Answer key, practice problems 2000--week 2

1. (a) Mode (i) -- autosomal recessive -- will work if the female is heterozygous (+/c) and male is homozygous recessive c/c where "+" denotes the normal allele and "c" denotes the allele for cut-wing--in this case, the recessive allele; the designation "+/c" indicates that the animal is heterozygous--it has the "+" allele on one homolog and the "c" allele on the other). [The female must be heterozygous; it couldn't be homozygous recessive, because it shows the normal phenotype, and it couldn't be homozygous dominant, or the progeny would all be heterozygous and therefore show the dominant phenotype.]

Mode (ii) -- autosomal dominant -- would work also, if the female is homozygous recessive (+/+) and the male is heterozygous (C/+). Note that here, allele C is dominant and causes the cut wing phenotype. Using the same logic as above, the male has to be heterozygous.

Mode (iii) -- X-linked recessive -- will work if the female is heterozygous and the male carries the recessive, cut-wing allele (c) on his sole X chromosome. The female cannot be homozygous normal because then she would have only normal-winged offspring (heterozygous females, or males with the normal X).

Mode (iv) -- X-linked dominant -- is not possible because if the male carried the dominant allele (C) on his sole X chromosome, all his daughters would inherit that chromosome and would therefore be affected.

The three modes that are consistent with the results are shown below. Note that for (i) and (ii), sex of the animal is independent of the cut wing phenotype, giving males and females in equal proportions for both cut wing and normal phenotypes.
(b) Because males and females from this cross show different phenotype ratios, autosomal modes of inheritance can be ruled out. [To confirm: if the trait were autosomal recessive, the normal x normal cross would be heterozygote x heterozygote, giving a 3:1 ratio of normal : cut wing in the progeny, with males and females in equal numbers; if the cut wing phenotype were autosomal dominant, the normal females and males would all be homozygous normal, as would their progeny.] Therefore, the only mode of inheritance that is consistent with the results is X-linked recessive. The parents and F1 progeny genotypes were as shown in cross (iii) above. The F1 x F1 cross resulting progeny are shown below:

(c) There are two types of normal-winged F2 females--homozygous and heterozygous (see cross above). Only the heterozygotes can give cut wing progeny when crossed with the cut wing F2 males (the homozygous normal females will have only normal-winged offspring, as the normal allele is dominant). Because the homozygous and heterozygous females are present in 1:1 ratio in the F2, half the crosses will give at least some cut-wing flies in the progeny. The cross that will give cut wing progeny (heterozygous females x cut wing males) is identical to the cross between the original parents (see cross iii above), and will give male and female, normal and cut wing flies in 1:1:1:1 ratio.

2. Based on the unequal phenotypes among males and females in the F2, we can conclude that the trait must be sex-linked (to confirm, we can go through and eliminate autosomal modes of inheritance as done in answer 1). It is not Y-linked, as the females and males both show the trait. That conclusion leaves us with two alternatives to consider: X-linked recessive and X-linked dominant. The X-linked recessive option can be ruled out as being inconsistent with the results--the female parent would have to be homozygous recessive, and all her male progeny would be bar-eyed, and all her female offspring would be normal. The only possibility that is consistent with the results is that the bar eye phenotype is X-linked dominant, with the female parent being heterozygous, as shown below.

3. Whether a trait is sex-linked or autosomal, one normally expects to see the same number of females as males in the progeny (all phenotypes added up). Here, we see that there are twice as many females as there are males, suggesting that half the males might be dying before adulthood. The sex-specificity of the hypothesized lethality (and the absence of the club-foot phenotype in males) would further suggest that the trait causes lethality in males. If the normal allele were dominant, the male parent (normal) would have only normal daughters (as they would inherit the allele from him). Therefore, the club foot phenotype must be dominant and the normal phenotype recessive. Presumably, the trait is lethal when homozygous--i.e., it is recessive with respect to lethality and dominant with respect to club foot phenotype. (The female must be heterozygous, else all her male progeny would inherit the allele and would therefore be dead.)

X-linked dominant with lethality in males (and perhaps lethality in females when homozygous, although we have no evidence for that)

4. Assuming that II-1 is homozygous normal (because the trait is rare), III-1 must have received the disease allele from his father, which is not normally possible for an X-linked allele (because he should receive an X only from his mother).

