Chapter 3
3.4. b: 0, c: ½
3.8. Panel A is anaphase I of meiosis, because the homologous chromosomes are paired. Panel B is anaphase II of meiosis, because the chromosome number has been reduced by half. Panel C is anaphase of mitosis, because the homologous chromosomes are not paired.
3.10: 5/6 expected to survive.
3.12: A) [0.5 x 0.5] + [0.5 x 0.5] + [0.5 x 0.5] = 0.75. B) 0.5 x 0.5 = 0.25
3.17: Note first that there is no chance of the X-linked color-blindness allele being passed to II-4, because he inherits his Y chromosome from his father (I-2) and his X chromosome from his normal mother (I-1). Because we are told that I-2 is the only possible source of the mutant gene, this means that there is no possibility that the male III-3 carries the mutant gene. (Moreover, III-3 is not color blind, and so his X chromosome must be nonmutant.) On the other hand, the female II-2 must be a carrier for the color-blindness allele because she inherits an X chromosome from her father (male I-2), whom we know carries the allele for color blindness. Given that II-2 is heterozygous, there is a probability of 1/2 that she transmits the allele to her daughter (III-2) and another probability of 1/2 that III-2 transmits the allele to the male IV-1. Hence, the total probability that IV-1 is color blind is 1 x 1/2 x 1/2 = 1/4.
3.19: There are 1000 progeny altogether, and the expected numbers are 562.5, 187.5, 187.5, and 62.5, respectively, from top to bottom. This chi-square test has 3 degrees of freedom (four classes of data); the final chi-square value is 3.6. The P value is about 0.33. The data give no reason to reject the hypothesis of 9:3:3:1 segregation.
Chapter 5
5.12: The mom has a Robertsonian translocation joining the long arm of Chr. 21 with the long arm of another acrocentric chromosome. The child has 46 chromosomes. This karyotype differs from the usual situation for trisomy 21 because the extra chromosome 21 is not a free chromosome.
Additional Question not in your textbook:
#1. An insect from a strain that is true-breeding for white eyes was crossed to an insect from a strain that is true-breeding for red eyes. All the F1s had white eyes. The F2s consisted of:
243 white eyed, 58 red eyed, 21 cream eyed.
Provide the genotypes for the parents, F1s and F2s of the cross.
Hint: epistasis!
How to solve this problem? First, start by analyzing the F2 phenotypic ratios:
Total progeny = 322
White = 243/322 ≈ 12/16
Red = 58/322 ≈ 3/16
Cream = 21/322 ≈ 1/16
The 16 in the denominator indicates that the F1 is segregating for two gene pairs, and thus that the F1 is a dihybrid. The F1 can arbitrarily be given the genotype of Aa Bb. The F2s can now be analyzed with respect to a modified 9:3:3:1 dihybrid ratio:
9 A_ B_
3 A_ bb
3 aa B_
1 aa bb
Thus,
P aa BB red x AA bb white
F1 Aa Bb white
F2 A_ _ _ white
aa B_ red
aa bb cream
