unit is transmission genetics Pangenesis, preformationism, acquired characters and blending inheritance were all disproved: know the test used against each Weismann (mouse tails, germ plasm) and Galton (rabbit blood transfusion) are the standard evidence Chromosome count follows centromeres, DNA molecule count follows chromatids: 12 chromosomes in G1 means 24 DNA molecules in G2 EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. State true or false and justify: A cell with 12 chromosomes in G1 will have 24 DNA molecules in G2. past paper style Answer True. In G1 each chromosome is a single chromatid, so 12 chromosomes contain 12 DNA molecules. S phase replicates every DNA molecule, so in G2 each of the 12 chromosomes has two sister chromatids and therefore 24 DNA molecules. The chromosome number is still 12 because the sister chromatids share one centromere. DSC-14 Unit 1 Notes | page 8 of 85 Q2. Differentiate between pangenesis and preformationism. past paper Answer Pangenesis: particles (gemmules) from every body part travel to the gonads and are passed to the embryo, so the embryo is built from the contributions of the whole body. Disproved by Galton's rabbit blood transfusion experiment. Preformationism: a complete miniature adult (homunculus) already exists in the egg or sperm and only grows in size. It implies no new structure forms in development and fails to explain inheritance from both parents. Key contrast: in pangenesis the embryo is assembled from particles gathered from the body, in preformationism it is already built and only enlarges. Q3. A student says: Weismann's experiment of cutting off mouse tails for 22 generations disproves Darwin's theory of pangenesis. Is this correct? hard Answer No. Weismann's experiment tests the inheritance of acquired characteristics (Lamarck): a tail lost during life is not passed on, so somatic changes do not enter the germ line. Pangenesis is a claim about gemmules carried from body parts to gonads and it was tested by Galton's blood transfusion experiment in rabbits. Notice too that a lost tail is an injury and not really a heritable trait, so the experiment is a weak test of even Lamarck's idea. The value of Weismann's work is the germ plasm concept: the germ line is separate from the soma. Q4. A cross between a red flowered and a white flowered plant gives pink F1. A student calls this blending inheritance. What single result from the F2 would prove him wrong? hard Answer In the F2 pure red and pure white plants reappear (in a 1:2:1 ratio). If the hereditary material had truly blended, the red and white determinants could not separate again, and every F2 plant would be pink or some intermediate shade. Reappearance of the original parental types shows the alleles stayed distinct. DSC-14 Unit 1 Notes | page 9 of 85 Q5. For a diploid organism with 2n = 8, give the number of chromosomes and DNA molecules at (a) G2, (b) anaphase of mitosis, (c) anaphase I of meiosis, (d) a mature gamete. hard Answer (a) G2: 8 chromosomes, 16 DNA molecules. (b) Mitotic anaphase: 16 chromosomes (sister chromatids split), 16 DNA molecules in total across the cell. (c) Anaphase I: the homologues separate but centromeres do not divide, so still 8 chromosomes in the cell (4 moving to each pole) and 16 DNA molecules. (d) Gamete: 4 chromosomes, 4 DNA molecules. DSC-14 Unit 1 Notes | page 10 of 85 1.2 Genetic variation Variation is the raw material of genetics. If all offspring were identical to their parents there would be nothing to study, and nothing for evolution to act on. Genetic variation means differences among individuals that are caused by differences in their DNA and can be inherited. Heritable and non-heritable variation Genetic (heritable) variation: different alleles or different combinations of alleles. Passed to offspring through the germ line. Environmental variation: differences caused by nutrition, temperature, disease, injury and so on. Not inherited. Identical twins raised apart show only this kind of difference. Most real traits are shaped by both: phenotype = genotype + environment (+ their interaction). Topic 14 (norm of reaction) makes this precise. Sources of genetic variation Source What happens Why it creates variation Mutation Change in DNA sequence (base The only source of brand new alleles. substitution, insertion or deletion), or in Everything else reshuffles existing alleles chromosome number or structure Independent Maternal and paternal chromosomes line A human can make 223 = 8,388,608 different assortment up at random at metaphase I gametes from chromosome assortment alone Crossing over Exchange between homologous chromatids Creates new allele combinations on a single in prophase I chromosome; makes the number of gamete types practically unlimited Random Any sperm can fuse with any egg Two parents can produce (223)2, about 7 x 1013 genetically different zygotes, even before fertilisation crossing over Gene flow Alleles move between populations Brings new alleles into a population Random change in allele frequency, strong Changes which alleles are present in a in small populations population (migration) Genetic drift DSC-14 Unit 1 Notes | page 11 of 85 Levels at which variation is measured Morphological: visible traits such as seed shape or flower colour (the ones Mendel scored). Biochemical: differences in proteins or enzymes, for example blood groups or enzyme activity. Molecular: differences in DNA itself, such as single nucleotide polymorphisms (SNPs), insertions and deletions, and copy number variants. Two unrelated humans differ at roughly 1 in 1000 nucleotides. Variation can also be discontinuous (clear cut classes) or continuous (a smooth range). That distinction is Topic 17. How many gamete types? The counting rule For an individual heterozygous at n genes on n different chromosome pairs: gamete types = 2^n F2 genotypes = 3^n F2 zygote combinations = 4^n n Gamete types 2^n F2 genotypes 3^n 1 2 3 2 4 9 3 8 27 5 32 243 10 1024 59,049 23 8,388,608 9.4 x 10^10 Key idea: Mutation makes new alleles. Recombination (assortment and crossing over) and random fertilisation shuffle them into new combinations. Selection and drift decide which combinations survive. DSC-14 Unit 1 Notes | page 12 of 85 QUICK RECAP Genetic variation is heritable, environmental variation is not; most traits have both Mutation is the only source of new alleles; assortment, crossing over and fertilisation only recombine existing ones n heterozygous genes on different chromosomes give 2^n gamete types and 3^n F2 genotypes Humans: 2^23 gametes from assortment alone, and about 7 x 10^13 possible zygotes per couple EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. A plant is heterozygous for 6 genes located on 6 different chromosome pairs. (a) How many gamete types can it make? (b) On selfing, how many different genotypes and phenotypes (complete dominance) appear in the offspring? numerical Answer (a) 26 = 64 gamete types. (b) Genotypes: 36 = 729. Phenotypes with complete dominance: 26 = 64. Q2. Two genetically identical clones of a plant are grown, one on rich soil and one on poor soil, and differ in height. Is this genetic variation? What does the difference tell you? concept Answer No, this is environmental variation. The genotypes are identical, so the height difference is caused by the environment. It also shows the genotype does not fix a single phenotype but allows a range (the norm of reaction, Topic 14). DSC-14 Unit 1 Notes | page 13 of 85 Q3. Why is mutation called the ultimate source of variation even though recombination produces far more new combinations each generation? hard Answer Recombination can only rearrange alleles that already exist. Without mutation, the pool of alleles would be fixed and would eventually run out of new combinations. Mutation is what creates new alleles in the first place, so it is the ultimate source, while recombination is the main day to day generator of new genotypes. Q4. An individual is AaBbCc where A and B are linked and lie very close together on one chromosome, and C is on another chromosome. Can you still use 23 = 8 as the number of gamete types? Explain. hard Answer The rule 2n only works for genes on different chromosomes (independent assortment). Here A and B are tightly linked, so the gametes AB and ab (parental types) are far more frequent than Ab and aB (rare recombinants). There are still up to 8 types possible in principle, but they are not equally frequent, so the 1:1:1:1... expectation fails. Linkage is Unit 3. DSC-14 Unit 1 Notes | page 14 of 85 1.3 Molecular basis of genetic information Mendel's "factors" were abstract. This topic covers the experiments that proved genes are made of nucleic acid, plus the modern molecular meaning of gene, allele and dominance. The structure of DNA itself (Watson and Crick, base pairing) is covered in your molecular biology notes. What genetic material must be able to do 1. Store information for building the organism. 2. Replicate accurately so it passes to daughter cells. 3. Express that information as proteins and RNAs. 4. Change rarely (mutation), giving the variation that evolution needs. Until the 1940s many biologists thought proteins were the better candidate because of their chemical variety (20 amino acids against only 4 nucleotides). Griffith's transformation experiment (1928) Frederick Griffith worked with Streptococcus pneumoniae. The S strain makes a polysaccharide capsule (smooth colonies) and is virulent. The R strain has no capsule (rough colonies) and is harmless. 1. Live S strain 2. Live R strain 3. Heat-killed S smooth, capsule rough, no capsule killed by heat mixed, then injected Mouse dies Mouse lives Mouse lives Mouse dies Live S bacteria found in blood 4. R + heat-killed S Live S bacteria found: R cells were transformed Griffith's four injections. The key result is panel 4: dead S cells plus live R cells kill the mouse, and live S bacteria are recovered. Griffith concluded that some "transforming principle" from the dead S cells changed live R cells into S cells, permanently and heritably. He did not know what the principle was. DSC-14 Unit 1 Notes | page 15 of 85 Why it was not simple reversion: the R strain Griffith used was derived from type II, and the dead S cells were type III. The bacteria recovered were type III S. A mutation back from R would have given type II S. So the R cells really took up something from the dead type III cells. Avery, MacLeod and McCarty (1944) They fractionated heat-killed S cells and tested each fraction for transforming ability, then treated the active extract with enzymes: Treatment of the extract Transformation? Meaning Protease (destroys protein) Still happens Protein is not the principle RNase (destroys RNA) Still happens RNA is not the principle DNase (destroys DNA) Lost DNA is the transforming principle Chemical analysis of the purified active Nitrogen to phosphorus ratio matched DNA, fraction not protein Many scientists still doubted it, because trace protein could not be ruled out and DNA was thought too simple. The Hershey and Chase experiment settled the argument. Hershey and Chase (1952) They used bacteriophage T2, which is only DNA and protein. Protein contains sulphur (cysteine, methionine) but almost no phosphorus. DNA contains phosphorus (in the backbone) but no sulphur. That gives two clean labels. DSC-14 Unit 1 Notes | page 16 of 85 Experiment 1: label the protein coat with 35S Phage protein coat labelled with 35S (sulphur is in protein, not in DNA) Phage attaches to E. coli and injects its genetic material into the cell Blender shakes off the empty coats from the bacterial surface Centrifuge: most 35S stays in the supernatant, bacterial pellet has little Blender shakes off the empty coats from the bacterial surface Centrifuge: most 32P is in the pellet, inside bacteria; new phages carry 32P Experiment 2: label the DNA with 32P Phage DNA labelled with 32P (phosphate is in DNA, not in protein) Phage attaches to E. coli and injects its genetic material into the cell Hershey and Chase. 