Question

One fourth of the progeny of an a/+ C. elegans hermaphrodite have the A phenotype (the rest are wild type). The same proportion of living progeny are produced from an a/+; b/+ hermaphrodite. In addition, the latter hermaphrodite produces several dead eggs (i.e., animals that died as embryos); numerically there are a few more dead eggs than A phenotype animals.

a. Describe the B phenotype (the phenotype produced by the b mutation) and say whether the mutation is recessive or dominant.

b. Explain why the ratio of A:WT animals would be the same for the progeny of a/+ and a/+; b/+. A Punnett Square will be useful.

c. Why are there slightly more dead eggs than animals with the A phenotype? How many more are there?

Can you please show the punnet square work and how you got the answer in detail thank you

Answer 1

ANSWER- The progeny of an a/+ C. elegans hermaphrodite are 1 +/+ : 2 a/+ : 1 a/a. 3/4th of the progeny are WT, 1/4th of the progeny have the A phenotype. Therefore, “a” must be a recessive allele.

A.From an a/+; b/+ C. elegans hermaphrodite, you obtain progeny where 1/4th of the live progeny have the A phenotype, but there are also a number of dead eggs. The dead eggs must arise from the presence of two “b” alleles, as demonstrated in the Punnett square on the uploaded photo-

		sperm	sperm	sperm	sperm
		+;+	a;+	+;b	a;b
eggs	+;+	+/+; +/+ WT	a/+;+/+ WT	+/+;b/+ WT	a/+;b/+ WT
eggs	a;+	a/+;+/+ wt	a/a;+/+ A	a/+;+/+ WT	a/a;b/+ A
eggs	+;b	+/+; b/+ WT	a/+;b/+ WT	+/+; b/b Dead	a/+;b/b Dead
eggs	a;b	a;+;b/+ WT	a/a; b/+ A	a/+;b/b Dead	a/a;b/b Dead

B.The progeny of an a/+ C. elegans hermaphrodite are 1 +/+ : 2 a/+ : 1 a/a. 3/4th of the progeny are WT, 1/4th of the progeny have the A phenotype.
The progeny of an a/+; b/+ C. elegans hermaphrodite are 9 WT: 3A: 4 dead
Of the 16 progeny from the Punnett square, only 12 will survive (9 WT, 3 A). Therefore, 3/12, or 1/4th of the surviving progeny have the A phenotype.

C.Though a and b are both recessive alleles, there are more dead eggs than A mutants because one of the A animals (a/a;b/b) dies due to the presence of two “b” alleles. Therefore, there will be 1/3 more dead eggs than A progeny.