hw2key - Genetics Homework #2 key 1. Suppose that a...

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Unformatted text preview: Genetics Homework #2 key 1. Suppose that a particular enzyme deficiency causes the Perturbed syndrome. A man has a perpetually perturbed sister, but he, his parents, his grandparents, and three other sisters usually remain calm. Discussions with relatives in his wife's family reveal that the syndrome is not likely to be present in her family, although some relatives recall that the brother of his wife's paternal grandmother was usually perturbed. Diagram and analyze the relevant pedigree to determine (a) the genetic basis for inheriting the perturbed trait; and (b) the highest probability that, if this couple has a child, the child will be perturbed (i.e. what is the chance of the worstcase scenario occurring?). a. (Pedigree too complicated to draw out, but this is an autosomal recessive trait.) b. For the couple in question to have an affected child, both must be carriers (A). The man has a 2/3 chance of being a carrier, since he has an affected sister. This tells us that both of his parents must be carriers. The probability of the wife being a carrier = 2/3 * * (chance that her grandmother is Aa and passed on a to her father, and the chance that her father passed on a to her). So there is a 2/3*1/2*1/2 =1/6 chance that the wife is a carrier * a 2/3 chance that the husband is a carrier so * 1/6 * 2/3 chance that kid will be perturbed= 1/36. 2. Genes L, N, and S are independently assorting and control the production of a purple pigment. a. Suppose that L, N, and S act in the following pathway: X L N Y Z S purple (Substances X, Y, and Z are all colorless.) The alternative alleles that give abnormal functioning of these genes are designated l, n, and s, respectively. A purple LLNNSS is crossed with a colorless llnnss to give a purple F1. The F1 is selfed. What proportion of the F2 individuals is colorless? In order to be purple using this pathway, individuals must have all three normal alleles: They must be LNS. The F1 is the trihybrid LlNnSs that when selfed, gives purple LNS individuals in *3/4*3/4 = 27/64 of the progeny. The remaining 1(27/64) of the progeny are colorless, having ll, nn, or ss genotypes. A 27 purple:37 colorless ratio is expected in the F2. b. Suppose instead that a different pathway is utilized. In it, the S allele produces an inhibitor that prevents the formation of purple by destroying the ability of N to carry out its function. The mechanism is as follows: Q purple S (inhibitor) (Substances Q and R are both colorless.) A colorless LLNNSS individual is crossed with a colorless llnnss, giving a colorless F1. The F1 is selfed to give an F2. What is the ratio of colorless to purple in the F2 individuals? In order to be purple using this pathway, individuals must have the L and N functions, but not the inhibitor function provided by S. They must be LNss. The chance of obtaining this genotype in the F2 is 3/4*3/4*1/4 = 9/64. The remaining 55/64 of the progeny will be colorless, so a 9 purple:55 colorless ratio is expected in the F2. c. How would you evaluate which of the biochemical pathways hypothesized in parts (a) and (b) is more likely? Here, the ratio of purple to colorless in the F2 can be used to distinguish between the two pathways. You can use a chisquare test to evaluate whether the F2 results fit either pathway. 3. Exploring Genomics exercises from Chapter 7 (page 193). Skip Exercise 1, part 4. Answer all other questions. Exercise I, part 2 (a) Stimulated by Retinoic Acid Gene 8 (STRA8); is located on human chromosome 7 (locus 7q33). (b) In mice, STRA8 expression in the testis occurs shortly after birth. In the developing ovary, Stra8 expression occurs just before initiation of meiosis, indicating a potential role for Stra8 in initiating meiosis and gamete development. Recall that in females, gamete formation through meiosis begins during embryogenesis. (c) As the gene name suggests, retinoic acid, a vitamin A derivative, is required for STRA8 expression and is a key regulator of meiotic initiation in developing ovaries. Exercise II, part 3 (a) chromosome 6 (b) The polypeptide sequence consists of 393 amino acids. (c) The Stra8 protein has a helix-loop-helix domain characteristic of many sequence-specific DNA-binding proteins, such as transcription factors, that regulate gene expression. (d) no (e) Embryonic ovaries from Stra8 / mice cultured in the presence of retinoic acid fail to undergo chromosome condensation during meiosis, thus preventing initiation of meiosis and oogenesis. L R N ...
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This note was uploaded on 09/12/2008 for the course BICD 100 taught by Professor Nehring during the Spring '08 term at UCSD.

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