Question

g Two linked genes, (A) and (B), are separated by 18 centiMorgans. A man with genotype Aa Bb marries a woman who is aa bb. The man's father was AA BB. What is the probability that their first child will both be ab/ab

Answer 1

Answer:

The probability that their first child will both be ab/ab = 41%.

Explanation:

Available data:

• Two linked genes  (A) and (B) ⇒ 18 centiMorgans apart
• Man ⇒  Aa Bb
• Man´s father ⇒ AA BB
• Woman ⇒ aabb

The recombination frequency is given by the distance between genes. We have to know that 1% of recombinations = 1 map unit = 1cm.

Recombination frequency = 0.18

Man: AaBb

Gametes) AB Parental

                 ab Parental

                 Ab Recombinant

                 aB recombinant

18 centi morgan = 18 % of recombination in total  

                            = % aB + % Ab  

                            = 9% aB + 9% Ab

       100% - 18% = 82% of parental in total  

                           = % of AB + % ab  

                           = 41% AB + 41% ab

The frequency for each gamete is:

AB 41%

ab 41%

Ab 9%

aB 9%

Cross: man x woman

Parental)          AaBb           x          aabb

Gametes) AB, Ab, aB, ab        ab, ab, ab, ab

Punnett square)     AB ( 41%)       Ab (9%)      aB (9%)       ab (41%)

                   ab      AaBb               Aabb          aaBb            aabb

                   ab      AaBb               Aabb          aaBb            aabb

                   ab      AaBb               Aabb          aaBb            aabb

                   ab      AaBb               Aabb          aaBb            aabb

F1) 41% of the progeny is expected to be AaBb

     9% of the progeny is expected to be Aabb

     9% of the progeny is expected to be aaBb

     41% of the progeny is expected to be aabb