35S labels the protein coat, 32P labels the DNA. Only 32P entered the bacteria and appeared in the next generation of phage. Result: the protein coat stayed outside, DNA entered. Since the infected bacteria made new phages with the same DNA and protein coat, the DNA carried the genetic information. Careful: the separation was not perfect (some 35S stayed with the cells, and some 32P was in the supernatant). The conclusion still stands because the bulk of each label went where the experiment predicted, and the progeny phage inherited 32P and almost no 35S. RNA as genetic material: TMV (Fraenkel-Conrat and Singer, 1957) Some viruses use RNA. They took apart tobacco mosaic virus (TMV) of two strains and made hybrid viruses with the RNA of one strain and the protein coat of another. The progeny always matched the strain that donated the RNA. So in TMV the RNA carries the genetic information, not the coat protein. From Mendel's factors to genes and alleles Garrod (1902) and Beadle and Tatum (1941): genes control enzymes. Beadle and Tatum irradiated Neurospora, found mutants that could not make a specific nutrient, and showed each mutant lacked one enzyme: the one gene one enzyme idea, later refined to one gene one polypeptide. Gene today: a DNA segment that encodes a polypeptide or a functional RNA, plus its regulatory sequences. Alleles are versions of the same gene that differ in DNA sequence (often by one nucleotide). DSC-14 Unit 1 Notes | page 17 of 85 Central dogma: DNA replicates, DNA is transcribed to RNA, RNA is translated to protein. Retroviruses add reverse transcription (RNA to DNA). Molecular explanation of dominance Most recessive alleles are loss of function alleles that make no working protein. A heterozygote (Rr) still has one working copy that makes about 50 percent of the normal enzyme, and that is usually enough for a normal phenotype. This is called haplosufficiency, and it is why dominance works. Mendel's round and wrinkled seeds: the wrinkled allele (r) has an insertion in the gene for starch branching enzyme I. Without the enzyme, seeds store less starch and more sugar, take in extra water, and shrivel when they dry. Rr seeds make enough enzyme to look round. Mendel's dwarf pea has a single amino acid change in an enzyme that makes the growth hormone gibberellin. QUICK RECAP Griffith (1928): a transforming principle from dead S cells converts live R cells; Avery et al. (1944): the principle is DNA (only DNase destroys it) Hershey and Chase (1952): 32P labelled DNA entered the bacteria, 35S labelled protein did not Fraenkel-Conrat and Singer (1957): in TMV the RNA, not the coat protein, is genetic material Gene = DNA that codes for a polypeptide or RNA; alleles = sequence variants; recessive alleles are usually loss of function and dominance reflects haplosufficiency EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper DSC-14 Unit 1 Notes | page 18 of 85 before opening the answer. Q1. Why did Hershey and Chase label protein with 35S and DNA with 32P and not the other way round, or with 14C or 3H? concept Answer The label must be unique to one molecule. Sulphur is present in some amino acids (cysteine, methionine) and absent from DNA. Phosphorus is present in the DNA backbone and is essentially absent from protein. Carbon and hydrogen are in both molecules, so they could not distinguish the two. Q2. In Griffith's experiment, a critic proposes that the living S cells recovered from the dead mouse simply arose by mutation of R cells. Give the evidence that rules this out. hard Answer Griffith's R strain came from type II bacteria, while the dead cells were type III S. The recovered live S bacteria were type III. A mutation in R (type II) cells would give type II S cells at best. So the type III capsule must have come from the dead type III S cells. Also, the mice injected with R alone or dead S alone survived. Q3. In the Avery, MacLeod and McCarty experiment, what would you predict if the transforming extract had been treated with (a) RNase, (b) protease, (c) DNase, (d) DNase and then the DNase was inactivated and fresh DNA from S cells was added back? hard Answer (a) Transformation occurs (RNA not required). (b) Transformation occurs (protein not required). (c) No transformation (DNA destroyed). (d) Transformation is restored, because the active molecule (DNA) is back. Part (d) is the strongest control, since it shows the loss in (c) was due to DNA and not a toxic side effect of the enzyme. DSC-14 Unit 1 Notes | page 19 of 85 Q4. A hybrid TMV is made from the coat protein of strain A and the RNA of strain B. It infects a tobacco leaf. What strain of virus, and what coat protein, will be found in the progeny? What does this show? hard Answer The progeny will be strain B in every respect, and they will carry the strain B coat protein, because the RNA of B carries the genetic instructions for making that coat protein. This shows RNA is the genetic material in TMV, and that the coat protein of the parent does not determine the progeny. Q5. A heterozygote Rr for the wrinkled seed gene has round seeds. Give a molecular explanation, and predict what would happen to the dominance relationship if the enzyme were needed in high amounts. hard Answer R makes working starch branching enzyme, r makes none. Rr makes about half the normal amount, which is still enough to store the normal amount of starch, so seeds are round (haplosufficiency, so R appears dominant). If a full dose of enzyme were needed for a normal phenotype, half the dose would give an intermediate phenotype and the allele would look incompletely dominant, not fully dominant. DSC-14 Unit 1 Notes | page 20 of 85 1.4 Mendel's work on transmission of traits Gregor Mendel, an Augustinian priest and teacher, crossed garden peas (Pisum sativum) for about eight years and published in 1866. His work was ignored until 1900, when de Vries, Correns and Tschermak independently rediscovered it. He is called the father of genetics because he did what nobody before had done: he counted offspring, worked in ratios, and studied one trait at a time. Why peas worked so well Many true breeding varieties with clear alternative forms of a trait (tall or dwarf, round or wrinkled). Flowers self fertilise naturally, but Mendel could remove the anthers (emasculation) and cross pollinate by hand. Short generation time and many seeds per plant, so ratios could be counted. He chose seven traits, each with only two contrasting forms and no intermediates. The seven characters and Mendel's actual F2 counts Character Dominant Recessive F2 dominant F2 recessive Ratio Seed shape Round Wrinkled 5474 1850 2.96 : 1 Seed colour Yellow Green 6022 2001 3.01 : 1 Seed coat colour Grey / brown White 705 224 3.15 : 1 Pod shape Inflated Constricted 882 299 2.95 : 1 Pod colour Green Yellow 428 152 2.82 : 1 Flower position Axial Terminal 651 207 3.14 : 1 Stem length Tall Dwarf 787 277 2.84 : 1 Your class notes also list flower colour as a character; in Mendel's own data, seed coat colour and flower colour go together (they are controlled by the same gene, a pleiotropic effect, Topic 12). DSC-14 Unit 1 Notes | page 21 of 85 Monohybrid cross and the law of dominance P generation: RR (round seed) x rr (wrinkled seed) gametes: all R gametes: all r F1: all Rr, all round F1 self pollinated (Rr x Rr) F1 gametes: 1/2 R and 1/2 r from each parent RR Rr Rr rr F2: 3 round : 1 wrinkled (phenotypic ratio 3:1) F2 genotypic ratio: 1 RR : 2 Rr : 1 rr Mendel's count: 5474 round : 1850 wrinkled (2.96 : 1) Monohybrid cross for seed shape. F1 shows only the dominant form; the recessive form reappears in F2 in a 3:1 ratio. Law of dominance: when two different alleles are present in an individual, only one (the dominant one) is expressed. The other is masked but not lost. Law of segregation (the law of purity of gametes): the two alleles of a gene separate from each other during gamete formation, so each gamete carries only one allele, with equal probability of each. In your class notes this is written as "two factors segregate during gamete formation in equal proportion". Punnett square for Tt x Tt (tall x tall heterozygotes): T t T TT Tt t Tt tt Tall = 3 Dwarf = 1 Exam trap: The 3:1 phenotypic ratio in F2 hides a 1:2:1 genotypic ratio. Two thirds of the tall F2 plants are heterozygous, so tall F2 plants cannot all breed true. That is why you need a test cross. DSC-14 Unit 1 Notes | page 22 of 85 Test cross, back cross and reciprocal cross Cross What is crossed Purpose Result for a monohybrid Test cross Individual of unknown Find the genotype of a If unknown is TT: all tall. If Tt: 1 tall genotype x dominant phenotype individual : 1 dwarf F1 hybrid x either Recover parental types, study F1 x recessive parent gives 1:1 parent allele transmission (same as a test cross). F1 x homozygous recessive Back cross dominant parent gives all dominant phenotype Reciprocal Swap which parent Test whether the trait depends Identical results for autosomal cross carries each trait on the parent's sex genes (extranuclear or sex linked inheritance) Test cross ratios to remember: monohybrid Aa x aa gives 1:1. Dihybrid AaBb x aabb gives 1:1:1:1. Trihybrid AaBbCc x aabbcc gives eight equal classes (1:1:1:1:1:1:1:1). A test cross ratio directly shows the gamete ratio of the tested parent. Dihybrid cross and the law of independent assortment Mendel crossed round yellow (RRYY) with wrinkled green (rryy). F1 was all round yellow (RrYy). Selfing F1 gave 556 seeds in the F2: F2 phenotype Observed Expected 9:3:3:1 Round yellow 315 312.75 Round green 108 104.25 Wrinkled yellow 101 104.25 Wrinkled green 32 34.75 ♀ / ♂ RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy DSC-14 Unit 1 Notes | page 23 of 85 Round Yellow = 9 Round Green = 3 Wrinkled Yellow = 3 Wrinkled Green = 1 Each trait taken alone still gives 3:1: round 12 : wrinkled 4 and yellow 12 : green 4, which matches your class notes. The two traits combine as if independent: (3:1) x (3:1) = 9:3:3:1. Law of independent assortment: when two or more pairs of alleles are considered, the alleles of one gene segregate into gametes independently of the alleles of other genes (this holds for genes on different chromosomes; genes very close together on one chromosome break the law, which is linkage, Unit 3). Trihybrid cross: the fork line (branching) method A 64 square Punnett square is slow and error prone. For independent genes, multiply the separate 3:1 ratios along a branching tree: 3/4 C_ A_ B_ C_ 27/64 A_ B_ cc 9/64 A_ bb C_ 9/64 A_ bb cc 3/64 aa B_ C_ 9/64 aa B_ cc 3/64 aa bb C_ 3/64 aa bb cc 1/64 1/4 cc 3/4 B_ 1/4 bb 3/4 C_ 3/4 A_ 1/4 cc AaBbCc x AaBbCc 3/4 C_ 1/4 aa 1/4 cc 3/4 B_ 1/4 bb 3/4 C_ 1/4 cc Fork line method for AaBbCc x AaBbCc. Multiply along each branch: for example A_ B_ C_ = 3/4 x 3/4 x 3/4 = 27/64. The eight phenotypic classes come out 27:9:9:9:3:3:3:1. General formulas for n independent genes Feature Monohybrid Dihybrid Trihybrid n genes Gamete types from F1 2 4 8 2^n F2 phenotypic classes 2 4 8 2^n DSC-14 Unit 1 Notes | page 24 of 85 Feature Monohybrid Dihybrid Trihybrid n genes F2 genotypic classes 3 9 27 3^n F2 combinations (Punnett squares) 4 16 64 4^n F2 phenotype ratio 3:1 9:3:3:1 27:9:9:9:3:3:3:1 (3:1)^n F2 fully homozygous recessive fraction 1/4 1/16 1/64 (1/4)^n Test cross classes and ratio 2 (1:1) 4 (1:1:1:1) 8 (equal) 2^n equal Was Mendel's data too good? R. A. Fisher (1936) reanalysed Mendel's counts and found them closer to the expected ratios than chance would usually allow. Possible reasons debated since then are unconscious bias in classifying doubtful seeds or plants, or selective reporting. Mendel's conclusions are confirmed by countless later experiments, so the law is not in doubt. Chi-square test: do the counts fit a ratio? chi-square = sum of [(observed - expected)^2 / expected] degrees of freedom = number of classes - 1 For the dihybrid counts above: (315 - 312.75)2/312.75 + (108 - 104.25)2/104.25 + (101 104.25)2/104.25 + (32 - 34.75)2/34.75 = 0.016 + 0.135 + 0.101 + 0.218 = 0.47. With 3 degrees of freedom the critical value at p = 0.05 is 7.81, so 0.47 is far below it: the data fit 9:3:3:1 (p is about 0.93). Reject the ratio only if your chi-square exceeds the critical value. The critical values at p = 0.05 are 3.84 (1 df), 5.99 (2 df), 7.81 (3 df), 9.49 (4 df). DSC-14 Unit 1 Notes | page 25 of 85 QUICK RECAP Mendel's laws: dominance, segregation (gametes carry one allele), independent assortment (genes on different chromosomes) Monohybrid F2: 3:1 phenotype and 1:2:1 genotype. Dihybrid F2: 9:3:3:1. Trihybrid F2: 27:9:9:9:3:3:3:1 Test cross gives the gamete ratio of the tested parent; 1:1 means heterozygous For n genes: 2^n gametes, 3^n genotypes, 4^n combinations; use the fork line method, not a giant Punnett square Chi-square (df = classes - 1) tests goodness of fit; a value below the critical value means the ratio fits EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. A tall pea plant of unknown genotype is test crossed. The progeny are 52 tall and 49 dwarf. What is the genotype of the plant, and what result would have shown it was homozygous? basic Answer Tt. A 1:1 test cross ratio means the unknown plant makes two gamete types (T and t) in equal numbers. If it were TT, every gamete would carry T and all progeny would be Tt, all tall. Q2. For the cross AaBbCcDd x aaBbccDd, what proportion of the offspring will show the dominant phenotype for A and C but the recessive phenotype for B and D? (All genes independent, complete dominance.) numerical Answer Handle each gene separately and multiply. Aa x aa: A_ = 1/2. Bb x Bb: bb = 1/4. Cc x cc: C_ = 1/2. Dd x Dd: dd = 1/4. Probability = 1/2 x 1/4 x 1/2 x 1/4 = 1/64. DSC-14 Unit 1 Notes | page 26 of 85 Q3. In the F2 of AaBbCc x AaBbCc find the probability of (a) AaBbCc, (b) A_ bb C_, (c) all three traits recessive, (d) showing at least one recessive trait. numerical Answer (a) 1/2 x 1/2 x 1/2 = 1/8. (b) 3/4 x 1/4 x 3/4 = 9/64. (c) 1/4 x 1/4 x 1/4 = 1/64. (d) Complement rule: 1 - P(all dominant) = 1 - 27/64 = 37/64. Q4. True or false and justify: A back cross is used to determine the genotype of the parent. past paper style Answer False. A back cross is a cross between an F1 hybrid and one of its parents. The cross used to determine an unknown genotype is a test cross (with a homozygous recessive). The two coincide only when the parent used in the back cross happens to be the homozygous recessive one. A back cross with the dominant parent gives no genotypic information, since all progeny look dominant. Q5. Mendel studied seven traits in a plant that has seven chromosome pairs, and every pair of traits assorted independently. Does that prove that each of his genes lay on a different chromosome? What condition would allow two genes on the same chromosome to still look independent? very hard Answer No. Independent assortment is observed if two genes are on different chromosomes or if they are on the same chromosome but so far apart (about 50 map units or more) that crossing over between them happens in about half the meioses, which gives 50 percent recombinants, the same as independent assortment. So absence of linkage does not prove different chromosomes. This is the basis of gene mapping in Unit 3. Q6. A dihybrid F2 gives 315 round yellow : 108 round green : 101 wrinkled yellow : 32 wrinkled green. A student says the round : wrinkled ratio is 423 : 133. Check that it fits 3:1 using chi-square. numerical Answer Total = 556, expected round = 417, wrinkled = 139. Chi-square = (423 - 417)2/417 + (133 139)2/139 = 0.086 + 