Sample questions for exam 2 -- answer key

1.	A certain X-linked dominant disease in humans occurs when there are more than 40 repeats of a CGG trinucleotide in a particular gene. The unique sequence outside the repeat region is as follows: 5' CAGTATGCA------(CGG)n-------ATGCGTAAT 3' 3' GTCATACGT------(GCC)n-------TACGCATTA 5'
	(a)	In the lab freezer, you find 5 primers of different sequences (listed below). If you wanted to use PCR to amplify the trinculeotide repeat region, which of these primers would you use? Primer: #1) 5'-GTCATACGT-3' #2) 5'-ATTACGCAT-3' -- "Reverse primer" primes leftward extension from the right end #3) 5'-ATGCGTAAT-3' #4) 5'-CAGTATGCA-3' -- "Forward primer" primes rightward extension from the left end #5) 5'-TGCATACTG-3'
	(b)	Shown below are the outline of a pedigree for the disease described above, and a representation of a gel showing PCR-amplified fragments detecting the number of CGG repeats. The DNA corresponding to each individual is directly below his or her place in the pedigree. Based on the information you have been given, fill in the pedigree to show the sex of the individual as well as the phenotype (affected vs. unaffected).

2.	Males of the worm C. elegans are XO -- they have one X chromosome and no Y chromosome, and their sperm have either one X or no X chromosome. You have identified recessive mutations nd-1 and nd-2 that cause X chromosome non-disjunction in every cell undergoing meiosis I and meiosis II, respectively. For all three questions below, assume that the eggs came from normal XX animals; also assume that XX sperm produce viable, non-male offspring.
	(a)	What fraction of a +/nd-1 male worm's progeny do you expect will be male (XO)? Each egg has an X chromosome; sperm have either one X or no X. Male progeny result from the fertilization of eggs by sperm lacking X chromosomes to make XO zygotes. Therefore, the fraction of a male's progeny that are also male is equal to the fraction of sperm that lack X chromosomes. For a normal male, half the sperm are O, so half his progeny will be male. Answer = 0.5
	(b)	What fraction of an nd-1/nd-1 male worm's progeny will be male (XO)? 0.5-- same as with (a), because meiosis-I non-disjunction does not affect an XO animal (there is no homolog to disjoin from in the first place).
	(c)	What fraction of an nd-2/nd-2 male worm's progeny will be male (XO)? Meiosis II nondisjunction results in three sperm with no X chromosomes, so the fraction of male progeny goes up to 0.75 (three out of four).

3.

The pedigree below shows the inheritance of an autosomal recessive disorder in humans. Shown below the pedigree (and lined up by individual) are the Southern blot data for two RFLP loci. One RFLP has alleles of either 7 kb or 5 kb and is revealed by digestion with the restriction enzyme BamHI; the other gives either a 4 kb or 3 kb EcoRI fragment. Is there evidence for linkage between the disease gene and either of the RFLPs? If so, which one(s)? Explain in ONE or TWO sentences. For the loci that you think are linked, which individuals are likely to be recombinants? The BamHI polymorphism is likely linked to the disease gene--the disease phenotype is mostly associated with the 5 kb BamHI allele. The parents can be represented as D 7/d 5 and d 5/d 5, where D and d are the wildtype and recessive disease gene alleles and 7 and 5 are the BamHI polymorphic forms. The children are mostly either D 7/d 5 (double heterozygotes) or d 5/d 5 (double homozygotes, affected). The exceptions are individuals II-3 and II-10 , who must be recombinants. In contrast, the EcoRI polymorphism shows no sign of linkage to the disease gene -- parental and recombinant classes are seen with equal frequency.

4.	Linkage analysis was done for a certain human disease gene with respect to polymorphic loci PL1, PL2, PL3, and PL4. Some of the resulting Lod scores are plotted. Each curve represents the lod score vs. recombination frequency distribution for a polymorphic site with respect to either the disease locus (D) or with respect to one of the other polymorphic sites. For example, Curve #1 shows the lod score vs. recombination frequency between the disease locus D and polymorphic site PL1. Construct the most likely linkage map that is consistent with the data shown. Your map should include all five loci, but they need not all be on the same linkage group.
	PL1 and PL2 show significant likelihood of linkage to the disease gene locus, while PL3 shows significant evidence against linkage at closer than ~28 cM, and shows evidence against linkage at all distances--it is probably not linked to D/d. PL4 shows extremely tight linkage to PL3, so it too is likely not linked to the disease locus D. The probable map distance between D and PL1 is about 5 cM (about 5% recombination), while the D-PL2 distance is about 10 cM. The likely distance between PL1 and PL2 is about 15 cM, which would place these two markers on either side of the disease locus D. Therefore, the map for the five loci can be represented as:

5.	Five recessive mutations on the right arm of the X chromosome of Drosophila cause stringy bristle (sg), lazy eyes (ly), tin body (tn), zillion eye (ze), and miniature wing (m) phenotypes. Two wildtype male flies were exposed to X-rays and then mated separately with (non-irradiated) females showing all five recessive traits (Mating 1 and Mating 2).
	(a)	Mating 1 produced normal females and sg ly tn ze m males in equal proportions. One of these phenotypically normal females, when mated to a male sibling, produced F2 progeny that consisted solely of normal and fully mutant phenotypes. Why was this result unexpected, and how do you explain it? (ONE or TWO sentences.) The F1 females are fully heterozygous; meiosis in these females should produce recombinant gametes. However, only the parental (NCO) allele combinations are seen, indicating that there has been a chromosomal rearrangement. Because the males are viable, a deletion can be ruled out. Translocations can also be ruled out by the all-or-none nature of the phenotype (if a portion of the chromosome had been translocated to another chromosome, we'd expect to see loss of a subset of the markers due to recombination). Therefore, the rearragement was probably an inversion--recombination in the inversion loop would lead to inviable gametes. The only productive gametes would be the non-recombinant ones.
	(b)	Mating 2 produced sg ly tn ze m males and ly tn females in equal proportions. Explain the result. (ONE sentence.) A deletion of ly+ and tn+ uncovered the recessive alleles in the female progeny. The phenotype of the males is as expected.

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