Question

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

Answer 1

Answer:

1. heterozygous yellow and star

2. 37

3. 1/8

4. 168

5. 1/4

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