Question

Suppose that goats have one gene that codes for color, where A is brown and a is white. The goats also have another gene that codes for height, where B is tall and b is short. If these two genes are unlinked, what is the probability that a cross between Aa Bb × Aa bb parents will produce two out of six offspring that are brown and short? Enter your answer as a decimal, rather than as a percentage.

waht is the Number ? Number:____.

Answer 1

The probability that a cross between Aa Bb × Aa bb parents will produce two out of six offspring that are brown and short is approximately 0.322.

The genes for the two traits are unlinked, hence they will be assorted independently in mendelian fashion. Thus, the probability of the brown and short offspring is calculated from the punnet square given below:

Gametes	AB	Ab	aB	ab
Ab	AABb= brown, tall	AAbb = brown, short	AaBb = brown, tall	Aabb = brown, short
Ab	AABb= brown, tall	AAbb = brown, short	AaBb = brown, tall	Aabb = brown, short
ab	AaBb = brown, tall	Aabb = brown, short	aaBb = white, tall	aabb = white, short
ab	AaBb = brown, tall	Aabb = brown, short	aaBb = white, tall	aabb = white, short

Thus, the probability of an offspring being brown and short is 6/16, therefore, for two offspring to be brown and short, the probability will be (6/16)², and the probability of not brown and short offspring will be (10/16), therefore for four offspring, the probability of not being brown and short will be (10/16)⁴. Also, out of 6 offspring, there are 15 possible ways that 2 offsprings can be brown and short (calculated by taking combination from the formula 6C2 = 6!/2!(6-2)! =15). Therefore, the probability of exactly 2 offspring being brown and short will be 15 x (6/16)² x (10/16)⁴ = 0.322 approximately.