Bi 122 Midterm Answer Key Problem 1 a) Assuming no crossing

Bi 122 Midterm Answer Key Problem 1 a) Assuming no crossing over, the mother could generate 2^87 types of gametes. The father can generate 2^88 types of gametes (an extra factor of 2 due to the presence of both X and Y chromosomes). (3 points) b) This could create 2^(88+87) = 2^175 different offspring. (2 points) c) The genes are no longer linked, and therefore the mother can now generate 4^87 types of gametes, and the father can generate 4^87 * 2 types of gametes. They can now generate 4^(87*2) * 2 = 4^174 * 2 different offspring. (5 points) Problem 2 a) telomeres are shorter in the 25 year old unicorn (2 points) b) 1. remove nucleus from a horse egg (4 points) 2. extract the nucleus from a somatic cell of a unicorn of either sex 3. fuse the nucleus into the egg 4. implant embryo into a female horse 5. repeat until surrogate mother gives birth to a unicorn progeny, which will be the same sex as the donor of the nucleus c) The progeny still have short telomeres and the same DNA damage as the somatic cell had. ← this was the answer we were looking for given questions 2a and 2d, but the following are also acceptable. (2 points) The unicorns may not live until they are sexually mature because of their damaged DNA. The progeny still have the mitochondrial DNA of their horse mother, which could cause problems later on. Also, there will be very little genetic diversity among first generation progeny (clones). This sort of inbreeding might create a lot of individuals homozygous for deleterious recessive alleles. d) temporarily activate telomerase in a culture of somatic cells of the unicorn you wish to clone, then use these in the procedure outline in problem 2 this can restore the length of the telomeres so the unicorn progeny don’t receive prematurely aged chromosomes (2 points) Problem 3 a) A pair of homologous chromosomes, one from the father, one from the mother (1 points) b) Three chromatids (two are sisters, one is separate) of the X chromosome are in the oocyte. One chromatid is in polar body (2 points) c) Three chromatids in the second polar body, two chromatid in the zygote (3 points) d) Embryo: diploid at chromosome 21, phenotypically normal (4 points) Problem 4 a)
In order to produce aaBbCcDDeeFfGGhhii in a selfcross, the parent must be of the genotype Aa Bb Cc ?? Ee Ff ?? ?? Ii The probability that the parent has a genotype of this form is ½ * ½ * 1 * 1 * 1 * 1 * ½ = ⅛. Now, for only these parents, we have the following probabilities for F2 offspring: Genotype F2 individual inherits: Probability of inheriting it from Aa Bb Cc ?? Ee Ff ?? ?? Ii aa ¼ Bb ½ Cc ½ DD 1 * ¼ + ¼ * ½ = ⅜ ee ¼ Ff ½ GG 1 * ½ + ¼ * ½ = ⅝ hh ¼ * ½ + 1 * ½ = ⅝ ii ¼ So the proportion of all F2 progeny with the desired genotype will be: ⅛ * ¼ * ½ * ½ * ⅜ * ¼ * ½ * ⅝ * ⅝ * ¼ = 75/2097152 (4 points) b) Hay’s speculation: the pathway is white → brown → black gene R takes white to brown, gene L takes brown to black anything with rr is white if a chicken has at least one R allele and one L allele, it’s black everything else is brown pure breeding white chickens must be rr LL (we’ll see later why it’s LL) pure breeding brown chickens must be RR ll then black progeny are RrLl RL Rl rL rl RL RRLL RRLl RrLL RrLl black black black black Rl RRLl RRll RrLl Rrll black brown black brown rL RrLl RrLl rrLL rrLl black black white white rl rrll Rrll rrLl rrll black brown white white So this gives expected ratios of 9/16 black offspring, 3/16 brown offspring, and 4/16 white offspring. Another speculation: codominance or incomplete dominance, single gene G GG → brown Gg → black gg → white Mate two brown chickens. Professor Hay’s hypothesis predicts that all offspring will be brown chickens. The other hypothesis predicts that some of them will be white. (3 points) c) The values seen match the predicted values for epistasis exactly, so chi^2 = 0. The expected values for the second speculation are 40 brown (GG), 80 black (Gg), and 40 white (gg). The actual values are 30, 90 and 40, respectively. There are 3 distinct phenotypes, so the number of degrees of freedom is 31 = 2. Chisquare test for the second speculation gives 3.750, which corresponds to a P value (~0.15) much larger than 0.05, so we cannot reject the one gene hypothesis. No, the chisquare test does not help us distinguish between these 2 possibilities. (3 points) Problem 5 Grading for this particular problem was lenient. Any logical explanation of the observed inheritance pattern that shows Bl, Br, and Y are alleles of the same gene (with Bl dominant to Br and Y, Br dominant to Y, and Y recessive to both) and W is a separate gene where a recessive allele (w) causes white (or no pigment) when homozygous. In this way the W gene is epistatic to the color gene. Problem 6 A) (2 points) There are four genes involved in leaf size control. Using complementationtest logic, the mutationgene combinations are: 1, 7, and 9; 2, 4, and 5; 3; 8 and 10… and 