Genetics 371B, Autumn 1999 Answer key -- Exam 2

1.	(25 pts total) You are a technician at a forensics company. Juliet hires you to determine the paternity of her child. There are two potential fathers of the child: Alex and Hugo. You obtain DNA samples from Juliet, her child, Alex and Hugo.
	(a)	You use PCR to genotype Juliet, her child, Alex, and Hugo at the polymorphic locus PS6 . The PS6 locus is shown below as a gray box, along with 10 base pairs of flanking sequence on each side of the PS6 locus. Which set of primers shown below would you use to amplify the PS6 locus? Circle one primer set. (5 pts) Primer set 1: 5'-GTACAGTGAC-3' 5'-CAGTCGAATC-3' Primer set 2: <---- this is the correct set 5'-GTACAGTGAC-3' 5'-GATTCGACTG-3' Primer set 3: 5'-CATGTCACTG-3' 5'-GTCAGCTTAG-3' Primer set 4: 5'-CAGTGACATG-3' 5'-CAGTCGAATC-3'
	(b)	You use gel electrophoresis to separate the DNA products obtained from each of the PCR reactions. The gel is depicted on the right. PS6 is a microsatellite repeat locus with repeat units of two (2) base pairs in length. Using this information, determine the genotype (number of repeats in each allele) of each person at PS6. (12 pts) Juliet: __20, 40____   Alex ___10, 30___ Child __10, 40____   Hugo ___40, 40___
	(c)	Of Alex and Hugo, who cannot be the child's father? Explain in ONE sentence. (8 pts) Hugo -- the identity of the mother is not in question, and it's clear that the child received the 100 bp allele from the mom (Juliet). So the dad must have the 40 bp allele, and Hugo does not have that allele.

2.	(25 pts total) In nematodes such as C. elegans, XX animals are hermaphrodites and XO animals are male. There is no Y chromosome. Dosage compensation is achieved by 50% reduction in gene transcription on both copies of the X chromosomes of hermaphrodites. In males (XO), gene transcription on the X chromosome remains normal.
	(a)	Would you expect to see Barr bodies in hermaphrodite cells? Explain in ONE sentence. (5 pts) No -- since worms use a reduction in transcription (rather than a shut-off of transcription) to accomplish dosage compensation, Barr bodies will not form.
	(b)	An X-linked nematode gene called wormy (wrm) is responsible for production of the enzyme W. A recessive, mutated allele (wrm-) of the gene fails to produce enzyme W. In males with a normal (wrm+) allele, 400 units of enzyme W are produced. How many units of enzyme W would you expect to be produced in each of the following animals? (12 pts) (i) heterozygous hermaphrodites (wrm+/wrm -) 200 units. (ii) homozygous normal hermaphrodites (wrm+/wrm+ ) 400 units. Explain in ONE sentence how you arrived at your answers above. The point of doing dosage compensation is that overall gene expression levels should be the same in animals with one X as in animals in two Xs -- so the amount of enzyme made in wildtype XX animals should be the same as in wildtype XO animals (400 units), with each copy in the hermaphrodite contributing only 200 units. In wrm+/wrm- heterozygotes, there is only one functional copy of the gene, so there should be only 200 units of enzyme overall.
	(c)	(8 pts) Mice have a wrm gene that is X-linked also. Normal male mice produce 10 units of enzyme W per cell. You have a tissue culture cell line derived from a single cell from an adult female mouse of the genotype (wrm+/wrm-). How many units of enzyme W would you expect to be produced per cell by this cell line if: (i) the X chromosome carrying the wrm- allele is late-replicating? 10 units (the mutated allele is silenced, so the normal is active). (ii) the X chromosome carrying the wrm+ allele is late-replicating? 0 units (the wildtype allele has been silenced).