0.259 = 0.35. With 1 df the critical value is 3.84, so it fits 3:1. DSC-14 Unit 1 Notes | page 27 of 85 1.5 Principles of inheritance and chromosome theory Mendel's factors were invisible. In 1902, two scientists working separately noticed that the way chromosomes behave in meiosis is exactly what Mendel's factors would need to do. This became the chromosome theory of inheritance: genes are carried on chromosomes. Sutton and Boveri (1902) Walter Sutton studied meiosis in grasshoppers (Brachystola) and Theodor Boveri studied sea urchin embryos. They saw a parallel: Mendel's factors (genes) Chromosomes Present in pairs in the body, one from each Homologous chromosomes come in pairs, one from each parent parent The two alleles separate during gamete Homologues separate in anaphase I of meiosis formation (segregation) Each gamete gets one allele of each gene Each gamete gets one chromosome of each homologous pair Different genes assort independently Different homologous pairs orient independently at metaphase I Fertilisation restores the pair Fertilisation restores the diploid number by combining chromosomes from both parents Alignment I Alignment II A a A a B b b B pole 1 pole 2 pole 1 Gametes: AB and ab pole 2 Gametes: Ab and aB Both alignments are equally likely, so the four gamete types come out 1 AB : 1 Ab : 1 aB : 1 ab Blue = chromosomes from one parent, orange = from the other parent Independent assortment as chromosome behaviour. Which homologue faces which pole is random for each pair, so both alignments occur with equal probability. DSC-14 Unit 1 Notes | page 28 of 85 Why the parallel needed proof A parallel is not a proof. There were two problems: chromosomes are far fewer than genes (so each chromosome must carry many genes, which raises the issue of linkage), and nobody had shown a specific gene on a specific chromosome. Four lines of evidence closed the gap. 1. Carothers (1913): grasshopper heteromorphic pair Eleanor Carothers found a grasshopper with one pair of homologues that were visibly different in shape, and watched how this pair and the X chromosome assorted at anaphase I. The two members of different pairs assorted randomly, giving direct visual proof of independent assortment. 2. Blakeslee (1922): Datura trisomics Albert Blakeslee found that each extra chromosome in the jimson weed (Datura) gave a plant with its own characteristic appearance (twelve different trisomics for the twelve chromosomes). So different chromosomes carry different sets of genes. 3. Morgan (1910): white eyes in Drosophila Thomas Hunt Morgan found a white eyed male in a red eyed Drosophila stock. The wild type allele is written w+ (red) and the mutant is w (white). The gene lies on the X chromosome and there is no matching gene on Y. Cross 1: red eyed female (X+X+) x white eyed male (XwY) Xw Y X+ X+Xw red female X+Y red male X+ X+Xw red female X+Y red male F1: all red eyed (females X+Xw, males X+Y). Crossing F1 with each other gives the F2 (X+Xw x X+Y): X+ Y X+ X+X+ red female X+Y red male Xw X+Xw red female XwY white male F2: 3 red : 1 white, and all the white eyed flies are males. The trait follows the X chromosome, so it must be on it. The reciprocal cross (white female x red male) gives red daughters and white DSC-14 Unit 1 Notes | page 29 of 85 sons, which is different from the first cross. Different results in reciprocal crosses point to sex linkage. 4. Bridges (1913 to 1916): non-disjunction Calvin Bridges gave the proof that could not be argued with. He crossed white eyed females (XwXw) with red eyed males (X+Y). Expected: red daughters, white sons. Nearly all offspring were like that, but about 1 in 2000 were exceptions: white eyed daughters and red eyed sons (and the red eyed sons were sterile). Bridges proposed that the two X chromosomes of the mother failed to separate in meiosis (primary non-disjunction), giving eggs with XX or with no X (O). Egg from X w X w mother Sperm X + (from father) Sperm Y (from father) XwXw (two X) XwXwX+: triple X, usually dies XwXwY: white eyed female, fertile O (no X) X+O: red eyed male, sterile YO: no X, dies Bridges then examined the chromosomes of the exceptional females under the microscope and saw XXY, exactly as predicted. A specific trait (eye colour, a genetic event) matched a specific chromosome event (non-disjunction). No other observation supported the chromosome theory so directly. Two more points Bridges made: in Drosophila the Y chromosome does not determine maleness (XXY is female, X0 is male), so the sex is decided by the number of X chromosomes relative to autosomes (Unit 4). And the XXY females can undergo secondary non-disjunction, about 4 percent, which produced more exceptional offspring, as predicted. Other confirmations Stern (1931) and Creighton and McClintock (1931) showed that genetic recombination goes with a visible physical exchange between chromosomes (Unit 3). Your class notes also list Alfred Blakeslee (1922) and Eleanor Carothers (1913) among the supporters, along with Sutton and Boveri (1902) and Morgan. What chromosome theory says, in summary Genes are located on chromosomes at specific positions called loci. Homologous chromosomes carry corresponding genes, one inherited from each parent. DSC-14 Unit 1 Notes | page 30 of 85 Separation of homologues in meiosis explains the law of segregation. Different chromosome pairs assort independently, which explains the law of independent assortment. Crossing over produces genetic recombination. Fertilisation restores the diploid number by combining chromosomes from both parents. QUICK RECAP Sutton and Boveri (1902): chromosomes behave exactly as Mendel's factors do in meiosis and fertilisation Carothers (1913), Blakeslee (1922), Morgan (1910) and Bridges (1913 to 16) gave the experimental proof Morgan: white eye is X linked; the reciprocal crosses differ, and all F2 white flies are male Bridges: exceptional white females and sterile red males came from non-disjunction in the mother; the microscope confirmed XXY EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. After T. H. Morgan's experiments on eye colour in Drosophila, Bridges' exceptions gave strong proof of the chromosomal theory of inheritance. Which other scientists' work gave support to this theory? Explain in detail. past paper Answer A full answer covers: Sutton and Boveri (1902), who observed that chromosomes behave in meiosis exactly as Mendel's factors do; Carothers (1913), who showed independent assortment in a grasshopper with a visibly heteromorphic chromosome pair; Blakeslee (1922), whose Datura trisomics each had a distinct phenotype for each extra chromosome; and Morgan (1910), who assigned the white eye gene to the X. Add Stern (1931) and Creighton and McClintock (1931), who showed a physical exchange of chromosome segments parallels genetic recombination. Then explain Bridges' non-disjunction as the finishing proof: the genetic exceptions (white daughters, red sons) and the cytological finding (XXY) matched. DSC-14 Unit 1 Notes | page 31 of 85 Q2. A white eyed Drosophila female is crossed with a red eyed male. One of the daughters is white eyed. What are the two possible explanations, and how would you tell them apart? hard Answer (a) Non-disjunction in the mother gave an XX egg fertilised by a Y sperm, producing an XwXwY white daughter. (b) A new mutation in the father's X+ sperm (X+ to Xw) produced XwXw. To tell them apart, look at the chromosomes: XXY means non-disjunction. Or test her offspring: an XXY female gives certain unusual progeny including XXY and XYY types, with a characteristic 4 percent of exceptional offspring. Mutation would give normal chromosomes and normal progeny. Non-disjunction is far more frequent (about 1 in 2000), which is why it explains Bridges' exceptions. Q3. Explain why the red eyed exceptional males in Bridges' cross were sterile, whereas the white eyed exceptional females were fertile. very hard Answer The red males are X+O: they have no Y chromosome. In Drosophila the Y is not needed for maleness but contains genes needed for sperm development, so an X0 male is sterile. The white females are XwXwY: they have two X chromosomes so they are female, and the extra Y does not affect their fertility (they can make eggs). Q4. Give two observations that show the Drosophila white eye gene is on the X chromosome and not on an autosome. concept Answer (1) The reciprocal crosses give different results: red female x white male gives all red F1, while white female x red male gives red daughters and white sons. Autosomal genes give the same result both ways. (2) In the F2 of the first cross, all the white eyed flies are male, so the trait is tied to sex, and therefore to the sex chromosome. The F2 male to female pattern matches X linked inheritance and the expected 3:1 red:white. DSC-14 Unit 1 Notes | page 32 of 85 Q5. Why is it not enough to say that chromosomes 'behave like' Mendel's factors? What extra kind of evidence was needed? concept Answer A similarity is only a correlation. What was needed was an experiment linking a specific gene to a specific chromosome. Morgan (a sex linked gene follows the X) and especially Bridges (an abnormal chromosome event, non-disjunction, accompanied by an abnormal inheritance of a gene) did this. Blakeslee also linked specific chromosomes to specific phenotypes. DSC-14 Unit 1 Notes | page 33 of 85 1.6 Laws of probability Every genetic cross is a chance event. Each gamete that fertilises is picked at random, so the ratios you see (3:1, 9:3:3:1) are probabilities, not guarantees. Learning three rules lets you answer any cross without a Punnett square. Basic idea probability = number of times an event happens / total number of possible outco It runs from 0 (impossible) to 1 (certain). For an Aa parent, P(A gamete) = 1/2 and P(a gamete) = 1/2. For Aa x Aa: P(aa) = 1/4, P(Aa) = 1/2, P(AA) = 1/4. Chance has no memory. Each fertilisation is independent. If two carrier parents already have an albino child, the chance the next child is albino is still 1/4, not lower and not higher. Rule 1: the addition rule (either or) Use it when either one event or another can occur and the events are mutually exclusive (they cannot happen together in the same trial). P(A or B) = P(A) + P(B) (mutually exclusive) P(A or B) = P(A) + P(B) - P(A and B) (not mutually exclusive) Example: Aa x Aa. P(offspring shows the dominant phenotype) = P(AA) + P(Aa) = 1/4 + 1/2 = 3/4. Rule 2: the multiplication rule (both, together) Use it when two or more independent events must occur together. P(A and B) = P(A) x P(B) (independent events) Example 1: Ss x Ss, so P(ss) = 1/4. For two children to both be ss: 1/4 x 1/4 = 1/16 = 6.25%. Example 2: Two genes on different chromosomes. AaBb x AaBb, P(aabb) = 1/4 x 1/4 = 1/16. This is why the 9:3:3:1 dihybrid ratio works: it is the product of two 3:1 ratios. Example 3 (all loci homozygous): AaBbCcDdEe x AaBbCcDdEe. At each locus, P(homozygous) = 1/2. For all 5 loci: (1/2)5 = 1/32. The chance that a zygote is not homozygous at all five loci is 1 - DSC-14 Unit 1 Notes | page 34 of 85 1/32 = 31/32. And P(homozygous dominant at all 5) = (1/4)5 = 1/1024. The complement rule When the question says "at least one", count the opposite and subtract: P(at least one) = 1 P(none). For four children of carrier parents, P(at least one albino) = 1 - (3/4)4 = 1 - 81/256 = 175/256 = about 68.4%. Rule 3: the binomial expansion (order does not matter) Use it when you ask for a particular mix of outcomes in a family, for example 4 girls and 2 boys, in any birth order. Let p = probability of one outcome, q = the other (q = 1 - p), and n = family size. (p + q)^n = sum of terms P(s of one kind and t of the other) = n! / (s! t!) x p^s x q^t where s + t = n The expansion for n = 5, for example: p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5. The coefficients are the numbers of birth orders (1, 5, 10, 10, 5, 1: row of Pascal's triangle). Worked example: six children, at least four girls p = girl = 1/2, q = boy = 1/2, n = 6. Probability of exactly s girls = 6!/(s!(6-s)!) x (1/2)6 = C(6,s)/64. Girls Boys Coefficient n!/(s!t!) Probability 6 0 1 1/64 5 1 6 6/64 4 2 15 15/64 3 3 20 20/64 2 4 15 15/64 1 5 6 6/64 0 6 1 1/64 At least four girls = P(4) + P(5) + P(6), using the addition rule on the three exclusive outcomes: 15/64 + 6/64 + 1/64 = 22/64 = 11/32 = 0.344 (34.4%). DSC-14 Unit 1 Notes | page 35 of 85 Worked example: a family with 2 albino and 3 normal children Two carriers (Aa x Aa): p(albino) = 1/4, q(normal) = 3/4. For n = 5 with 2 albino and 3 normal in any order: 5! / (2! 3!) x (1/4)^2 x (3/4)^3 = 10 x 1/16 x 27/64 = 270/1024 = 0.264 The two numbers in the bracket, 1/4 and 3/4, are the single-birth probabilities. The coefficient 10 counts the different birth orders. Factorial method (for more than two categories) P = n! / (a! b! c! ...) x p1^a x p2^b x p3^c ... For a dihybrid family, the four phenotypes have probabilities 9/16, 3/16, 3/16 and 1/16, and you extend the same formula with more categories. Exam trap: Do not use the binomial formula when the events are ordered. "First child albino, second normal" is a single ordered outcome: 1/4 x 3/4, with no coefficient. Only "one albino and one normal in any order" needs the coefficient 2. QUICK RECAP Addition rule: either or, exclusive events, add probabilities Multiplication rule: independent events together, multiply At least one: 1 minus the probability of none Binomial: n!