Two gene loci, A and B, are unlinked (and thus assort independently), and allelesA and B are dominant over alleles a and b. Indicate the probabilities of producing the following:a. An AB gamete from an AABb individual? b. An AaBb zygote from a cross AaBb * AABB? c. An AB phenotype from a cross AaBb * AaBb? d. An AB phenotype from a cross aabb * AABB? e. An aB phenotype from a cross AaBb * AaBB?

Question

Biology

Pirate

24 January, 21:45

Two gene loci, A and B, are unlinked (and thus assort independently), and allelesA and B are dominant over alleles a and b. Indicate the probabilities of producing the following:a. An AB gamete from an AABb individual? b. An AaBb zygote from a cross AaBb * AABB? c. An AB phenotype from a cross AaBb * AaBb? d. An AB phenotype from a cross aabb * AABB? e. An aB phenotype from a cross AaBb * AaBB?

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Charisma · Answer 1

a) 1/2

b) 1/4

c) 9/16

d) 1

e) 1/4

Explanation:

This crosses involves two genes A and B that are unlinked, hence, can undergo recombination and assort into gametes independently.

a) an AABb individual will produce two types of gametes after undergoing meiosis in accordance with Mendel's law of independent assortment. The two different combinations of gametes are AB and Ab. Hence, the probability of producing an AB gamete is 1/2.

b) In a cross between AaBb * AABB, the following gamete combinations can be formed by each parent

AaBb: AB, Ab, aB, ab

AABB: AB, AB, AB, AB

Crossing these gamete combinations using a punnet square will result in 16 possible offsprings with only four (4) having a AaBb genotype i. e. heterozygous dominant for both genes. Hence, the probability will be 4/16 or 1/4.

c) In a cross between AaBb * AaBb i. e. two heterozygous parents, gametes AB, Ab, aB, and ab will possibly be produced by each parent. Performing a punnet square will result in 9 out of 16 offsprings having an AB phenotype i. e. dominant for both genes. Hence, the probability will be 9/16.

d) In a cross between aabb * AABB homozygous genotypes, AABB will produce only AB gametes which aabb will produce only ab gametes. Hence, all the offsprings will have a heterozygous genotype (AaBb), which are dominant for both genes since alleles A and B are dominant over alleles a and b respectively. Therefore, the probability of having a AB phenotype is 16/16 or 1.

e) In a cross between a AaBb * AaBB, possible gamete combinations (AB, Ab, aB, ab) and (AB, AB, aB, aB) will be produced respectively by each parent. Involving them in a punnet square will result in 16 offsprings with only four (4) having an (aB) phenotype i. e. offsprings aaBB, aaBB, aaBb and aaBb. Hence, the probability is 4/16 or 1/4.