3.	(14 pts total) H.J. Muller's experiment to estimate the mutation frequency in Drosophila yielded a variety of strains of flies that carry recessive lethal mutations. In addition, his crosses yielded some strains that carry mutations that do NOT cause lethality when recessive but do cause other phenotypes. You are interested in the process of wing formation, so you identify a strain of Muller's that carries a non-lethal X-chromosome mutation that causes crumpled wings. The crosses shown depict Muller's cross that generated the crumpled wings (cwi ) mutation. [NOTE: w+ and w are the same white gene alleles that you heard about in lecture.]
	(a)	(4 pts) Identify all of the flies in the above diagram that would exhibit the crumpled wings phenotype if the mutation is DOMINANT (use the identifying numbers shown next to the flies): Flies 3, 5, 7 (any animal that has the mutation). Identify all the flies that would exhibit crumpled wings if the mutation is RECESSIVE: Fly # 7 (this fly is hemizygous for the mutated allele, so the mutant phenotype will be expressed).
	(b)	You determine that the crumpled wings mutation (cwi) is recessive. To map cwi, you cross crumpled winged males to females of 7 different strains that carry different deletions of known position on the X-chromosome. The cross and the phenotypes of the resulting female progeny are shown below. Where is the cwi gene? Mark its location (with an arrow) on the chromosome below the figure. [The vertical grid is provided for your convenience. Remember that gaps in the chromosomes represent regions of the chromosome that are deleted.] (10 pts) The indicated region is the only portion that is missing in ALL flies that show the cwi phenotype -- that must be where the cwi gene is located.

4.	(18 pts total) The gene for an autosomal dominant disease maps to chromosome 21. Based on pedigree analysis of two separate, large families that have a history of this disease, Lod score values for odds of linkage of the disease gene to each of two polymorphic loci (PS1 and PS2) were calculated (q = % recombination; D = disease gene):
	(a)	What are the loci and map distances for which there is conclusive evidence (at least 95% confidence) for or against linkage to the disease gene? (12 pts) Lod scores derived from separate pedigrees can be added up -- so conclusive evidence is available for both polymorphic loci. PS1 is tightly linked to the disease gene, with conclusive evidence in favor of linkage at 10 cM or less. For PS2, there is conclusive evidence against linkage at a map distance of 10 cM or less.
	(b)	You heard in lecture about cloning of the normal allele of the Niemann-Pick gene by rescue of the disease phenotype in tissue culture cells. Why would a similar strategy NOT work in the case of the disease gene described in (a)? -- i.e., why could the normal allele of the gene not be cloned by rescue of the disease phenotype in cultured cells? (Assume that tissue culture cells showing the disease phenotype are available.) ONE sentence, please! (6 pts) The disease phenotype is dominant, so introducing the wildtype allele will not rescue the phenotype.

5.	(18 pts total) In each of the two families described in Question 4, a Down syndrome baby was born to an unaffected couple. In both cases, the parental genotypes at PS1 are {17,20} (mom) and {6,6} (dad). Neither child showed evidence of chromosomal translocation or mosaicism -- so in each case, Down syndrome could have resulted either from meiotic nondisjunction of chromosome 21 in one of the parents, or from mitotic nondisjunction in the first cleavage in the zygote (child). Genotyping of the two babies at PS1 revealed unambiguously that Down syndrome in Baby #1 was the result of meiotic nondisjunction in the mom. In Baby #2, the genotype did not allow an unambiguous conclusion to be made.
	(a)	What was the genotype of Baby #1 at PS1? (4 pts) {17,20,6} -- the child received two PS1 alleles from the mom. If there was mitotic nondisjunction, or ND in the dad, there would be only two types of alleles in the child (see below).
	(b)	List the possible genotypes of Baby #2 at PS1. (8 pts) {17,17,6} {20,20,6} {17,6,6} {20,6,6} Any of these could result from meiotic nondisjunction in the mom (in meiosis II) or in the dad, or by mitotic nondisjunction in the zygote. Any of these could also result from meiosis I ND in themom, provided there had been an odd number of crossovers between PS1 and the centromere.
	(c)	For Baby #1, why is it not possible to say with absolute certainty whether nondisjunction occurred in Meiosis I or in Meiosis II? (ONE sentence!) (6 pts) Nondisjunction in meiosis I in the mom could give this genotype {17,20,6} if there had been an even number of crossovers (or no crossovers) between PS1 and the centromere; but ND-II could also give this genotype if there had been an odd number of crossovers between the centromere and PS1, such that the chromatids that failed to disjoin had different alleles on them.

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