Biology, 25.06.2021 01:30 oofoofoof1

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

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Answer from: zachisthequeen2002

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

Explanation:

Available data:

Two linked genes (A) and (B) ⇒ 18 centiMorgans apartMan ⇒ Aa BbMan´s father ⇒ AA BBWoman ⇒ 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

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

Answer from: Quest

hoped i

explanation:

Answer from: Quest

answer: 21

step-by-step explanation:

divide 630 by 30. 30 goes into 630 21 times

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