NAME Gene 603 Exam III November 30, 2,001

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1 NAME Gene 603 Exam III November 30, 2,001 I.. Suppose that a single nucleotide polymorphism exists in the centromere of chromosome 5 in Neurospora crassa. Use as many of the "ordered asci" as needed in figure A below to show the expected segregation pattern(s) by shading spores with one of the "alleles". FIG A The ws mutation (white spore) is about 8 map units from the centromere of LG-5. Use as many of the "ordered asci" as needed in figure B below to show expected segregation pattern(s), with expected frequency, from a cross between a ws mutant and a ws + strain (darken ascospores with the + allele; equivalent arrangements need not be shown). FIG B % II. What is the basis for the precision seen in crossing over (recombination)? (a drawing may be the easiest way to answer) heteroduplex formed by base pairing of matching DNA sequences serves as the initiation of crossing over

2 How does the precision relate to gene conversion? Gene conversion results from mismatch repair in heteroduplex regions. III. Pairwise crosses were made on E. coli B between 4 rii-a mutants and I rii-b mutant. The number of plaques produced when 10 6 progeny from each cross were spread in lawns of E. coli K, are shown in the table below: Pair A1 A2 A3 A4 B1 A A A A B1 20 Which mutations are point mutations? How do you know? A1 and B1; they revert (+ progeny when "selfed) Make a map that shows relative positions of deletions and point mutations. A1 B1 A2 A3 A4 Use a picture to describe the test and results that led to the classification of some mutants as rii-a and others as rii-b Complementation in E. coli K A x B x ppa ppb if all gene products made, host cell is lysed IV. Rank the following situations/aberrations in the order from most likely (1) to least likely (4) to have a child with trisomy 21: 2

3 1 21 years old; 45 XX T years old; 46 XY 3 21 years old; 45 XY T years old; 46 XX V. Show that q 2 + pqf = q 2 (1- F) +qf : substituting 1-q for p gives: q 2 + (1-q) q F = q 2 +qf - q 2 F = q 2 (1-F) + qf a) Suppose that a homozygous recessive condition that prevents reproduction occurs 1 in 250,000 births. What are the "gene" fequencies. q= Square root of 4 X 10-6 = 2 X 10-3 (0.002) p = b) Why/How is this recessive allele maintained in the population? in heterozygotes with selection balanced by mutation. c) Calculate F for the individual indicated by an arrow in the pedigree diagrammed below. (1/2) 5 d) What is the probability that the child would have the condition described above? (= q 2 from random mating plus pqf from inbreeding) = 6.6 X 10-5 VI. In alfalfa, which is an autoteraploid, presence of at least one P allele results in purple flowers. A cross between white and purple flowered plants gave 1/2 purple:1/2 white. A) Give the genotypes of the purple and white parents. Pppp X pppp B) What ratio(s) would be expected in progeny of a self-pollination if the purpleflowered plant? 3/4 th purple and 1/24 th white VII Shown in diagrammatic form are the gene arrangements of chromosomes I and II in 4 populations of Drosophila. Assume that each population is 3

4 homozygous for the arrangement shown, and that the "primative" organization is that found in population 1. Population Chr. 2 A B C D E F G Population Chr. 2 A B C D E F G Population Chr. 2 A B C D E F G Population 4 A B Chr C D E F G A) Starting with population 1, show how stepwise aberrations could during the development of new or subspecies. (ex. 1 -> 2 -> 3 -> 4) 1 -> 3 -> 2 -> 4 B. Define the event that occurred in each step 1-3 paracentric inversion; 3-2 pericentric inversion 2-4 translocation C. Show synapsis of heterozygotes from crosses between the following populations: 1 X 3 2 X C D E F G B A B A C D E F G 4