5. (i) The disease is probably not autosomal recessive--there are several instances where people marrying into the family have affected children; the people marrying in would all have to be heterozygotes, an improbably scenario.
(ii) The pedigree is fully consistent with autosomal dominant where I-1 is heterozygous and 1-2 is homozygous normal, as is everyone marrying into the family.
(iii) X-linked recessive can be ruled out, because affected females have unaffected fathers (e.g., II-1, IV-3).
(iv) X-linked dominant can be ruled out also, because affected men have unaffected daughters (who would inherit the X chromosome carrying the dominant disease allele from the father).--e.g., II-5.
(v, vii) Males and females are affected, so the disease is not Y-linked or sex-limited.
(vi) With sex-influenced inheritance, there are two possibilities--dominant in males and recessive in females, or dominant in females and recessive in males. Affected women have unaffected sons (e.g., I-1 and II-3), so it cannot be recessive in women and dominant in men. Likewise, affected men have unaffected daughters (e.g., II-5 and III-6) so it cannot be dominant in women and recessive in men.

6. Although five genes are listed, only three of the five are heterozygous, so the number of gamete classes that can be produced by independent assortment is (2 x 2 x 2) = 8.

7. To get none of the paternal centromeres in one egg, the disposition of homologs during meiosis has to be such that all maternally derived homologs go to one pole of the spindle and all paternally derived homologs go to the other pole. The probability that none of the paternal centromeres will end up at one particular pole = the probability all four maternal centromeres will end up one particular pole of the spindle is (1/2)(1/2)(1/2)(1/2) = 1/16. However, they have two possible poles to choose from; since they could all go to either of two poles, the probability that they will end up at the same pole (any one of the two poles) = 1/16 + 1/16 = 1/8. In general, the probability that all maternal (or all paternal) centromeres will end up at the same pole is (1/2)n-1 where n is the haploid chromosome number.

8. (i)
 
(ii)
 
(iii)

9.

10. Because this is a heterozygote x heterozygote cross (normal = dominant, albino = recessive), we expect to see normal and albino children in 3:1 ratio-- i.e., the probability of a normal child is 3/4, and the probability of an albino child is 1/4.
(a)  (1/2)5 = 1/32
(b) The probability of the outcome described = (3/4)(3/4)(1/4)(1/4)(1/4) = 9/1024
(c) The probability of 2 normal and 3 albino children in any order can be calculated using binomial expansion. Let a = p(albino) = 1/4 and b = p(normal) = 3/4; since there are five children, the equation to use is:

The term representing the probability of 3 albino and 2 normal children is 10a3b2. Substituting the values of a and b, we get:

p(3 albino, 2 normal) =

10(1/4)3(3/4)2 = 45/512 = 0.088
(d) The probability that all five will be normal is:

(3/4)5 = 243/1024 = 0.237

11. (a) Because there was only one class of F1 progeny, the parents were probably homozygous for all traits, and the F1 heterozygous, and the phenotypes of the F1 must be the dominant phenotypes. Thus, the cross can be represented as shown below:

(b) Because we are assuming independent assortment, we can treat each trait separately -- it is a heterozygous x heterozygous cross for each trait, for which we can expect a 3:1 ratio of dominant:recessive phenotypes in the progeny. The phenotype ratios can be determined by drawing a Punnett square (a very large Punnett square) or by drawing a branch diagram, as shown. The branch diagram gives both the phenotypes and proportions of the progeny, obtained by following each branch and multiplying the fractions. The phenotypes and ratios so obtained are shown in the table.

(c) Remember that the F2 consists of a 1:2:1 genotype ratio of homozygous dominant:heterozygous:homozygous recessive offspring. Therefore, true-breeders -- i.e., homozygotes (whether dominant or recessive) -- will constitute 1/4 of the progeny for each trait. Both homozygous varieties described are expected to be found in the F2 progeny (all combinations of phenotypes are expected, as we have assumed independent assortment), and each variety is expected to constitute (1/4)(1/4)(1/4)(1/4) = 1/256 of the progeny.

12. Answer is in the back of the lecture notes