Question

1. What is the probability that each of the following pairs of parents will produce the indicated offspring? Assume independent assortment of all gene pairs. Hint: remember to do the calculations for each allele separately. For example, what is the probability that AA x aa will produce Aa?

A. What is the probability that a cross of a being s with genotypes AA and aa (AA x aa) producing a child with the following genotypes? (Use fractions!)

AA

aa

Aa

B. Now let’s look at the probability of dihybrid crosses. What is the probability of the following offspring from a cross of AABB x aabb. Again use fractions when appropriate.

AABB?

aabb?

AAbb

AaBb

C. Now for trihybrid crosses. What is the probability of the following offspring from the crosses indicated?

a. AABBCC x aabbcc → AaBbCc

b. AABbCc x AaBbCc → AAbbCC

  

c. AaBbCc x AaBbCc → AaBbCc

d. aaBbCC x AABbcc → AaBbCc

Answer 1

Answer:

A).

AA x aa ---- parents

Aa -------- progeny.

Probability of AA = 0

Probability of Aa = 1

Probability of aa = 0

B).

AABB x aabb -= produces only one type of progeny that is AaBb.

AABB = 0

aabb = 0

AAbb =0

AaBb = 1

C).

a).

AABBCC x aabbcc = AaBbCc = 100%

b).

AABbCc x AaBbCc -- parents

AAxAa = AA(1/2)&Aa(1/2)

Bb x Bb = BB (1/4), Bb (1/2) & bb (1/4)

Cc x Cc = CC (1/4), Cc(1/2), cc (1/4)

Probability of AAbbCC = 1/2*1/4*1/4= 1/32

c).

AaBbCc x AaBbCc -- parents

Probability of AaBbCc = 1/2*1/2*1/2= 1/8

d).

àaBbCC x AABbcc ---- parents

aa x AA = Aa (1)

Bb x Bb = BB (1/4), Bb (1/2) & bb (1/4)

CC x cc = Cc (1)

Probability of AaBbCc = 1*1/2*1= 1/2