Question

EXAM TOMORROW! PLEASE HELP! The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes? Show your work.

(1)aabbccdd
(2) AaBbCcDd
(3) AABBCCDD
(4) AaBBccDd,

Answer 1

Probability of heterozygote is given by 2/4 which is 1/2 (Zz)and Probability of homozygote is given by 1/4 (ZZ or zz)Hence any thing in format AA or aa will be given probability as 1/4 and anything in the format as Aa will be assigned probability as 1/2.Now,
1) aabbccdd = 1/4*1/4*1/4*1/4= 1/256
2) AaBbCcDd= 1/2*1/2*1/2*1/2= 1/16
3) AABBCCDD= 1/4*1/4*1/4*1/4= 1/256
4) AaBBccDd= 1/2*1/4*1/4*1/2= 1/64

Answer 2

You can use the Punnett square or the shortcut method. Either way, it is given that in any circumstance that the probability of a heterozygote (Aa) is 2/4 or simply 1/2 and the probability of a homozygote (AA) or (aa) is ¼.

a. aabbccdd = (¼)(¼)(¼)(¼) =1/256

b. AaBbCcDd = (1/2) (1/2) (1/2) (1/2) = 1/16

c. AABBCCDD = (¼)(¼)(¼)(¼) = 1/256

d. AaBBccDd =(1/2)(¼)(¼)(1/2) = 1/64