Match each dihybrid parental cross genotype of independently assorting genes with its expe… — JAMB UTME

Match each dihybrid parental cross genotype of independently assorting genes with its expected offspring phenotypic ratio.

$AaBb \times AaBb$ $9 : 3 : 3 : 1$
$AaBb \times aabb$ $1 : 1 : 1 : 1$
$AaBb \times Aabb$ $3 : 3 : 1 : 1$
$AABb \times aaBb$ $3 : 1$

Cevap

The correct matches pair

AaBb \times AaBb

with

9:3:3:1

AaBb \times aabb

with

1:1:1:1

AaBb \times Aabb

with

3:3:1:1

, and

AABb \times aaBb

with

3:1

Each parent cross is accurately matched to its offspring phenotypic distribution by applying the product rule of probability to independently assorting gene pairs:

AaBb \times AaBb

produces

9:3:3:1

AaBb \times aabb

produces

1:1:1:1

AaBb \times Aabb

produces

3:3:1:1

, and

AABb \times aaBb

produces

3:1

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Analyze the dihybrid self-cross

AaBb \times AaBb

Combining independent monohybrid crosses

(3 \text{ dominant} : 1 \text{ recessive}) \times (3 \text{ dominant} : 1 \text{ recessive})

produces the phenotypic ratio

9 : 3 : 3 : 1

According to Mendel's Law of Independent Assortment, the segregation of alleles for one gene occurs independently of the segregation of alleles for another gene.

Analyze the dihybrid test cross

AaBb \times aabb

The heterozygous parent produces four distinct gamete types (

AB, Ab, aB, ab

) in equal frequencies of

25\%

each, while the homozygous recessive parent produces only

ab

gametes, yielding a

1:1:1:1

phenotypic ratio.

Test cross phenotypic ratios directly mirror the gametic frequencies produced by the heterozygous parent.

Analyze the cross

AaBb \times Aabb

For gene A (

Aa \times Aa

), expected probabilities are

\frac{3}{4}

dominant (

A\_

) and

\frac{1}{4}

recessive (

aa

). For gene B (

Bb \times bb

), expected probabilities are

\frac{1}{2}

dominant (

B\_

) and

\frac{1}{2}

recessive (

bb

). Multiplying probabilities gives

\frac{3}{8} A\_B\_ : \frac{3}{8} A\_bb : \frac{1}{8} aaB\_ : \frac{1}{8} aabb

, simplified as

3:3:1:1

Applying the product rule of probability for independent genetic events.

Analyze the cross

AABb \times aaBb

For gene A (

AA \times aa

100\%

of offspring express the dominant phenotype (

Aa

). For gene B (

Bb \times Bb

\frac{3}{4}

express the dominant phenotype (

B\_

) and

\frac{1}{4}

express the recessive phenotype (

bb

). The total offspring ratio simplifies to

3:1

Since trait A displays zero phenotypic variation in the progeny, the overall phenotypic ratio is determined solely by the segregation of trait B.

Anahtar Kavram

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios

Soru

Cevap

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Anahtar Kavram

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