# Answers to Mendelian genetics questions BI164 Spring, 2007

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2 5. The allele for taillessness is dominant. A cross of tailless x tailless mice produces more tailless mice than wildtype mice. The ratio of 2:1 brings to mind lethal genes. We know from the above information that tailless mice do not breed true. The expected ratio of three tailless mice to one wildtype mouse is not observed. The most likely explanation is that a homozygote for the tailless allele is a lethal combination. The surviving tailless mice do not have that lethal combination because they are heterozygous. The heterozygotes outnumber the wildtype mice 2:1, as expected. 6. The chance that a child will be a girl or a boy is independent of other offspring. The probability of the next child being a boy is 1/2 in both cases. 7. Calculate the probability of obtaining: a. An Aa BB Cc zygote from a cross of Aa Bb Cc x Aa Bb Cc Chance of Aa 2/4 Chance of BB 1/4 Chance of Cc - 2/4 Since these genes assort independently, the overall probability is the product of the three individual probabilities = 1/2 x 1/4 x 1/2 = 1/16 b. An Aa BB cc zygote from a cross of aa BB Cc x AA bb CC Chance of Aa 1/1 Chance of BB 0/1 Chance of cc - 0/1 Therefore, for two reasons, there is no way to realize the Aa BB cc genotype in this cross. c. An A B C phenotype from a cross of Aa Bb CC x Aa Bb cc Chances of A_ - 3/4 Chances of B_ - 3/4 Chances of C_ - 1/1 Overall probability - 3/4 x 3/4 x 1 = 9/16 d. An a b c phenotype from a cross of AA BB CC x AA BB CC There is no source of recessive alleles for any of the loci and therefore the phenotype cannot be produced.

3 8. The results suggest a simple two-allele, single locus system. The results conform to the expected 3:1 ratio of phenotypes for a monohybrid cross. The only true-breeding purple flowers will be the homozygotes (PP). They make up a third of the purple flowers. All of the white-flowered progeny will be true breeding. 9. Dihybrid cross problems. a. PP ss x pp SS 100% purple 100% spiny 100% purple flowers, spiny pods b. Pp SS x pp ss 50% purple 100% spiny 1/2 purple flowers, spiny pods 50% white 100% spiny 1/2 white flowers, spiny pods c. Pp Ss x Pp SS 75% purple 100% spiny 3/4 purple flowers, spiny pods 25% white 100% spiny 1/4 white flowers, spiny pods d. Pp Ss x Pp ss 75% purple 50% spiny 3/8 purple flowers, spiny pods 50% smooth 3/8 purple flowers, smooth pods 25% white 50% spine 1/8 white flowers, spiny pods 50% smooth 1/8 white flowers, smooth pods

4 e. Pp Ss x Pp Ss 75% purple 75% spiny 9/16 purple flowers, spiny pods 25% smooth 3/16 purple flowers, smooth pods 25% white 75% spiny 3/16 white flowers, spiny pods 25% smooth 1/16 white flowers, smooth pods f. Pp Ss x pp ss 50% purple 50%% spiny 1/4 purple flowers, spiny pods 50% smooth 1/4 purple flowers, smooth pods 50% white 50%% spiny 1/4 white flowers, spiny pods 50% smooth 1/4 white flowers, smooth pods 10. Autosomal dominant If we assume I-2 is heterozygous for the trait, we can find no part of the pedigree that cannot be explained by inheritance as an autosomal dominant. Autosomal recessive We can reject this mechanism because II-1 and II-2 produced a son without the trait. X-linked dominant If this mechanism is correct, the grandfather I-1 must be hemizygous for the recessive allele and the grandmother I-2 must be heterozygous because their children include both a male and female without the trait. II-2 must have married a woman heterozygous for the trait (II-1) because they produce sons with and without the trait. II-6 is hemizygous for the trait. His wife (II-7) must be homozygous recessive. All of their daughters must inherit a dominant allele from the father and the sons will lack the trait since they must inherit their only allele from their mother. X-linked recessive We can reject this mechanism because female I-1 or II-1 could only have sons with the trait. II-5, III-1 and III-4 lack the trait so we know the trait cannot be an X-linked recessive. In summary, we have two possible mechanisms of inheritance for this trait. We cannot distinguish between them with the present information.

5 However, the fact that every daughter of a man with the trait also has the trait indicates that X-linked dominance is the more likely mechanism. With autosomal dominance, one would expect half of the daughters of fathers II-2 and II-6 to have the trait and half to lack it. 11. ABO blood type questions a. An O child from the marriage of two A individuals This result is possible if both parents are heterozygotes (I A i) b. An O child from the marriage of an A to a B Also possible if both parents are heterozygous (I B i and I A i) c. An AB child from the marriage of an A to an O Not possible. The parent with O blood cannot contribute the B allele needed. d. An O child from the marriage of an AB to an A Not possible. The O child must be ii and the AB parent has no i allele to contribute. e. An A child from the marriage of an AB to a B Possible if the B parent is heterozygous (I B i). 12. For each of the first cousins, one parent had a sibling with the disease. We know that their grandparents both had to be heterozygous for the PKU gene since neither had the disease but one of their children did have the disease. Therefore the two siblings (II-2 and II-4) have two chances in three they have the disease (since they are not homozygous recessive, there are three genotypes homozygous dominant, heterozygous with PKU gene inherited from the mother, heterozygous with the PKU gene inherited from the father). The chance that each first cousin (III-1 and III-2) inherited the PKU gene from her/his parent is 1:2 by Mendel s Law of Segregation. So, the probability that each first cousin is a carrier for the PKU allele is 2/3 x 1/2 = 1/3 When the two cousins marry, four genotypes are possible but only one will cause the disease (homozygous recessive). So, the overall probability is: Probability III-1 has the PKU gene x probability that III-2 has the PKU gene x probability of producing a homozygous recessive = 1/3 x 1/3 x 1/4 = 1/36

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