3) A mouse is going to have six offspring. What is the probability that these si

Question

3) A mouse is going to have six offspring. What is the probability that these six offspring ill consist of two males and four females? A) 6/64 (-0.094) B) 20/64 (-0.31) C) 15/64 (-0.23) D) 1/2 E) 49/64 (-0.77) 4) Suppose an individual has the following genotype for five traits: Gg HH Tt yy Dd. After meiosis, what is C) 1/16 the probability that one randomly chosen gamete will have the genotype G Htyd? E) 1/2 A) 1/8 B) 1 D) 1/4 5) Consider five traits for the following cross between two individuals: Aa Bb DD tt Pp x AA bb Dd Tt Pp C) 1/64 What is the probability of obtaining an offspring with the following genotype: Aa bb DD tt pp? A) I/128 B) 1/32 D) 1/16

Explanation / Answer

4.a heterozygous alleles would contribute to 2 different gametes,

Gg forms 2 gametes Gand g

HH forms 2 gametes but both H

tT forms 2 gametes T and t

yy forms 2 gametes but both y

Dd forms 2 gametes D and d.

Since yy and HH form similar gametes. They are not going to contribute to the difference of the overall genotype as they are the same. Now, look at the other 3. Each one has 2 possibilities.

Therefore 32 =9 gametes will be formed.

now the probability of receiving G H t y d is-

50% or 1/2 gets G gamete.

100% gets H gamete

50% or 1/ 2 gets t gamete

100% will get y gamete

50% or 1/2 gets d gamete.

Overall probability is-1/2 x 1 x 1/2 x 1 x 1/2 =1/8

5. Aa Bb DD tt Pp x AA bb Dd Tt Pp

Aa x AA

Probability of Aa 2/4 or 1/2

Bb x bb

probability of getting bb-1/4 or 1/2

DD X Dd

probability of getting DD -1/4

tt x Tt

probability of getting tt -2/4 or 1/2

Pp x Pp

pp

probability of getting pp -1/4

Thus the total probability of getting this genotype i.e Aa bb DD tt pp is-

1/2 x1/2 x 1/4 x1/2 x 1/4 = 1/128.

A a A AA Aa A AA Aa

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