Question

35. (a) Write the phenotypes of the offsprings in the following cross:-
(i) RrYy x rryy
(ii) RRYy x rrYy
(b) In a cross between plants bearing purple flowers with round seeds and those bearing white flowers with wrinkled
seeds the following observations were made
F1 plants ---- All having purple flowers and round seeds
F2 plants -----1152 plants
(i) Calculate the number of plants with white flowers round seeds and white flowers wrinkled seeds.
Give reason for the appearance of new combination of characters in F2 progeny in the above cross.​

Answer 1

Answer:

A cross is made between homozygous yellow round seeds(YYRR) to another homozygous green wrinkled seeds(yyrr). In the F1 generation, all plants are heterozygous yellow round(YyrRr). When this plant is subjected to self pollination, in the F2 generation, 9 plants are Yellow round, 3 plants are yellow wrinkled, 3 plants are green round and 1 plant is green wrinkled are formed.

The dihybrid phenotypic ratio is 9:3:3:1.

So, the correct option is ‘9:3:3:1’.

solution

Explanation:

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