/(s!t!) p^s q^t; the coefficient counts birth orders Chance has no memory: earlier children do not change the next one EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper DSC-14 Unit 1 Notes | page 36 of 85 before opening the answer. Q1. Consider a family of six children. What is the probability that at least four will be girls? past paper Answer Use the binomial with p = q = 1/2, n = 6. P(4 girls) = 15/64, P(5 girls) = 6/64, P(6 girls) = 1/64. Add: 22/64 = 11/32 (about 34.4%). Q2. Two parents are both heterozygous carriers for phenylketonuria (autosomal recessive). They plan five children. Find the probability of (a) exactly two affected children, (b) at least one affected child, (c) the first two affected and the last three unaffected, in that order. numerical Answer P(affected) = 1/4, P(unaffected) = 3/4. (a) 5!/(2!3!) x (1/4)2 x (3/4)3 = 10 x 27/1024 = 270/1024 = 0.264. (b) 1 - (3/4)5 = 1 - 243/1024 = 781/1024 = 0.763. (c) A single ordered outcome, so no coefficient: (1/4)2 x (3/4)3 = 27/1024 = 0.026. Q3. AaBbCcDdEe x AaBbCcDdEe (all independent). What is the probability that the zygote is homozygous at all five loci? What is the probability it is not homozygous at all five loci? What is the probability it is homozygous dominant at all five loci? numerical Answer Homozygous at one locus = 1/2. For five: (1/2)5 = 1/32. Not homozygous at all five: 1 - 1/32 = 31/32. Homozygous dominant at one locus = 1/4, so at all five: (1/4)5 = 1/1024. Q4. For a cross AaBbCc x AaBbCc, find the probability that the offspring shows the dominant phenotype for exactly two of the three traits. hard Answer P(dominant for one trait) = 3/4 and P(recessive) = 1/4. Exactly two out of three dominant: 3!/(2!1!) x (3/4)2 x (1/4) = 3 x 9/64 = 27/64. The coefficient 3 counts which two traits are dominant (AB, AC or BC). DSC-14 Unit 1 Notes | page 37 of 85 Q5. A couple have three children, all daughters. A relative says the next child is more likely to be a boy 'to balance things out'. Is this correct? Give the probability of a fourth daughter and of four daughters in a row from the start. concept Answer No. Each birth is independent, so P(fourth daughter) = 1/2 regardless of the earlier births. The probability of four daughters in a row from the start is (1/2)4 = 1/16, but once three daughters have happened that is no longer the relevant probability. Mixing these two is the gambler's fallacy. Q6. In Ss x Ss (S = short hair, s = long hair), what is the probability that among three offspring at least one has long hair? What is the probability that exactly two are short haired? numerical Answer P(long, ss) = 1/4, P(short) = 3/4. At least one long: 1 - (3/4)3 = 1 - 27/64 = 37/64. Exactly two short and one long: 3 x (3/4)2 x (1/4) = 27/64. DSC-14 Unit 1 Notes | page 38 of 85 1.7 Incomplete dominance Mendel's law of dominance says one allele hides the other. With incomplete dominance neither allele fully hides the other, so the heterozygote has a phenotype intermediate between the two homozygotes. This is not a failure of segregation: the alleles still separate in gametes and come back unchanged in the next generation. Classic example: snapdragon (Antirrhinum majus) flower colour Red (RR) x white (R'R') gives all pink (RR') in F1. Selfing the pink F1 gives red, pink and white in a 1 : 2 : 1 ratio. Because each genotype has its own phenotype, the phenotypic ratio equals the genotypic ratio. P generation: RR (red) x R'R' (white) F1: all RR' (pink) RR RR' RR' R'R' F2: 1 red : 2 pink : 1 white (1:2:1, phenotype ratio = genotype ratio) Incomplete dominance in snapdragon. Red and white flowers reappear in F2, which proves the alleles did not blend. Do not confuse with blending inheritance: in blending the pink would breed true pink and the red and white could never reappear. Here the pink is only the heterozygote. Selfing it regenerates red and white, so the alleles stayed separate and intact. Molecular basis The R allele makes a working enzyme that produces red pigment. R' makes none (or a defective one). The heterozygote makes about half the normal amount of pigment, and this dosage effect produces the intermediate colour. The single functional allele is not enough for the full phenotype (unlike haplosufficiency in complete dominance, Topic 3). DSC-14 Unit 1 Notes | page 39 of 85 Other examples Four o'clock plant (Mirabilis jalapa): red x white gives pink, discovered by Correns. Andalusian fowl: black x splashed white gives bluish grey (blue) birds. Blue x blue gives 1 black : 2 blue : 1 splashed white, so blue Andalusians can never breed true. Familial hypercholesterolaemia (human): homozygotes have no working LDL receptors and very high cholesterol, heterozygotes have about half the receptors and moderately high cholesterol, normal homozygotes have normal levels. Useful crosses to memorise (snapdragon) Cross Offspring RR x R'R' (red x white) All RR' pink RR' x RR' (pink x pink) 1 red : 2 pink : 1 white RR' x RR (pink x red) 1 red : 1 pink RR' x R'R' (pink x white) 1 pink : 1 white RR x RR (red x red) All red Test cross of pink 1 pink : 1 white QUICK RECAP Heterozygote is intermediate; phenotypic ratio = genotypic ratio = 1 : 2 : 1 in F2 Segregation still holds; alleles are not blended and reappear in F2 Caused by a dosage effect: half the gene product gives an intermediate phenotype Blue Andalusian fowl and pink snapdragons cannot be bred true EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper DSC-14 Unit 1 Notes | page 40 of 85 before opening the answer. Q1. Explain incomplete dominance with a suitable example. Does it violate any of Mendel's laws? past paper Answer Incomplete dominance: the heterozygote has a phenotype intermediate between the homozygotes, because neither allele completely masks the other. Example: red x white snapdragons give pink F1, and pink x pink gives 1 red : 2 pink : 1 white. It violates the law of dominance (no allele is dominant) but not the law of segregation: the alleles separate in gametes and reappear in F2 in the same 1 : 2 : 1 pattern. Independent assortment is also not affected. Q2. A breeder wants to produce only blue Andalusian chickens. Which cross should she use, and why can she never establish a pure breeding blue line? hard Answer Cross black x splashed white every generation (BB x B'B' gives all BB' blue). Blue birds are heterozygous, so blue x blue always gives 1 black : 2 blue : 1 white, and only half the offspring are blue. A pure breeding line would need homozygotes, but the homozygotes are black or white, never blue. Q3. In snapdragon, flower colour is incompletely dominant (RR red, RR' pink, R'R' white) and leaf width is controlled by N (broad, completely dominant) and n (narrow). Two plants RR'Nn are crossed. Give the F1 phenotypic ratio. hard Answer Multiply a 1 : 2 : 1 colour ratio by a 3 : 1 leaf ratio: 3 red broad : 6 pink broad : 3 white broad : 1 red narrow : 2 pink narrow : 1 white narrow. There are 6 phenotypic classes (not 4), because incomplete dominance doubles the number of classes for that gene. DSC-14 Unit 1 Notes | page 41 of 85 Q4. Familial hypercholesterolaemia shows three phenotypes: normal, moderately high, and very high cholesterol. Two parents both with moderately high cholesterol have a child. What is the chance of a child with very high cholesterol, and what genotype ratio is expected? concept Answer Both parents are heterozygotes (Hh). The cross is Hh x Hh, giving 1 HH (normal) : 2 Hh (moderately high) : 1 hh (very high). The chance of a very high cholesterol child is 1/4. The genotype ratio and the phenotype ratio are the same, 1 : 2 : 1. Q5. Pink snapdragons are crossed with red snapdragons and 200 offspring are raised. How many of each phenotype do you expect? If you instead cross pink x pink and get 800 seeds, how many will grow into pink plants? numerical Answer Pink x red: 1 red : 1 pink, so 100 red and 100 pink. Pink x pink: 1 : 2 : 1, so pink is 1/2 of 800 = 400 pink plants (200 red, 200 white). DSC-14 Unit 1 Notes | page 42 of 85 1.8 Codominance In codominance the heterozygote shows the phenotypes of both alleles fully and separately, at the same time. Nothing is blended and nothing is hidden. Compare this with incomplete dominance, where the heterozygote shows a new intermediate phenotype. Examples MN blood group: LMLM makes only M antigen, LNLN only N antigen, and LMLN makes both M and N antigens on the same red cell. Cross LMLN x LMLN gives 1 M : 2 MN : 1 N. AB blood group: IAIB individuals have both A and B antigens on the red cells. Roan cattle (Shorthorn): red x white gives roan calves, whose coat has both red and white hairs. It is not pink, and the two colours are separate. Roan x roan gives 1 red : 2 roan : 1 white. Sickle cell heterozygote: a person HbAHbS makes both normal and sickle haemoglobin. Complete, incomplete, codominance: how to tell them apart Type Heterozygote phenotype Example F2 phenotype ratio (monohybrid) Complete Looks like the dominant Pea seed shape 3:1 dominance homozygote Incomplete Intermediate, new phenotype Snapdragon pink 1:2:1 Both parental phenotypes fully MN blood group, 1:2:1 expressed together roan cattle Heterozygote is fitter than either Sickle cell and homozygote malaria dominance Codominance Overdominance not a ratio, a fitness effect Key idea: Whether an allele is called dominant, incompletely dominant or codominant depends on which level of the phenotype you examine. Sickle cell disease is recessive at the level of anaemia, incompletely dominant at the level of red cell shape (some sickling under low oxygen), and codominant at the level of haemoglobin protein (both types made). The gene is the same; the description changes with the level. DSC-14 Unit 1 Notes | page 43 of 85 Crosses to know Roan x roan R (red) W (white) R RR red RW roan W RW roan WW white Roan x red gives 1 red : 1 roan. Roan x white gives 1 roan : 1 white. The only cross that gives all roan is red x white. QUICK RECAP Codominance: both alleles show fully and separately in the heterozygote (MN, AB, roan) Incomplete dominance gives a blend like phenotype; codominance gives both phenotypes side by side Both give a 1 : 2 : 1 ratio, so you must look at the heterozygote phenotype to decide The type of dominance can change with the phenotype level examined (sickle cell) EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. Distinguish between incomplete dominance and codominance using two examples each. concept Answer Incomplete dominance: the heterozygote shows an intermediate, blended phenotype because the single functional allele gives half a dose of product (pink snapdragon, Andalusian blue fowl). Codominance: the heterozygote shows the phenotypes of both alleles distinctly, because both alleles make working products which are both detectable (MN blood group: both antigens on the same cell; roan cattle: red and white hairs side by side). Both give a 1 : 2 : 1 F2 ratio, so the difference is in the heterozygote phenotype. DSC-14 Unit 1 Notes | page 44 of 85 Q2. A woman with blood group M has a child with blood group MN. Which blood groups are possible for the father? Can a man with blood group N be ruled out as the father? hard Answer The child got LM from the mother, so the child's LN came from the father. The father must have at least one LN allele, so his group is N or MN. A man with group N cannot be ruled out; a man with group M is excluded. Q3. Sickle cell disease is called a recessive disease, yet a heterozygote makes both normal and sickle haemoglobin. How can it be both recessive and codominant? Also explain why the allele is common in malarial regions. hard Answer The description depends on the level. At the organism level the disease (severe anaemia) only shows in HbSHbS, so it is recessive. At the protein level both haemoglobins are made in the heterozygote, so the alleles are codominant. The carrier (heterozygote) is more resistant to malaria than HbAHbA people and does not suffer the disease as HbSHbS people do, which is heterozygote advantage (overdominance) and keeps the allele common. Q4. Roan cattle are crossed with red cattle. Give the ratio of offspring. What cross would you use to produce only roan calves? concept Answer Roan (RW) x red (RR): 1 red : 1 roan. Only red x white gives all roan calves, since roan animals are heterozygotes and always segregate. DSC-14 Unit 1 Notes | page 45 of 85 1.9 Multiple alleles An individual carries at most two alleles of a gene (diploid), but a population can have many. A gene with three or more alleles in the population is a multiple allelic gene. The alleles arise by different mutations in the same gene and are usually ranked in a dominance series. Counting genotypes For a gene with n alleles: total genotypes = n(n + 1)/2, homozygous = n, heterozygous = n(n 1)/2. Alleles (n) Genotypes Homozygous Heterozygous 2 3 2 1 3 (ABO) 6 3 3 4 (rabbit C series) 10 4 6 12 78 12 66 Coat colour in rabbits (C series) Allele Name Phenotype Dominance C Full colour (agouti) Wild type, dark grey brown Dominant to all cch Chinchilla Silvery grey, pigment reduced Dominant to ch and c ch Himalayan White body with dark ears, nose, feet and tail Dominant to c c Albino White with pink eyes, no pigment Recessive to all Dominance series: C > cch > ch > c. A full colour rabbit can be CC, Ccch, Cch or Cc, and only cc is albino. Himalayan is discussed again in Topic 14 (temperature sensitivity). Other examples Ducks: Restricted (MR) > Mallard (M) > Dusky (md). MRMR, MRM and MRmd are Restricted; MM and Mmd are both Mallard; mdmd is Dusky. Drosophila eye colour (white locus): w+ (red) > wch (cherry) > w (white), and several more. Plant self incompatibility (S alleles): dozens of alleles stop self and near relative fertilisation. DSC-14 Unit 1 Notes | page 46 of 85 ABO blood groups (three alleles, codominance and dominance together) The I gene encodes a glycosyltransferase that modifies the H antigen on red cells. IA adds Nacetylgalactosamine (making A antigen), IB adds galactose (making B antigen), and i makes a non-functional enzyme (O). IA and IB are codominant to each other and both dominant to i. Blood group Genotypes Antigens on RBC Antibodies in serum A IAIA or IAi A anti-B B IBIB