Pea plant with round yellow seeds (RRYY) is crossed with another pea plant having wrinkled green seeds (rryy). $F_{2}$ progeny consists of
(i) Heterozygous for both shape and colour of seeds
(ii) Dominant for colour and recessive for shape
(iii) Homozygous for both shape and colour
(iv) Heterozygous for seed shape only

i-2, ii-4, iii-3, iv-8

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i-4, ii-2, iii-3, iv-8

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i-3, ii-4, iii-8, iv-2

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i-4, ii-3, iii-2, iv-8.

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Solution

The correct option is D i-4, ii-3, iii-2, iv-8.

i)Heterozygous for both shape and colour of seeds – 4

ii)Dominant for colour and recessive for shape – 3

iii)Homozygous for both shape and colour – 2

iv)Heterozygous for seed shape only – 8
So, the correct option is ‘i-4, ii-3, iii-2, iv-8’.

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Q. The pea plant with round shaped seeds with yellow colour seed (RRYY) crossed with another pea plant showing wrinkled shape with green cotyledons (rryy). The results of progeny in

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i) Heterozygous for both shape and colour of seeds ......... number.
ii) Dominant for colour, but recessive for shape ........ number.
iii) Homozygous for both shape and colour of seeds ......... number.
iv) Heterozygous for seed shape only ........ number.

Question 11
Two pea plants one with round green seeds (RRyy) and another with wrinkled yellow (rrYY) seeds produce F1 progeny that have round, yellow (RrYy) seeds. When F1 plants are selfed, the F2 progeny will have a new combination of characters. Choose the new combination of the following.
(i) Round, yellow
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(iii) Wrinkled, yellow
(iv)Wrinkled, green
(a) (i) and (ii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)

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