# Dr. Young. Genetics Problems Set #1 Answer Key

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1 BIOL276 Dr. Young Name Due Genetics Problems Set #1 Answer Key For problems in genetics, if no particular order is specified, you can assume that a specific order is not required. 1. What is the probability of drawing one card at random from a deck and obtaining the following? (A deck has 52 cards. Each card has a suit spade, heart, diamond, or club and a number [1-10] or a face jack, queen, king. Thus, there are 13 cards of each suit and four cards of each number or face.) Express your answers as both a fraction and a percentage. a. The queen of diamonds P(queen) = 4/52, and P(diamond) = 13/52. Using the Multiplication Rule, (4/52)(13/52) = 1/52 = = 1.9% b. Any heart P(heart) = 13/52 =.25 = 25% c. A 10 (any suit) P(10 of any suit) = P(10) = 4/52 =.077 = 7.7% d. Any even-numbered card There are 5 even-numbered cards per suit (2, 4, 6, 8, 10), and there are 4 suits. Therefore, (4)(5)/52 = 20/52 =.38 = 38% e. An even-numbered spade There are only five even-numbered cards that are also spades. Therefore, 5/52 = 9.6% 2. In a family of seven children, what is the probability of obtaining the following numbers of boys and girls? a. All boys (!) 7 = 1/128 b. All children of the same sex The children could be all boys or all girls: (!) 7 chance of being all boys and (!) 7 chance of being all girls. 1/ /128 = 2/128 or 1/64 chance of being either all boys or all girls. Parts c e require the use of the binomial expansion. Let a equal the probability of being a girl and b equal the probability of being a boy. The probabilities of a and b are!. (a + b) 7 = a 7 + 7a 6 b + 21a 5 b a 4 b a 3 b a 2 b 5 + 7ab 6 + b 7 c. Six girls and one boy The probability for part (c) is provided for by the term 7a 6 b. Because the probabilities of a and b are!, then the overall probability is 7(!) 6 (!) = 7/128. d. Four boys and three girls This probability is provided for by the term 35a 3 b 4. The overall probability is 35(!) 3 (!) 4 = 35/128. e. Four girls and three boys Using the term 35a 4 b 3, we see that the overall probability is 35(!) 4 (!) 3 = 35/128.

2 3. Two gene loci, A and B, are unlinked (and thus assort independently), and alleles A and B are dominant over alleles a and b. Indicate the probabilities of producing the following. a. An AB gamete from an AaBb individual? 1/4 b. An AB gamete from an AABb individual? 1/2 c. An AABB zygote from a cross AaBb! AaBb? 1/16 d. An AaBb zygote from a cross AaBb! AABB? 1/4 e. An Aabb zygote from a cross AaBb! AAbb? 1/4 f. An AB phenotype from a cross AaBb! AaBb? 9/16 g. An AB phenotype from a cross aabb! AABB? 1 (100%) h. An ab phenotype from a cross AaBb! AaBB? 1/4 4. A man has either an AaBB or AABb genotype with equal probability. a. Assume these genes are unlinked (i.e., assorting independently). What is the overall probability that the man will produce an Ab gamete? Only the AABb genotype can produce Ab gametes, and the expected probability of Ab gametes produced will be 1/2. Because the parent has an equal probability (50%) of having either genotype (AaBB or AABb), the final probability of the parent producing an Ab gamete is (1/2) " (1/2) = 1/4 = 0.25 = 25%. b. What is the probability of the man producing an AB gamete? For genotype AaBB, the probability of generating an AB gamete is 1/2. For genotype AABb, the probability of generating an AB gamete is 1/2. Again, because both genotypes have equal probabilities, the final probability of the parent generating an AB gamete is (1/2) " (1/2) + (1/2) " (1/2) = 2/4 = 0.5 = 50%. 5. In snapdragons, the allele for red flowers is incompletely dominant over the allele for white flowers, and thus heterozygotes have pink flowers. What ratios of snapdragon flower colors would you expect to see among progeny generated from the following crosses? a. red! white 100% pink b. red! pink 50% pink; 50% red c. white! pink 50% pink; 50% white d. white! white 100% white e. pink! pink 25% red; 50% pink; 25% white f. red! red 100% red