or IBi B anti-A AB I AI B A and B none O ii none (H only) anti-A and anti-B Transfusion logic: donor red cells must not carry an antigen that the recipient's serum has antibodies against. O red cells are a universal donor type and AB recipients are universal recipients (for red cells). Inheritance tips: A x B can give A, B, AB and O (if both parents are heterozygous IAi x IBi). AB x O can only give A or B. O x O can only give O (except for the Bombay phenotype, Topic 11). Two AB parents cannot have an O child. QUICK RECAP Multiple alleles: more than two alleles in a population, but any individual has only two n alleles give n(n+1)/2 genotypes Rabbit: C > c^ch > c^h > c; duck: M^R > M > m^d; ABO: I^A = I^B > i ABO is both codominance (I^A and I^B) and dominance (both over i) EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper DSC-14 Unit 1 Notes | page 47 of 85 before opening the answer. Q1. State true or false and justify: In the multiple allelic series Restricted, Mallard, Dusky (MR > M > md), the MM and Mmd genotypes show different phenotypes. past paper Answer False. M is dominant over md, so the heterozygote Mmd has the Mallard phenotype exactly like MM. Only mdmd shows Dusky. Q2. A full colour rabbit crossed with an albino produced 1 full colour : 1 chinchilla. What is the genotype of the full colour parent? hard Answer The albino (cc) gives only c. The chinchilla offspring must be cchc, so the full colour parent carries cch, and the full colour offspring carries C. The full colour parent is Ccch. (Cch would have given full colour and Himalayan, and CC would have given all full colour.) Q3. A couple with blood groups A and B have four children with blood groups A, B, AB and O. Give the genotypes of all six people. concept Answer The O child (ii) must have received i from each parent, so the parents are IAi (A) and IBi (B). Children: A = IAi, B = IBi, AB = IAIB, O = ii. Q4. A child has blood group O. The mother has group A. Group B and group AB men are named as possible fathers. Which can be excluded? hard Answer The child (ii) received i from the mother, who is therefore IAi. The father must also give i, so he must be O, A (IAi) or B (IBi). An AB man cannot be the father because he has no i allele. A group B man (genotype IBi) is still possible. DSC-14 Unit 1 Notes | page 48 of 85 Q5. How many genotypes exist for a gene with 12 alleles? How many are heterozygous? Why does a population have more alleles than an individual? numerical Answer Genotypes = 12 x 13 / 2 = 78. Heterozygous = 12 x 11 / 2 = 66 (and 12 homozygous). An individual is diploid and can carry at most two alleles, but the population is made of many individuals, each carrying a different pair. DSC-14 Unit 1 Notes | page 49 of 85 1.10 Lethal alleles A lethal allele causes death of the organism, usually in the zygote or embryo, sometimes later. Lethals change the expected Mendelian ratios because the affected genotype never shows up among the survivors. Recessive lethal: yellow mice (Cuenot, 1905) Yellow coat is dominant to agouti (wild type). Yellow mice never breed true. Yellow x yellow gives 2 yellow : 1 agouti instead of 3 : 1. Yellow x agouti gives 1 : 1. AYa x AYa AY a AY AYAY dies in the uterus AYa yellow a AYa yellow aa agouti Explanation: AYAY embryos die early in development (Castle and Little showed the missing class as dead embryos). The allele AY is dominant for coat colour but recessive for lethality: this is a pleiotropic allele (Topic 12). Litters from yellow x yellow are about a quarter smaller than litters from yellow x agouti. More examples Manx cats: the M allele causes a short or missing tail. MM is lethal in embryos. Manx x Manx gives 2 Manx : 1 normal-tailed. Creeper fowl: short legs and wings. Creeper x creeper gives 2 creeper : 1 normal and one quarter of the eggs fail to hatch. Achondroplasia (human dwarfism): heterozygotes are dwarfs, homozygotes die. Tay-Sachs disease, cystic fibrosis: recessive lethals (or semi-lethals) in humans. Huntington disease: a dominant lethal allele with late onset, so carriers usually reproduce before symptoms. Dominant lethal vs recessive lethal Effect appears Dominant lethal Recessive lethal Heterozygote (a single copy kills) Homozygote only in DSC-14 Unit 1 Notes | page 50 of 85 Carriers Dominant lethal Recessive lethal None survive, unless late onset Heterozygous carriers are normal (or nearly) and can pass it on Persistence in Removed quickly by selection; kept population only if late onset (Huntington) or by Hidden in carriers, so persists at low frequency new mutation Effect on ratios Example The affected heterozygous class is Carrier x carrier: one quarter of zygotes (the removed, so an affected line usually homozygotes) die. If the allele also has a dominant ends within one generation visible effect, survivors show 2 : 1 (yellow mice) Huntington disease (late onset), some Tay-Sachs disease; the AYAY yellow mouse Drosophila mutations Other types Conditional lethal: kills only in certain conditions, for example temperature sensitive mutants that die at 37 degrees C but live at 25 degrees C. Semi-lethal (sub-vital): kills part of the homozygous class. Gametic lethal: the gamete carrying the allele fails to function. Balanced lethal system: in Drosophila stocks carrying two different recessive lethals on the two homologues (for example Curly, Cy, on one and Plum, Pm, on the other), only the heterozygote survives. The stock breeds true as a heterozygote because both types of homozygote die. Lethals with two genes: the 6 : 2 : 3 : 1 ratio If AY (yellow, lethal when homozygous) and a second gene B (black) / b (brown) assort independently, then AaBb x AaBb gives 4/16 dead embryos (AA). The 12 survivors are in the ratio: Dead (AA) = 4 F1 gametes AB Ab aB ab AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb Aabb aB AaBB AaBb aaBB aaBb ab AaBb Aabb aaBb aabb Yellow, B_ = 6 Yellow, bb = 2 DSC-14 Unit 1 Notes | page 51 of 85 Agouti, B_ = 3 Agouti, bb = 1 Survivors: 6 yellow black : 2 yellow brown : 3 agouti black : 1 agouti brown. This 6 : 2 : 3 : 1 is a modified 9 : 3 : 3 : 1 in which the class containing AA is lost. It is a recognisable sign of a lethal. QUICK RECAP Lethal alleles remove a genotype from the survivors, so ratios change (3 : 1 becomes 2 : 1) Yellow mice, Manx cats and creeper fowl: 2 : 1 with the homozygote dead Dominant lethals are removed fast unless late onset; recessive lethals hide in carriers A^Y is dominant for colour and recessive for lethality (pleiotropy); 6:2:3:1 signals a lethal in a dihybrid EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. Differentiate between dominant lethal and recessive lethal genes. past paper Answer See the table above. In short: a dominant lethal kills the heterozygote, so it is eliminated in one generation unless it acts late in life (Huntington disease) or appears by new mutation. A recessive lethal kills only the homozygote, so heterozygous carriers survive and pass it on, which lets it persist in a population. Ratio effect: the lethal homozygote class is missing, so carrier x carrier of a recessive lethal with a visible dominant effect gives 2 : 1 (yellow mice). Q2. Yellow mice crossed with yellow mice give 2 yellow : 1 agouti, and litters average 6 pups. Yellow x agouti litters average 8 pups. Explain both observations. What fraction of the zygotes from yellow x yellow die? hard Answer Yellow mice are heterozygous (AYa). Yellow x yellow gives 1 AYAY : 2 AYa : 1 aa. The AYAY embryos die in the uterus, so the survivors are 2 yellow : 1 agouti. One quarter of the zygotes die, so litters fall to 3/4 of 8 = 6. Yellow x agouti (AYa x aa) gives no lethal class, so litters are the full 8, in a 1 : 1 ratio. DSC-14 Unit 1 Notes | page 52 of 85 Q3. In Drosophila, Curly wings (Cy) is dominant and Cy/Cy is lethal. A Curly fly is crossed with another Curly fly. What are the offspring? Then, a stock has two different recessive lethals on the two homologues (l1/l2). Why does the stock breed true? hard Answer Curly x Curly (Cy/+ x Cy/+) gives 2 Curly : 1 wild type, because Cy/Cy dies. A stock with l1/l2: selfing gives 1 l1/l1 (dies) : 2 l1/l2 : 1 l2/l2 (dies), and only the l1/l2 heterozygotes survive. The stock therefore always contains the same genotype and appears to breed true. It is a balanced lethal system. Q4. A dihybrid AaBb x AaBb is studied where AA is embryonic lethal. What is the phenotypic ratio among survivors, and how many embryos die out of 640 zygotes? numerical Answer Survivors ratio 6 : 2 : 3 : 1 (see the table). The lethal class is 4/16 = 1/4 of zygotes, so 640 x 1/4 = 160 embryos die and 480 survive. Q5. Huntington disease is caused by a dominant lethal allele, yet it has not been eliminated from human populations. Explain. concept Answer The disease appears late in life (typically 35 to 50 years), usually after people have already had children. Selection cannot remove an allele that acts after reproduction, so the allele is passed on to the next generation before it kills. New mutations also add to the pool. DSC-14 Unit 1 Notes | page 53 of 85 1.11 Epistasis and modified ratios Epistasis (William Bateson, 1907) is an interaction between two different genes in which the alleles of one gene mask or modify the expression of the alleles of another gene. The masking gene is epistatic; the gene that is masked is hypostatic. Epistasis is not dominance. Dominance is an interaction between the two alleles of one gene (A over a). Epistasis is an interaction between alleles of different genes (a effect at gene A hides gene B). Why the ratios change The genes still assort independently, so the F2 still has 16 combinations (9 : 3 : 3 : 1 genotype classes). Epistasis only makes some of the four classes look the same, so they merge. That is why every epistatic ratio adds to 16. The six standard ratios F2 Name Example Rule ratio Test cross (AaBb x aabb) 9:3:3 No epistasis :1 9:7 Pea seed colour Independent genes 1:1:1:1 1:3 and shape Complementary genes Sweet pea flower Both A_ and B_ needed for the (duplicate recessive) colour; albino snails trait; aa or bb gives the other phenotype 9:3:4 Recessive epistasis Labrador coat aa masks B_ and bb 1:1:2 A_ masks B and b 2:1:1 3:1 colour; mouse coat colour 12 : 3 : Dominant epistasis 1 15 : 1 13 : 3 Summer squash fruit colour Duplicate dominant Shepherd's purse A_ or B_ (or both) give the genes seed capsule shape same trait; only aabb differs Dominant and recessive Poultry plumage A dominant inhibitor plus a epistasis (inhibitory) colour recessive masking allele DSC-14 Unit 1 Notes | page 54 of 85 3:1 F2 Name Example Rule Test cross ratio (AaBb x aabb) 9:6:1 Duplicate interaction Squash fruit shape (additive) A_B_ one type, A_bb or aaB_ 1:2:1 intermediate, aabb third type How to recognise an epistatic ratio in the exam: add the numbers. If they total 16 and are not 9 : 3 : 3 : 1, it is a modified dihybrid ratio. Then match the pattern: 9 : 7 (two classes), 9 : 3 : 4 (three classes, the last is 4), 12 : 3 : 1 (first is 12), 15 : 1 (two classes, first is 15), 13 : 3, 9 : 6 : 1. Pathways behind the ratios 9:7 complementary genes (sweet pea) Colourless precursor 1 enzyme C Colourless precursor 2 enzyme P Purple pigment Gene C makes enzyme 1, gene P makes enzyme 2. If either enzyme is missing (cc or pp) there is no pigment. 9:3:4 recessive epistasis (Labrador coat) Pigment precursor gene B Black (B_) or brown (bb) pigment gene E Pigment deposited in hair (E_) Gene E controls deposition. ee dogs are yellow whatever their B genotype, so ee hides B and b. 12:3:1 dominant epistasis (summer squash) Colourless precursor yellow enzyme Yellow pigment (Y_) yy: no change Green (yy) pigment W_ blocks pigment formation altogether: fruit is white. ww lets Y act: yellow (Y_) or green (yy). Three biochemical pathways that produce epistatic ratios. Epistasis usually means two genes acting in the same pathway. 