6 is ambiguous. Five genes were also accepted due to #6. If adequate reasoning was provided for additional genes (more than 5), that may have been accepted as well because of the ambiguities with number 6. B) (2 point) Mutation 6 very likely contains multiple mutations. C) Thrown out, due to an incorrect pset question. D) (2 points) He must have been carrying the B isoform (A is linked to the disease, and he could not have obtained an X chromosome from the father). E) (2 points) Nondisjunction in both parents has occurred, in meiosis I for both of them (or in M2 for the mother), in order to generate a son with Klinefelter’s with XXXY. (He must have three X chromosomes to have all three isoforms). F) (2 points) X chromosome inactivation allows the daughter to have the disease in certain parts of her body (at random). Problem 7 A) (4 points) The only possible set of genotypes for the parents would be: bn+bn+;shsh;or+or+, and bnbn;sh+sh+;oror. (the parents’ genotypes will occur the most frequently in the gametes of the F1 generation). Leniency was employed. B) (3 points) The genetic map would show the genes in the following order: bn, or, sh. Bn and or are spaced apart by 20 map units, while or and sh are spaced apart by 30 map units. Credit is given if done correctly based on answer to part A. C) (3 points) bn and sh are separated by 38cM, so after recombination, there will be 0.5* 38% =19% chance that the gametes will be homozygous mutated. Full credit is given if they used the recombination frequency obtained in part B. Problem 8 A) The number of alleles are not equal. (2 point) B) Gene conversion. See diagram. (2 point) C) gene conversion, along with two crossing over events. (see diagram below). (2 points) D) Meiosis I. (two crossing over events). (1 points) E) See diagrams. (3 points) Problem 9 a) Phe/F has DNA sequence TTT, TTC, and Cys/C has DNA sequence TGT, TGC. No information about codon bias. Also, as the mutation rate is given per nucleotide, there is ½ chance that the mutation will revert (see lecture slides). We also assume that that mutation will randomly change one nucleotide to another (so if the starting nucleotide is A, it can change to T, C, G with equal probability), and the mutations are dominant. So TTT> TGT= (1108)⅓*½ (108)(1108) TTT>TGC= (1108) ⅓ *½(108)⅓ *½(108) TTC>TGT=(1108) ⅓ *½(108)⅓ *½(108) TTC>TGC= (1108)⅓*½ (108)(1108) Sum all of these=3.33*109 Ser/S TCT, TCC, TCA, TCG to Pro/P, CCT, CCC, CCA, CCG TCT>CCT 2 codon changes TCT>CCC 2 codon changes TCT>CCA 2 codon changes TCT>CCG 2 codon changes TCC>CCT 2 codon changes TCC>CCC 1 codon change TCC>CCA 2 codon changes TCC>CCG 2 codon changes TCA>CCT 2 codon changes TCA>CCC 2 codon changes TCA>CCA 1 codon change TCA>CCG 2 codon changes TCG>CCT 2 codon changes TCG>CCC 2 codon changes TCG>CCA 2 codon changes TCG>CCG 1 codon change 3 1 codon change, 13 2 codon changes 1 codon change: (1108)⅓*½ (108)(1108) 2 codon change: (1108) ⅓ *½(108)⅓ *½(108) So 3*(1108)⅓*½ (108)(1108)+13*(1108) ⅓ *½(108)⅓ *½(108)=4.99*109 Val/V has GTT, GTC, GTA, GTG and Gly/G GGT, GGC, GGA, GGG GTT>GGT 1 codon change GTT>GGC 2 codon changes GTT>GGA 2 codon changes GTT>GGG 2 codon changes GTC>GGT 2 codon changes GTC>GGC 1 codon change GTC>GGA 2 codon changes GTC>GGG 2 codon changes GTA>GGT 2 codon changes GTA>GGC 2 codon changes GTA>GGA 1 codon change GTA>GGG 2 codon changes GTG>GGT 2 codon changes GTG>GGC 2 codon changes GTG>GGA 2 codon changes GTG>GGG 1 codon change 4 1 codon change, 12 2 codon changes 1 codon change: (1108)⅓*½ (108)(1108) 2 codon change: (1108) ⅓ *½(108)⅓ *½(108) So 4*(1108)⅓*½ (108)(1108)+12*(1108) ⅓ *½(108)⅓ *½(108)=6.66*109 4.99*109 *6.66*109=3.3233^1017 +1 for using the 108 probability, +1 for codon change, +1 for ½ conversion, +1 for ⅓ correct conversion, +1 for correct answer. b) Three mutations: at protein position 989, 1016, and 1534. The first two are quite close, and so we can say recombination between those two are negligible. The last two mutations are far. How far? 1534*31016*3=1554 bp apart. Given 10 mu=17*106 bp, then they are ~1/1000 % chance of recombination>1/100,000 +1 for ignoring/explaining position 989, +1 for bp conversion, +1 for understanding map units = %, +1 for correct answer. +1 free point. Problem 10 a) X inactivation Mitochondrial DNA Chimeras Mutations> different genotypes in each cell b) Passed down from father to son, and it’s always dominant because there is only one Y chromosome. Therefore, any harmful mutations are apparent and more likely will have negative consequences. Also, not many genes on the Y chromosome. Problem 11 a) 1 O 4 GXC 3 RHME 2 ABF F is the last gene, integrated on bacterial chromosome, etc. +1 correct order, +2 correct origin, +1 direction, +1 explanation b) Any good experiment was awarded full credit. An example would be to use conjugation and see if the phenotype was passed to the recipient cell (don’t interrupt). If it has not, then the gene mutation is in the chromosome.

Download Report