Question

Y-: yellow S-: star yy: black ss: starless

All the sneetches want their children to have stars on their bellies. The combination of alleles that makes a black, starless sneetch is embryonic lethal. Y and S are two independently assorting autosomal genes. A true breeding black star bellied sneetch mated with a true breeding yellow starless sneetch and produced 20 F1 sneetches. The F1 then mated with each other and produced 300 F2 progeny.

1. What is the F1 phenotype?

2. Among 300 F2 sneetches, how many of them are black star bellied sneetches?

3. What proportion of these black star F2 would be true breeding (homozygous)?

4. Among 300 F2 sneetches, how many of them are yellow star bellied sneetches?

5.What proportion of these yellow star F2 would be true breeding?

Answer 1

1. heterozygous yellow and star

2. 37

3. 1/8

4. 168

5. 1/4

Step-by-step explanation:

Given ,

In f1 generation a cross is made between a true breeding black star bellied sneetch mated with a true breeding yellow starless sneetch

yySS x YYss

It is taken as - Y (yellow) is dominant over y (black)

and S (star) is dominant over s (starless)

1. F1 Generation

Genotype of parents yySS X YYss

gametes - yS, yS, Ys, Ys

All 16 offspring will have genotype YySs

phenotype would be heterozygous yellow and star

2. F2 generation cross

YySs X YySs

YS Ys yS ys

YS YYSS YYSs YySS YySy

Ys YYSs YYss YySs Yyss

yS YySS YySs yySS yySs

ys YySs Yyss yySs yyss

Genotype of offspring are –

YYSS – 1

YYSs – 2

YySS – 2

YySs – 4

YYss- 1

Yyss- 2

yySS – 1

yySs- 1

yyss- 1

2. Out of 16, 2 are black star bellied sneetches . Which means only 1/8 are black star bellied sneetches

So out of 300, 37 are black star bellied sneetches

3. Only 2 out of 16 are true breeding. i.e 1/8

4. 9 out of 16 are yellow star bellied sneetches, so out of 300, 168 are yellow star bellied sneetches

5. 4 out of 16 are true breeding yellow. Thus, ¼ are true breeding