9 : 7, complementary genes Bateson and Punnett (1906): two true breeding white sweet pea lines (CCpp x ccPP) give purple F1 (CcPp). The F2 is 9 purple : 7 white. Purple needs a dominant allele at both genes, so each gene "complements" the other. Any plant with cc or pp is white. DSC-14 Unit 1 Notes | page 55 of 85 F1 gametes AB Ab aB ab AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb Aabb aB AaBB AaBb aaBB aaBb ab AaBb Aabb aaBb aabb Coloured (A_B_) = 9 White = 7 9 : 3 : 4, recessive epistasis In Labradors, gene B decides the pigment: B_ black, bb brown (chocolate). Gene E decides whether pigment is deposited in the hair: E_ allows it, ee blocks it so the dog is yellow whatever its B genotype. Thus ee is epistatic to B. F1 gametes EB Eb eB eb EB EEBB EEBb EeBB EeBb Eb EEBb EEbb EeBb Eebb eB EeBB EeBb eeBB eeBb eb EeBb Eebb eeBb eebb Black (E_B_) = 9 Chocolate (E_bb) = 3 Yellow (ee) = 4 12 : 3 : 1, dominant epistasis In summer squash, W_ (white) blocks pigment altogether and is epistatic to Y. In ww plants, Y_ gives yellow fruit and yy gives green. F1 gametes WY Wy wY wy WY WWYY WWYy WwYY WwYy Wy WWYy WWyy WwYy Wwyy wY WwYY WwYy wwYY wwYy wy WwYy Wwyy wwYy wwyy White (W_) = 12 Yellow (wwY_) = 3 DSC-14 Unit 1 Notes | page 56 of 85 Green (wwyy) = 1 15 : 1, duplicate dominant genes In shepherd's purse (Capsella), a dominant allele at either of two genes is enough to give a triangular seed capsule. Only aabb plants have the ovoid capsule. F1 gametes AB Ab aB ab AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb Aabb aB AaBB AaBb aaBB aaBb ab AaBb Aabb aaBb aabb Triangular (A_ or B_) = 15 Ovoid (aabb) = 1 13 : 3, dominant and recessive epistasis In poultry, a dominant inhibitor I blocks colour, and the recessive cc also prevents colour. Cross White Leghorn (IICC) x White Wyandotte (iicc). F1 (IiCc) is white, and F2 gives 13 white : 3 coloured. Only ii C_ birds are coloured. F1 gametes IC Ic iC ic IC IICC IICc IiCC IiCc Ic IICc IIcc IiCc Iicc iC IiCC IiCc iiCC iiCc ic IiCc Iicc iiCc iicc White = 13 Coloured (iiC_) = 3 9 : 6 : 1, duplicate interaction In squash fruit shape, A_B_ gives disc shape, either one dominant allele alone (A_bb or aaB_) gives sphere, and aabb gives long. The two genes add up. F1 gametes AB Ab aB ab AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb Aabb aB AaBB AaBb aaBB aaBb ab AaBb Aabb aaBb aabb Disc (A_B_) = 9 Sphere (A_bb or aaB_) = 6 DSC-14 Unit 1 Notes | page 57 of 85 Long (aabb) = 1 The Bombay phenotype: epistasis in human blood groups The ABO antigens are built on a base structure called H substance. The H gene (encoding the enzyme FUT1) makes H substance from a precursor. Then IA and IB modify it into A or B antigen, while i leaves it unchanged (O). IA allele adds N-acetylgalactosamine A antigen: blood group A H gene (FUT1) Precursor substance H substance (H antigen) IB allele adds galactose B antigen: blood group B hh: blocked no H substance i allele adds nothing H antigen only: blood group O An hh person has no H substance, so IA and IB cannot act. Blood typing shows O (the Bombay phenotype). The H gene is upstream of the ABO gene. A person homozygous hh cannot make H substance, so the ABO alleles have nothing to act on. Most people are HH or Hh. A rare person who is hh makes no H substance at all, so whatever ABO alleles they carry, their red cells carry no A, B or H antigen and they type as O. This was first described in Bombay (now Mumbai) in 1952. It is very rare in most populations but more frequent in parts of India. A striking consequence. A Bombay woman with genotype IBIB hh types as O. Married to a normal O man (ii HH), all her children are IBi Hh, and they are blood group B. Two "O" parents have B children, which looks impossible until you know about epistasis. The Bombay woman was a hidden B. Genotype Ordinary O Bombay (hh) ii with at least one H allele hh, with any ABO genotype (HH or Hh) H antigen on red Present Absent anti-A and anti-B anti-A, anti-B and anti-H Can receive blood O (and in emergency, O Only another Bombay (hh) donor. Ordinary O blood has H from only) antigen, which would be attacked cells Antibodies in serum DSC-14 Unit 1 Notes | page 58 of 85 Inheritance Ordinary O Bombay (hh) O x O gives only O Bombay x O can give A, B or AB children QUICK RECAP Epistasis: one gene masks or modifies another gene; dominance is between alleles of the same gene Six modified ratios: 9:7, 9:3:4, 12:3:1, 15:1, 13:3, 9:6:1, and each still totals 16 Test cross ratios differ and help identify the type (1:3, 1:1:2, 2:1:1, 3:1, 3:1, 1:2:1) Bombay phenotype: hh blocks H substance, so ABO alleles cannot act, and the person types as O; needs Bombay donors EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. What is epistasis? Explain it with a suitable example. past paper Answer Definition: Epistasis is an interaction between alleles of different genes in which the allele(s) of one gene (the epistatic gene) mask or suppress the expression of the alleles of another gene (the hypostatic gene). It differs from dominance, which is between alleles of the same gene. Example (Labrador coat colour, recessive epistasis): gene B controls pigment (B_ black, bb brown), gene E controls deposition of pigment (E_ deposits, ee does not). BbEe x BbEe gives 9 black (B_E_) : 3 chocolate (bbE_) : 4 yellow (B_ee and bbee). Yellow dogs carry black or brown alleles, but ee hides them. Show the 16 box Punnett square and the modified ratio. Mention that it is a modification of 9 : 3 : 3 : 1 because the last two classes merge. DSC-14 Unit 1 Notes | page 59 of 85 Q2. Two albino snails are crossed and their F1 progeny (AaBb) are pigmented. On selfing F1, the F2 shows 135 pigmented snails and 105 albino snails. What is the underlying genetic phenomenon in this cross? Mention the likely biochemical pathway needed to form the pigment. past paper Answer The total is 240 = 15 x 16, so the ratio is 135 : 105 = 9 : 7. This is complementary gene action (duplicate recessive epistasis). The two albino parents are aaBB x AAbb, each mutant in a different step of the pathway, so their F1 (AaBb) has a working copy of both genes and is pigmented. Pathway: colourless precursor 1 --(enzyme A, from gene A)--> colourless intermediate -(enzyme B, from gene B)--> pigment. Pigment needs both enzymes, so only A_B_ snails (9/16) are pigmented, and any snail with aa or bb (7/16) is albino. Q3. Differentiate between blood group O and Bombay phenotype. past paper Answer See the comparison table above. Key points: ordinary O (ii) has the H antigen and its red cells are not agglutinated by anti-H; Bombay (hh) individuals lack H substance because the H gene is nonfunctional, so their cells lack A, B and H antigen even if they carry IA or IB. Serum of Bombay individuals has anti-A, anti-B and anti-H, so they can only receive blood from another Bombay donor. The Bombay phenotype is an example of recessive epistasis (hh is epistatic to the ABO gene) and it can give apparent violations of ABO inheritance. Q4. Two true breeding white flowered sweet pea lines are crossed and F1 is purple. F2 has purple and white flowers. (a) Give the parent genotypes. (b) What F2 ratio is expected? (c) What is the test cross ratio of the F1? hard Answer (a) CCpp x ccPP. (b) 9 purple : 7 white. (c) F1 CcPp x ccpp gives CcPp (purple) : Ccpp : ccPp : ccpp (all white) = 1 purple : 3 white. DSC-14 Unit 1 Notes | page 60 of 85 Q5. In Labradors, a black male (BbEe) is crossed with a yellow female (bbee). What phenotypes and ratios appear in the puppies? A second cross is BbEe x BbEe and gives 320 puppies: 175 black, 66 chocolate and 79 yellow. Test the fit to 9 : 3 : 4 (critical chisquare for 2 df at 5 percent is 5.99). numerical Answer First cross: BbEe, Bbee, bbEe, bbee = black : yellow : chocolate : yellow = 1 black : 1 chocolate : 2 yellow. Second cross: expected 180 black, 60 chocolate, 80 yellow. Chi-square = 25/180 + 36/60 + 1/80 = 0.139 + 0.600 + 0.0125 = 0.75. This is much less than 5.99, so the data fit 9 : 3 : 4 (recessive epistasis). Q6. A cross gives an F2 ratio of 12 white : 3 yellow : 1 green in squash. (a) Name the type of interaction. (b) What is the genotype of the F1 and the parents? (c) Give the test cross ratio. hard Answer (a) Dominant epistasis: W_ is epistatic to Y. (b) F1 is WwYy (white). True breeding parents can be WWYY (white) x wwyy (green), or WWyy (white) x wwYY (yellow). (c) WwYy x wwyy gives WwYy (white), Wwyy (white), wwYy (yellow), wwyy (green) = 2 white : 1 yellow : 1 green. Q7. A Bombay woman needs a transfusion. Which donors can she safely use, and why? Can her blood be given to a group O patient? hard Answer Only Bombay (hh) donors. Her serum has anti-H, and ordinary O, A, B and AB red cells all carry H antigen (O has H itself, A and B are built on H), so they would be destroyed. Her own cells lack H, A and B antigens, so they can be given safely to almost anyone (no antigen to attack), including a group O patient, in terms of red cell antigens. Compatibility still needs to be checked for other blood groups (Rh). DSC-14 Unit 1 Notes | page 61 of 85 1.12 Pleiotropy Pleiotropy (Ludwig Plate, 1910) is the ability of a single gene to affect several different traits. It is the opposite of polygenic inheritance (many genes, one trait). Usually pleiotropy is a result of the gene product having several roles or being used in several tissues. Example: phenylketonuria (PKU) Intellectual disability Fair skin and hair PAH gene mutation (phenylketonuria) Seizures Musty body odour Phenylalanine hydroxylase lost Eczema Small head size One gene, one biochemical defect, many unrelated looking traits A single mutation in the phenylalanine hydroxylase gene has many effects because phenylalanine and tyrosine feed into several pathways. PAH is the enzyme that converts phenylalanine to tyrosine. Without it phenylalanine builds up, which damages the developing brain (intellectual disability, seizures, small head). Less tyrosine reaches the melanin pathway, so PKU patients tend to have fair skin and hair. Excess phenylalanine leaves the body as by-products that give a musty odour, and eczema is common. More examples Gene Effects FBN1 (Marfan Fibrillin defect: tall stature, long fingers, lens dislocation in the eye, aortic weakness syndrome) HBB (sickle cell) Sickled cells: anaemia, pain crises, organ damage, spleen damage, and malaria resistance in carriers CFTR (cystic fibrosis) Thick mucus in lungs, pancreatic damage, salty sweat, male infertility AY in mice Yellow coat colour (dominant) and embryonic death when homozygous (recessive) DSC-14 Unit 1 Notes | page 62 of 85 Gene Effects white in Drosophila White eyes, and effects on sperm storage and behaviour Mendel's gene A in Purple flower, grey seed coat and red axils at the leaf all follow together peas Primary and secondary pleiotropy Primary: the gene product directly affects several processes (a protein used in several tissues). Secondary: one direct effect causes a chain of consequences (PKU: build up of phenylalanine leads to damage in several organs). Pleiotropy, polygeny and epistasis compared Pleiotropy Polygenic inheritance Epistasis One Many Two (or more) Traits affected Many traits One trait One trait (masked) Sign in data Unrelated traits always travel Continuous variation, bell Modified ratio adding to together curve 16 PKU, Marfan Height, skin colour Labrador coat, Bombay Genes involved Example Key idea: Ask two questions. How many genes? How many traits? One gene, many traits is pleiotropy. Many genes, one trait is polygenic. Two genes with one masking the other is epistasis. QUICK RECAP Pleiotropy: one gene, several phenotypic effects (PKU, Marfan, sickle cell, A^Y mice) Usually because the gene product is used in many tissues or pathways Primary pleiotropy is direct, secondary pleiotropy is a chain of consequences Do not confuse with polygenic inheritance (many genes) or epistasis (gene interaction) DSC-14 Unit 1 Notes | page 63 of 85 EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. Write a short note on pleiotropy. past paper Answer Definition (Plate, 1910). Mechanism: the gene product is involved in more than one pathway or in more than one tissue. Examples: PKU (mental retardation, fair skin, eczema, odour), Marfan syndrome (FBN1: skeletal, eye and heart effects), sickle cell (anaemia, organ damage, malaria resistance). Mention the yellow mouse (AY gives yellow coat and lethality). Distinguish from polygeny and epistasis. Types: primary and secondary. Importance: medically, pleiotropy explains syndromes; in evolution, it means selection on one trait changes others. Q2. Identify the phenomenon (pleiotropy, polygenic inheritance or epistasis) in each: (a) a single mutation in FBN1 causes tall stature, lens dislocation and aortic weakness; (b) human height depends on hundreds of loci; (c) ee blocks pigment deposition in Labradors whatever the genotype at B; (d) mutation in a gene for a channel protein causes both deafness and heart defects. concept Answer (a) Pleiotropy. (b) Polygenic inheritance. (c) Epistasis. (d) Pleiotropy (one gene, two organs). Q3. How does the 2 : 1 ratio from yellow x yellow mice reveal pleiotropy of the AY allele? hard Answer AY has two effects at once: dominant effect on coat colour (heterozygote is yellow) and a recessive effect on survival (homozygote dies). One gene, two phenotypic effects with different dominance relationships, so it is pleiotropic. The 2 : 1 ratio comes from losing the homozygous class, showing that an allele can be dominant for one trait and recessive for another. DSC-14 Unit 1 Notes | page 64 of 85 Q4. Why do people with PKU often have lighter skin and hair than their siblings, although the gene defect is in an enzyme that has nothing directly to do with pigment? hard Answer PAH converts phenylalanine to tyrosine. Tyrosine is the starting material for melanin, so PKU patients have low tyrosine, and melanin production is reduced, giving fair skin and hair. It is a case of secondary pleiotropy: one enzyme block leads to consequences in a different pathway. DSC-14 Unit 1 Notes | page 65 of 85 1.13 Penetrance and expressivity A genotype does not always give its phenotype in every individual, and when it does, it does not always give it to the same degree. Two terms describe this. Penetrance: the percentage of individuals with a given genotype who show the associated phenotype at all. It is an all-or-none property of a population. Expressivity: the degree or intensity to which a phenotype is expressed in individuals who do show it. It is a matter of severity in each person. Penetrance (%) = (number showing the phenotype / number with the genotype) x 10 Penetrance Complete penetrance: every individual with the genotype shows the trait (100 percent). Incomplete (reduced) penetrance: some genotype carriers show no trait. Example: polydactyly (extra digits) is caused by a dominant allele P, but not every PP or Pp person has extra fingers; in one family a parent with the allele showed no extra digits yet passed the allele to a child who did. Other examples: retinoblastoma (about 90 percent), BRCA1 mutations and breast cancer (well below 100 percent, and dependent on age). I Aa aa shows the trait no trait II Aa affected aa Aa, no trait aa (spouse) has the allele, no trait (non-penetrant) III Aa affected aa A dominant trait that skips a generation. II-2 has the allele but does not show the trait (non-penetrance) and still passes it to III-1. Expressivity Variable expressivity: people with the same genotype show the trait at different levels. Examples: polydactyly (one extra finger in one person, extra digits on hands and feet in another), DSC-14 Unit 1 Notes | page 66 of 85 neurofibromatosis type 1 (from a few skin