5 D. What factors would influence fertility in crosses between populations 3 and 4? Predict the approximate frequency of functional gametes in the hybrid. There will be a pericentric inversion and a translocation involving the same chromosomes; thus crossovers in the loop (if it can form) and alternate versus adjacent segregation will be factors. At most 1/4 th of the gametes would be expected to include full sets of the genome. VIII. Equilibrium for a balanced polymorphism (heterozygote advantage) has been established where the frequency of allele A is 0.4 and f(a) = 0.6. Fill in the chart below, showing fitness in relative terms where selection against the recessive is 0.6. Genotype AA Aa aa Frequency Fitness (W) Suppose that as in the case of Malaria, selection against the AA genotype can be relaxed. What will the gene frequencies be (theoretical) when equilibrium is re-established and what force(s) will drive the change? This will be when mutation is balanced by selection against aa: it will occur at q eq = (µ/0.6) 0.5 ; driven by selection What if selection is relaxed against both homozygotes? Then it will on;y be forward vs reverse mutation: q eq = µ/µ+v IX. You find a pregnant female mouse from an inbred strain that can swim for only a few minutes before drowning or rescue. When her pups -2 females and 2 males- are born, they exhibit the same "rapid tiring" trait (rt). Write the crosses that you would make and the results that would allow you to determine if the trait is caused by a) an autosomal recessive gene, b) a sexlinked recessive gene, c) cytoplasmic inheritance. A) rt female by RT male -> all RT; F 2 -> 3 Rt: 1 rt; same if start with reciprocal cross B) rt female by RT male -> all sons rt; all daughters RT ; RT female by rt male -> all RT progeny; sib matings -> C) rt female by RT male ->all rt progeny RT female by rt male -> all RT progeny Which of the cases could be confused with maternal effect? How would you test for this possibility? 5

6 C could be if stopped with just the reciprocal cross; continue sib matings to next generation to see if all progeny like mother or if some show segregation. X. Like peas, rice is naturally self-fertilized, so that it is relatively easy to develop pure lines. When a true breeding variety that flowers in 52 days was crossed to another that flowers in 80 days, the average days to flowering in the F 1 was 66 days. In the F 2, the average was still 66 but rare plants were seen to flower at 48 days and 84 days. Assume 9 genes are segregating in the F 1. 1) What is the value of each contributing allele? = 36 days for 18 alleles so 2 days per contributing allele What assumptions were made to derive this answer? all alleles have same effect and the true extremes were detected 2) What fraction of the F 2 plants occur at the extremes? (1/4) 9 3) Give potential genotypes for the 52 and 80 day parents, and the F 1 : 52: A'A' BB CC DD EE FF GG HH II 80: AA B'B' C'C' D'D' E'E' F'F' G'G' H'H' I'I' F 1 : A'A B'B C'C D'D E'E F'F G'G H'H I'I 4) Is there transgressive segregation in this F 2? Explain Yes: F2 extremes go beyond parents 5) If the variance in the F 1 was 4 and in the F 2 it was 24, what is H 2? Vp=24; Ve= $ so Vg = 20 and H 2 = 20/24 6) Design an experiment to measure h 2. Select all F2 above or below a specific cutoff and measure the regression of the mean in the next generation. XI What elements of a transposable element are critical for transposition? Long terminal repeats must be present; possibly transposase and integrase too. 6

7 Provide a plausible explanation as to why some transposons are active only during DNA replication. One strand is no longer methylated immediately after replication, which could allow a promnoter usually silenced by methylation to permit transcription. XII. Compare the organization of genes that encode light (κ) and heavy antibody chains In addition to similar V and J repeats, the H chain genes also have a set of Diversity regions and have multiple Constant regions that show class switching. XIII. Even if donors were plentiful, it could still be extremely difficult to find a replacement kidney for successful transplant. Why? At least 4 HLA (MHC) genes are expressed to produce cell surface antigens and each gene has many different alleles; any different allele in the transplanted organ can lead to rejection XIV. Why is Caenorhabditis elegans such a good model organism for studying developmental genetics? The adult worm has only about 1, 000 cells, each of which can be seen, targeted by laser "surgery" and followed through all cell divisions; it has some useful genetic tools including selfing, RNAi to silence genes, very short life cycle, producing quite a few progeny per mating. 7