Match each dihybrid parental cross genotype involving independently assorting genes with i… — JAMB UTME

Match each dihybrid parental cross genotype involving independently assorting genes with its corresponding expected phenotypic ratio in the offspring.

$AaBb \times AaBb$ $9 : 3 : 3 : 1$
$AaBb \times aabb$ $1 : 1 : 1 : 1$
$AaBB \times AaBB$ $3 : 1$
$AABB \times aabb$ 100% dominant for both traits ( $1 : 0$ )

Answer

The correct pairings match

AaBb \times AaBb

9:3:3:1

AaBb \times aabb

1:1:1:1

AaBB \times AaBB

3:1

, and

AABB \times aabb

to 100% dominant for both traits.

Each cross results from independent assortment during meiosis. Selfing a double heterozygote (

AaBb \times AaBb

) yields 9:3:3:1; testcrossing a double heterozygote (

AaBb \times aabb

) yields 1:1:1:1; selfing a single heterozygote with one homozygous locus (

AaBB \times AaBB

) yields a 3:1 ratio; and crossing true-breeding dominant and recessive lines (

AABB \times aabb

) produces uniform offspring exhibiting both dominant traits.

Step-by-Step Solution

Determine gamete types produced by each parent genotype

AaBb

produces 4 gamete types (

AB, Ab, aB, ab

aabb

produces 1 type (

ab

AaBB

produces 2 types (

AB, aB

), and

AABB

produces 1 type (

AB

Mendel's Law of Independent Assortment states that gene pairs segregate independently during gamete formation.

Combine gametes to calculate phenotypic probabilities for each cross

Combining gametes yields the classic dihybrid phenotypic ratios for each cross type.

The phenotypic ratio depends on allele interactions and dominance relationships across both gene loci.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Cross Ratios

Question

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Step-by-Step Solution

Key Concept

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