spots to serious tumours), Marfan syndrome (mild to life threatening), eye size in Drosophila eyeless (from small eyes to no eyes). Causes Modifier genes elsewhere in the genome. Environmental factors (diet, temperature, exposure). Age (Huntington disease and BRCA1 risk increase with age). Epigenetic differences and chance events in development. Penetrance vs expressivity Penetrance Expressivity Question it answers Does the genotype show the trait at all? How strongly does it show? Level Population, a percentage Individual, a degree of severity Not fully seen when Carrier has no trait (non-penetrant) Trait is mild in one person, severe in another Example Polydactyly skipping a generation Polydactyly of one finger vs several digits Exam trap: A trait can have 100 percent penetrance and still show variable expressivity (neurofibromatosis type 1). And a trait with low penetrance can be uniform whenever it is shown. QUICK RECAP Penetrance: percent of genotype carriers who show the phenotype at all Expressivity: degree to which the phenotype is shown by those who have it A dominant trait can skip a generation when a carrier is non-penetrant Modifier genes, environment and age cause variation in both EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper DSC-14 Unit 1 Notes | page 67 of 85 before opening the answer. Q1. Differentiate between penetrance and expressivity using polydactyly as an example. past paper Answer Penetrance is the proportion of people carrying the polydactyly allele who show extra digits at all. If 80 of 100 carriers show it, penetrance is 80 percent. The other 20 are carriers with normal hands (non-penetrant). Expressivity is how much extra digit each affected person has: a small extra nub of skin on one hand in one person, extra fully formed digits on all four limbs in another. Penetrance = presence or absence; expressivity = degree. Q2. A dominant allele has 75 percent penetrance. What is the probability that a child shows the trait if the parents are (a) Aa x aa, (b) Aa x Aa? numerical Answer (a) P(child has A) = 1/2, then 75 percent penetrant: 1/2 x 0.75 = 0.375 (3/8). (b) P(child has A_) = 3/4, so 3/4 x 0.75 = 0.5625 (9/16). Q3. Parents Aa (unaffected, non-penetrant) x aa have four children. Penetrance is 80 percent. What is the probability that none of the four shows the trait? hard Answer P(a child shows the trait) = 1/2 x 0.80 = 0.40, so P(does not) = 0.60. For four independent children: 0.604 = 0.13. Point to note: a child not showing the trait may still be Aa, so 'no affected children' does not mean 'no carriers'. Q4. A pedigree of a dominant disease shows an unaffected individual whose parent and child are both affected. How can this be explained, and what is this individual's genotype? concept Answer Incomplete penetrance. The unaffected individual carries the dominant allele (Aa) but did not express it. Since both the parent and the child are affected, the allele must have passed through this person. Genotype: Aa (obligate carrier). It shows why a dominant trait can appear to skip a generation. DSC-14 Unit 1 Notes | page 68 of 85 Q5. Can a genotype show 100 percent penetrance and variable expressivity at the same time? Give an example. hard Answer Yes. In neurofibromatosis type 1 practically every carrier shows some feature (full penetrance), but the severity ranges from a few skin spots to serious tumours (variable expressivity). The two measures are independent. DSC-14 Unit 1 Notes | page 69 of 85 1.14 Norm of reaction A genotype does not fix a phenotype. The norm of reaction is the range of phenotypes that a single genotype produces across a range of environments. The idea was developed by Woltereck (1909) from work on Daphnia. Reading a norm of reaction graph Plot the phenotype (y axis) against the environment (x axis) for each genotype. Each line is the norm of reaction of one genotype. Parallel lines: no interaction Crossing lines: interaction G2 G1 G1 Phenotype value Phenotype value G3 G2 G3 Environment 1 Environment 2 Environment 1 Environment 2 Left: the lines are parallel, so genotype and environment act separately. Right: the lines cross or diverge, so the effect of the environment depends on the genotype (genotype by environment interaction). A flat line means the genotype is insensitive to the environment (buffered). Parallel lines mean the environment shifts all genotypes in the same way (no interaction). Lines that cross or diverge show a genotype by environment interaction: the best genotype in one environment is not the best in another. Examples Himalayan rabbit: the allele ch makes a temperature sensitive form of tyrosinase (the pigment enzyme) that works only at low temperature. Ears, nose, feet and tail (cooler) are dark, the warm body is white. If a patch of white fur is removed and the area is kept cold, the new fur grows black. Kept at high temperature (above about 30 degrees C) the whole rabbit stays white. Primula sinensis: red flowers at 20 degrees C, white flowers at 30 to 35 degrees C for the same genotype. DSC-14 Unit 1 Notes | page 70 of 85 Achillea (yarrow): cuttings (clones) of one plant grown at low, medium and high altitude showed different growth patterns, and no single genotype was best at all altitudes. Hydrangea: flower colour depends on soil acidity (blue in acid soil, pink in alkaline soil). Drosophila: the number of eye facets in Bar mutants and wing length in vestigial flies change with the rearing temperature. Human PKU: the same pp genotype gives severe intellectual disability on a normal diet, but a nearly normal phenotype on a low phenylalanine diet. How it is measured You need many individuals of the same genotype raised in different environments: clones or cuttings of plants, inbred lines, or identical twins reared apart. Mendelian crosses cannot do this because each offspring has a different genotype. Why it matters Genes give a range of possible phenotypes, not a single outcome ("genes are not destiny"). Heritability (Topic 17) is not a fixed property of a trait: it depends on the range of environments in the population. It is the basis of the difference between penetrance and phenocopy (Topic 15). QUICK RECAP Norm of reaction: the range of phenotypes one genotype produces across environments Parallel lines: no genotype by environment interaction; crossing lines: interaction Himalayan rabbit, Primula, Achillea, Hydrangea and PKU diet are the standard examples Measured using clones, inbred lines or identical twins so genotype is held constant EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper DSC-14 Unit 1 Notes | page 71 of 85 before opening the answer. Q1. Explain the norm of reaction. Give one plant and one animal example. past paper Answer The norm of reaction is the range of phenotypes a genotype produces across different environments (represented by a line on a phenotype versus environment graph). Plant: Primula sinensis, red at 20 degrees C and white at 30 to 35 degrees C for the same genotype. Animal: the Himalayan rabbit, where the temperature sensitive tyrosinase gives dark fur only in cold body parts. Mention that the genotype specifies a range and the environment selects a point in it. Q2. Mean plant height (cm) of three genotypes measured at 15 degrees C and 25 degrees C: G1: 10 and 20; G2: 15 and 25; G3: 30 and 20. Is there a genotype by environment interaction? Which genotype is best in each environment? numerical Answer G1 and G2 both rise by 10, so they are parallel. G3 falls by 10, so its line crosses those of G1 and G2. This is a genotype by environment interaction. At 15 degrees C the best is G3 (30), at 25 degrees C the best is G2 (25). No genotype is best in both environments. Q3. A Himalayan rabbit is raised at 35 degrees C throughout its life. What is its coat like? What happens if a patch of fur is shaved from its back and the rabbit is kept at 5 degrees C? hard Answer At 35 degrees C the temperature sensitive tyrosinase is inactive, so no pigment forms and the whole coat is white, including the ears and tail. On the shaved back kept at 5 degrees C the enzyme is active, so the regrown fur is black. This shows that the phenotype depends on the environment and the genotype is unchanged. DSC-14 Unit 1 Notes | page 72 of 85 Q4. How would you experimentally determine the norm of reaction of a plant genotype and why can it not be done with F2 seeds of a cross? hard Answer Take cuttings (clones) of one plant, or use a completely inbred line, so all individuals have the same genotype, and grow samples in several controlled environments. Measure the trait. F2 seeds cannot be used because F2 plants are genetically different from one another, so differences between them mix up genotype and environment effects. Q5. Distinguish the norm of reaction from penetrance. concept Answer Norm of reaction: the complete range of phenotypes one genotype shows across environments (a continuous relationship, environment specific). Penetrance: the percentage of individuals with the genotype who show the phenotype at all, in a given population and environment. Norm of reaction describes how a phenotype changes with environment; penetrance is a single number describing whether it appears. DSC-14 Unit 1 Notes | page 73 of 85 1.15 Phenocopy A phenocopy is a phenotype produced by the environment that mimics the phenotype of a specific genotype (a gene mutation), even though the individual does not carry that genotype. The word was introduced by Richard Goldschmidt in 1935. The key fact is that a phenocopy is not inherited: the offspring of a phenocopy are normal. Examples Drosophila heat shock: Goldschmidt (and others) exposed Drosophila pupae to a heat shock at a sensitive stage and obtained adult flies that looked like known mutants (for example with changed wing veins or abnormal abdomen), but their offspring reared normally were wild type. Thalidomide babies: the drug taken in early pregnancy caused short or missing limbs (phocomelia), resembling a rare recessive genetic disorder. The next generation is not affected. German measles (rubella) in early pregnancy: deafness, cataract and heart defects that resemble inherited disorders. Iodine deficiency goitre and rickets due to vitamin D deficiency can mimic inherited conditions. How to tell a phenocopy from a genetic condition 1. Breeding test: raise offspring in a normal environment. Genetic conditions come back, phenocopies do not. 2. Pedigree: a genetic disorder follows a Mendelian pattern in the family; a phenocopy occurs as an isolated case. 3. History: look for an exposure (drug, infection, radiation, diet) at the critical stage of development. 4. Timing: phenocopies are only produced if the environmental agent acts at a specific sensitive period. Related terms Term Meaning Phenocopy Environmentally produced phenotype that mimics a genetic phenotype; not inherited DSC-14 Unit 1 Notes | page 74 of 85 Term Meaning Genocopy Different genotypes (mutations in different genes) that produce the same phenotype Mutant phenotype Genetically caused phenotype that is inherited Norm of reaction Range of phenotypes of one genotype across environments Link to Topic 14: A phenocopy is an extreme point on a wild type genotype's norm of reaction. The wild type genotype, under an unusual environment, gives a phenotype that another genotype gives in the normal environment. QUICK RECAP Phenocopy: an environmentally caused phenotype that looks like a genetic mutant Not inherited; offspring raised in the normal environment are normal Examples: heat shock phenocopies in Drosophila, thalidomide limbs, rubella defects Distinguish from genetic conditions by breeding tests, pedigree and exposure history EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. What is a phenocopy? Give two examples. past paper style Answer A phenotype caused by the environment (drug, temperature, infection) that mimics the phenotype produced by a mutant gene, but which is not heritable. Examples: (1) heat shock of Drosophila pupae giving flies with wing abnormalities like those of known mutants, whose progeny are normal; (2) thalidomide induced limb reduction resembling a genetic phocomelia; (3) rubella induced defects. DSC-14 Unit 1 Notes | page 75 of 85 Q2. Wild type Drosophila pupae are given a heat treatment and the adults show cross veinless wings like the mutant crossveinless. When these adults are bred at normal temperature, all offspring have normal wings. Explain. hard Answer The heat treatment produced a phenocopy. The flies are genetically wild type, and the heat disturbed the developmental process at a sensitive stage, giving a phenotype that resembles the mutant. Because the genotype is unchanged, the offspring reared at normal temperature are wild type. A true crossveinless mutant would breed true. Q3. Two families each have a child with very short limbs. Suggest an approach to decide whether each is a genetic case or a phenocopy. hard Answer Check the pedigree for other affected relatives and a Mendelian pattern (genetic), and check the mother's history for thalidomide or other teratogens during the critical weeks of pregnancy (phenocopy). Then observe the affected child's offspring or relatives: a genetic form will recur, the phenocopy will not. Genetic testing for a known mutation can confirm. Q4. Differentiate between a phenocopy and a genocopy. concept Answer Phenocopy: one genotype (normal) but an environmentally induced abnormal phenotype that mimics a mutant. Genocopy: different genotypes (mutations in different genes) giving the same or similar phenotype; both are inherited (for example, different genes causing similar forms of retinitis pigmentosa or deafness). DSC-14 Unit 1 Notes | page 76 of 85 1.16 Polygenic inheritance Many traits (height, skin colour, weight, yield, intelligence test scores) do not fall into distinct classes. They vary continuously from one extreme to the other. Polygenic inheritance means that a trait is controlled by many genes, each with a small, similar and additive effect, and is further blurred by the environment. It is also called quantitative or multiple factor inheritance. Nilsson-Ehle's wheat (1909) Herman Nilsson-Ehle studied kernel colour in wheat, varying from dark red to white. One gene pair: red x white gives F1 intermediate and F2 3 red : 1 white (or 1 : 2 : 1 with incomplete dominance). Two gene pairs: dark red (R1R1R2R2) x white (r1r1r2r2) gives medium red F1 (R1r1R2r2). F2 has 5 shades in the ratio 1 : 4 : 6 : 4 : 1, and 15/16 are coloured to some degree, 1/16 are white (the 15 : 1 ratio when shades are ignored). Three gene pairs: 7 shades in 1 : 6 : 15 : 20 : 15 : 6 : 1 and 63 coloured : 1 white. Each red allele (R) adds the same amount of pigment and the effects add up, so what matters is the number of R alleles. 