In pea plants, the allele for axial flower position (A) is completely dominant over the al… — JAMB UTME

In pea plants, the allele for axial flower position ( $A$ ) is completely dominant over the allele for terminal flower position ( $a$ ). If two heterozygous axial-flowered plants ( $Aa$ ) are crossed, what is the expected genotypic ratio ( $AA : Aa : aa$ ) among their offspring according to Mendel's First Law?

$1 : 2 : 1$ Answer
B
$3 : 1$
C
$1 : 1$
D
$9 : 3 : 3 : 1$

Answer

The expected genotypic ratio among the offspring is

1 : 2 : 1

(

1\,AA : 2\,Aa : 1\,aa

In a monohybrid cross between two heterozygous individuals (

Aa \times Aa

), Mendel's Law of Segregation dictates that alleles segregate into gametes independently. Punnett square analysis gives

1/4\,AA

2/4\,Aa

, and

1/4\,aa

, producing an exact genotypic ratio of

1 : 2 : 1

Step-by-Step Solution

Identify parental genotypes and alleles produced

Both parents have the genotype

Aa

. Gametes produced by each parent are

A

and

a

in equal proportions (

1/2\,A

and

1/2\,a

Mendel's Law of Segregation states that allele pairs separate during gamete formation so that each gamete carries only one allele for each gene.

Construct a Punnett square for the cross

Aa \times Aa

Offspring combinations are:

AA

(

1/4

Aa

(

2/4

), and

aa

(

1/4

Random fertilization combines gametes from both parents to form offspring genotypes.

Determine the genotypic ratio

The ratio of genotypes is

1\,AA : 2\,Aa : 1\,aa

, which simplifies to

1 : 2 : 1

The question specifically asks for the ratio of genotypes (

AA : Aa : aa

), not physical appearances (phenotypes).

Key Concept

Mendel's Law of Segregation and Monohybrid Cross Genotypic Ratio

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