3 6. Two pea plants with purple flowers are crossed. Among the offspring, 63 have purple flowers, and 17 have white flowers. With a chi-square test, compare the observed numbers with a 3:1 ratio and determine the probability that the difference between observed and expected could be a result of chance. Plugging the numbers into the chi-square equation: (63 60) 2 /60 + (17 20) 2 /20 = = 0.6 (the chisquare statistic). Degrees of freedom = the # of classes (phenotypes) 1 = 2 1 = 1. Looking at the chi-square table for one degree of freedom and using the standard critical value of 5%, we do not reject the hypothesis and thus conclude that the observed differences between observed and expected progeny numbers are a result of chance. We can further conclude that both purple parents are heterozygous at the single gene locus controlling flower color (at least for purple and white). 7. In a plant species, you notice that purple and yellow leaf colors, as well as hairy and smooth stems, segregate. You cross a plant with purple leaves and hairy stems to a plant with yellow leaves and hairy stems, and generate the progeny indicated below: Class Phenotypes 1 68 yellow leaves, hairy stems 2 66 purple leaves, hairy stems 3 22 purple leaves, smooth stems 4 25 yellow leaves, smooth stems Total: 181 a. Indicate the most likely genotypes for each parent. Using P for the leaf color locus and S for the stem type locus: PpSs " ppss. b. Propose a hypothesis to explain the progeny results. Based on this hypothesis, what ratios are expected for each of the four classes of progeny? A possible hypothesis is that each trait is controlled by independently assorting single gene pairs, and that one allele in each pair exhibits complete dominance over the other. Class 1 = 3/8, class 2 = 3/8, class 3 = 1/8, class 4 = 1/8. a. Using the chi-square method, test your hypothesis and indicate whether you accept or reject it. (68 68) 2 /68 + (66 68) 2 /68 + (22 23) 2 /23 + (25 23) 2 /23 = This number (the chi-square statistic) is very small; therefore you confidently accept your hypothesis.

4 8. In animals, the inability to make the pigment melanin results in albinism, a recessive somatic condition. Two normal parents, who have decided to have three children, have a first child that is albino (genotype aa). Determine the following probabilities: a. That the second and third children are also albino. Because the first child is albino, both parents must be carriers for the recessive albino allele. The probability of each child having the albino genotype aa equals 1/4 = 0.25 = 25%. Using the multiplicative rule, the probability of the next two children being albino is (1/4) (1/4) = 1/16 = (6.25%). b. That the second and third children are normal. The probability of the couple having any child with normal pigmentation (genotypes AA or Aa) is 3/4 (75%). Therefore, using the multiplicative rule, the probability of the next two children having normal pigmentation is (3/4) (3/4) = 9/16 = = 56.3%. 9. In sailfin mollies (fish), gold color is due to an allele (g) that is recessive to the allele for normal color (G). A gold fish is crossed with a normal fish. Among the offspring, 88 are normal and 82 are gold. a. What are the most likely genotypes of the parents in this cross? Because we know that gold (g) is recessive to normal (G), we can figure out that the gold parent fish has the aa genotype. We also know that the normal parent fish must have at least one G allele, but because there are gold offspring (gg), we can predict that the normal parent must be heterozygous Gg In summary: gold fish (gg) " normal fish (Gg)!88 normal fish (Gg) 82 gold fish (gg) b. Assess the plausibility of your hypothesis by performing a chi-square test. If the parents are gold (gg) " normal (Gg), we would expect the offspring to have a phenotypic ratio of! gold to! normal. By using the goodness-of-fit chi-square test, we can determine if the deviations are likely to have been produced by chance. Phenotype Observed (O) Expected (E) (O E) 2 /E or (! 2 ) Gold Normal Total Degrees of freedom (df) = (number of phenotypic classes) 1. Because there are two phenotypic classes, the degrees of freedom (df) is 1. From the chi-square table, we can see that the calculated chisquare value falls between ( P of.9) and 0.51 ( P of.975). The probability is relatively high that differences between what we expected and what we observed was generated by chance and that our parents are as predicted, (gg " Gg).

5 10. In guinea pigs, the allele for black coat color (B) is dominant over the allele for white coat color (b). At an independently assorting locus, an allele for rough coat (R) is dominant over an allele for smooth coat (r). A guinea pig that is homozygous for black color and rough coat is crossed with a guinea pig that has a white and smooth coat. In a series of matings, the F 1 are crossed with guinea pigs having white, smooth coats. From these matings, the following phenotypes appear in the offspring: 24 black, rough guinea pigs; 26 black, smooth guinea pigs; 23 white, rough guinea pigs; and 5 white, smooth guinea pigs. a. Using a chi-square test, compare the observed numbers of progeny with those expected from the cross. Phenotype Observed (O) Expected (E) (O E) 2 /E or (! 2 ) Black, smooth Black, rough White, rough White, smooth Total Degrees of freedom = 4 1 = 3. The chi-square value falls below for a probability value less than.005 or 0.5 % that random chance produced the observed ratio of guinea pigs. b. What conclusions can you draw from the results of the chi-square test? From the chi-square value, we can see that it is unlikely that random variations produced the observed ratio. Some other phenomena must be acting. 11. (How s your logic?) Marta collects her laundry from the dryer and places it in a basket. She has nine pairs of socks in the load. If she randomly pulls socks from the basket, what is the probability that she will pull nine different socks from the basket before she pulls a sock that makes a pair? (18/18) x (16/17) x (14/16) x (12/15) x (10/14) x (8/13) x (6/12) x (4/11) x (2/10) = 10,321,920/980,179,200 = = 1.05%

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