2 gene pairs: 5 classes (1:4:6:4:1) 3 gene pairs: 7 classes (1:6:15:20:15:6:1) 6 20 4 15 4 15 6 1 6 1 1 0 1 2 3 4 0 1 1 2 3 4 5 6 x axis: number of additive (dark) alleles in the genotype, y axis: relative frequency out of 16 (left) and 64 (right) The more gene pairs involved, the smoother the bell shape and the rarer the two extremes Frequency of genotypes by number of additive alleles in the F2. The distribution is the binomial (Pascal triangle) and approaches a bell curve as gene number rises. DSC-14 Unit 1 Notes | page 77 of 85 Rules for equal additive genes Number of gene pairs = n Phenotypic classes in F2 = 2n + 1 Fraction of F2 like either extreme parent = (1/4)^n = 1 / 4^n F2 class frequencies = coefficients of (a + b)^(2n) Gene pairs n F2 classes Frequency of each extreme Ratio 1 3 1/4 1:2:1 2 5 1/16 1:4:6:4:1 3 7 1/64 1 : 6 : 15 : 20 : 15 : 6 : 1 4 9 1/256 1 : 8 : 28 : 56 : 70 : 56 : 28 : 8 : 1 Human skin colour Davenport (1913) proposed that human skin colour is controlled by about 3 gene pairs with additive effects. Today we know many more genes are involved (dozens of loci), each adding a small effect, and sun exposure changes the tan. A child of a very dark and a very light parent is usually intermediate, and the children of two intermediate parents range from very light to very dark. For a three gene model, two parents AaBbCc x AaBbCc give a child with exactly k dark alleles with probability C(6,k)/64: 1/64 for the lightest, 15/64 for two dark alleles, 20/64 for three (medium). Features of polygenic traits Continuous range of phenotypes; a bell shaped (normal) distribution in the population. Many genes, small equal additive effects; each individual gene follows Mendel's laws. Strong environmental influence (nutrition on height, sun on skin). Transgressive segregation: in the F2 some individuals are more extreme than either parent, because they collect more additive alleles than either parent had. F1 is uniform and intermediate; F2 is much more variable than F1 (genetic plus environmental variation, against environmental variation only in F1). Analysed with the mean, variance and heritability, and located by QTL mapping. DSC-14 Unit 1 Notes | page 78 of 85 Polygenic vs Mendelian traits Mendelian (oligogenic) trait Polygenic trait Number of genes One or two Many Effect of each gene Large Small, additive Phenotypes Distinct classes Continuous range Environment Small effect Large effect Analysis Ratios, Punnett squares Means, variances, heritability Example Pea seed shape Height, grain colour QUICK RECAP Polygenic trait: many genes with small, additive effects and continuous variation n gene pairs give 2n + 1 classes in F2 and each extreme is 1/4^n of the F2 Nilsson-Ehle wheat: 15 : 1 with two gene pairs (1:4:6:4:1 in shades), 63 : 1 with three Transgressive segregation: F2 individuals more extreme than either parent EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper before opening the answer. Q1. In a wheat cross between a dark red and a white variety the F2 shows 1 of every 256 plants as white. How many gene pairs control kernel colour, and how many colour classes are there in the F2? numerical Answer The fraction of each extreme is 1/4n. 1/256 = 1/44, so n = 4 gene pairs. The number of classes is 2n + 1 = 9. DSC-14 Unit 1 Notes | page 79 of 85 Q2. A plant with genotype AABBcc (each capital letter adds 1 unit of height) is crossed with aabbCC. (a) What is the F1 genotype and height class? (b) How many height classes appear in F2? (c) What fraction of the F2 is as tall as the F1? hard Answer (a) F1 is AaBbCc, with 3 additive alleles (medium). (b) 3 gene pairs, so 2n + 1 = 7 classes (0 to 6 additive alleles). (c) The class with 3 additive alleles is 20/64 = 5/16 of the F2 (about 31 percent). Q3. Two parents of medium skin colour are both AaBbCc (three additive gene pairs). What is the chance that a child is as dark as the darkest possible genotype, and what is the chance that the child has exactly two dark alleles? numerical Answer Darkest is AABBCC: (1/4)3 = 1/64. Exactly two dark alleles among 6: C(6,2)/64 = 15/64. Q4. Why is the F2 of a polygenic cross more variable than the F1? Why can F2 contain plants taller than the tallest parent? hard Answer The F1 individuals are all genetically identical, so the variation among them is environmental only. The F2 individuals differ genetically as well because the alleles have segregated and reassorted, so there is genetic plus environmental variation. Some F2 individuals collect more additive alleles than either parent had (transgressive segregation), so they can be more extreme than both parents. Q5. How would you distinguish a trait controlled by two genes with incomplete dominance from a polygenic trait, given only F2 data? hard Answer Count the F2 classes and their ratios. Two genes with additive alleles give five classes in a 1 : 4 : 6 : 4 : 1 ratio, and the extremes are each 1/16. In a truly polygenic trait (many genes), the number of classes is large, the extremes are very rare, the classes are hard to separate because environment blurs them, and the distribution looks like a smooth bell curve. DSC-14 Unit 1 Notes | page 80 of 85 1.17 Continuous vs discontinuous variation and heritability Continuous vs discontinuous variation Discontinuous variation: distinct classes A B AB Continuous variation: smooth bell curve O e.g. ABO blood group, pea seed shape e.g. human height, skin colour, grain yield Discontinuous traits fall into separate classes. Continuous traits form a smooth bell curve. Discontinuous (qualitative) Continuous (quantitative) Classes Distinct, no intermediates A continuous range, no clear classes Measured by Counting individuals in each class Measuring (cm, kg), with means and variances Genes One or few genes of large effect Many genes of small effect (polygenic) Environment Little influence Large influence Distribution Bar graph Bell shaped curve Examples ABO blood group, pea seed shape, sex Human height, skin colour, grain yield Exam trap: A bell curve does not prove many genes. A trait controlled by a single gene can also appear continuous if environmental variation is large, because the classes overlap. A few genes plus strong environmental effects can look continuous too. Heritability: how much of the variation is genetic? The phenotypic variance in a population (VP) can be divided into parts: DSC-14 Unit 1 Notes | page 81 of 85 VP phenotypic variance VG genetic variance VA additive VD dominance VE environmental VI epistatic Broad sense: H2 = VG / VP Narrow sense: h2 = VA / VP VP = VG + VE VG = VA + VD + VI Partitioning of variance. VP is total phenotypic variance in the population. VG = genetic variance = VA (additive, effects of alleles adding up) + VD (dominance, interaction between alleles at a locus) + VI (epistasis, interaction between loci). VE = environmental variance (and a genotype by environment term). Broad and narrow sense heritability Broad sense heritability (H 2 ) Narrow sense heritability (h 2 ) Formula H2 = VG / VP h2 = VA / VP Includes All genetic variance (additive, Only additive genetic variance dominance and epistatic) Range 0 to 1 0 to 1, always at most H2 Tells How much of phenotype variation is How much of the variation is passed on reliably from you due to genotype differences parents to offspring, and so how well selection works Use Comparing genetic and environmental Predicting response to selection in breeding variation Estimating heritability From variances: if VP = 100 and VE = 40 (measured in an inbred line, where VG = 0), then VG = 60 and H2 = 0.60. Realised heritability (selection experiment): h2 = R / S, where S is the selection differential (mean of selected parents minus mean of the population) and R is the response (mean of offspring minus mean of the original population). The breeder's equation is R = h2S. Parent offspring regression: the slope of offspring value on the mean of the two parents is h2 (on a single parent it is h2/2). DSC-14 Unit 1 Notes | page 82 of 85 Twin studies (Falconer): H2 is roughly 2 x (rMZ - rDZ), the difference between correlations in identical and non-identical twins. Example: population mean height 170 cm, selected parents' mean 180 cm (S = 10). Offspring mean = 174 cm, so R = 4. h^2 = R / S = 4 / 10 = 0.4 Misunderstandings to avoid Heritability describes variation in a population, not the degree to which an individual's trait is genetic. "Heritability of 0.8 for height" does not mean 80 percent of your height is from genes. It is specific to a population and an environment. Change the environment or the population (for example a uniform inbred line) and it changes. In an inbred line, VG = 0, so H2 = 0. High heritability does not mean a trait cannot be changed by the environment (height has high heritability yet rose with nutrition). A trait that is completely genetic can have low heritability. The number of fingers on a human hand is genetically determined (five), but the small variation among people is mostly caused by accidents, so heritability is near zero. QUICK RECAP Discontinuous variation gives classes (one or few genes); continuous variation gives a bell curve (many genes plus environment) VP = VG + VE and VG = VA + VD + VI Broad sense H^2 = VG/VP; narrow sense h^2 = VA/VP; realised h^2 = R/S Heritability is about population variance, not individuals, and depends on the environment and population EXAM QUESTIONS Written in the style of your teacher's paper: apply the concept, do not just recall it. Try each one on paper DSC-14 Unit 1 Notes | page 83 of 85 before opening the answer. Q1. Distinguish between broad sense and narrow sense heritability. past paper Answer Broad sense heritability H2 = VG/VP includes all genetic variance (additive, dominance and epistatic). Narrow sense heritability h2 = VA/VP includes only additive variance, the part that responds to selection because additive effects of alleles pass predictably from parent to offspring. Since VA is part of VG, h2 is never larger than H2. Broad sense measures the overall genetic contribution to variation, narrow sense predicts response to selection. Q2. In a population VP = 100 units, VE = 40, VD = 15 and VI = 5. Calculate VG, VA, H2 and h2. numerical Answer VG = VP - VE = 60. VA = VG - VD - VI = 60 - 15 - 5 = 40. H2 = 60/100 = 0.60. h2 = 40/100 = 0.40. Q3. Mean body weight of a flock is 2.0 kg. Birds with a mean of 2.6 kg are selected as parents and the offspring average 2.2 kg. Calculate the realised heritability, and predict the offspring mean if the selected parents had averaged 3.0 kg. numerical Answer S = 2.6 - 2.0 = 0.6 kg. R = 2.2 - 2.0 = 0.2 kg. h2 = R/S = 0.2/0.6 = 0.33. With parents at 3.0 kg, S = 1.0 kg, R = 0.33 x 1.0 = 0.33 kg, so the offspring mean is about 2.33 kg. Q4. Correlation of a trait in identical twins is 0.90 and in non-identical twins is 0.60. Estimate the broad sense heritability using Falconer's formula. What assumption is needed? hard Answer H2 = 2 x (0.90 - 0.60) = 0.60. It assumes identical and non-identical twins share their environments to the same degree (equal environments assumption), and that non-identical twins share half their genes. DSC-14 Unit 1 Notes | page 84 of 85 Q5. The number of fingers on a human hand is entirely genetically determined, yet its heritability is close to zero. Reconcile these statements. hard Answer Heritability measures the proportion of variation in the population that is due to genetic differences, not the extent to which a trait is genetically determined. Nearly everyone has the same genotype for five fingers, so genetic variance VG is almost zero. The rare differences (missing fingers) are mostly due to accidents, that is environmental variance. Hence H2 = VG/VP is close to zero even though the trait is entirely genetic in origin. Q6. State true or false and justify: A bell shaped distribution of a trait rules out control by a single gene. hard Answer False. If the environmental variance is large compared with the effect of the alleles, the phenotype ranges of the genotypes overlap, so even a single gene can give a continuous bell shaped distribution. Continuous variation is best evidence for polygenic control only when environmental influence is controlled. Q7. In a completely inbred line of plants heritability is found to be zero. Explain, and say what this tells you about the trait in a different, outbred population. hard Answer In an inbred line all plants have the same genotype, so genetic variance VG = 0 and any variation is environmental. Therefore H2 = 0/VP = 0. It says nothing about outbred populations: heritability is population specific, and the same trait can have high heritability in a genetically diverse population. Notes compiled for exam review, DSC-14 Unit 1. Built from your class notes, the syllabus and past papers. DSC-14 Unit 1 Notes | page 85 of 85