In maize plants, the allele for smooth kernels (S) is completely dominant over the allele … — JAMB UTME

In maize plants, the allele for smooth kernels ( $S$ ) is completely dominant over the allele for wrinkled kernels ( $s$ ). According to Mendel's Law of Segregation, a monohybrid cross between two heterozygous smooth-kernel maize plants ( $Ss \times Ss$ ) produces offspring with a genotypic ratio of $3:1$ .

Answer: Answer

Answer

The statement is False. A monohybrid cross between two heterozygous parents yields a genotypic ratio of

1:2:1

(

1\ SS : 2\ Ss : 1\ ss

) and a phenotypic ratio of

3:1

(3 smooth : 1 wrinkled).

Evaluating the statement as False is correct because Mendel's Law of Segregation dictates that crossing two monohybrids (

Ss \times Ss

) yields three distinct genotypes (

SS

Ss

, and

ss

) in a

1:2:1

ratio, while the

3:1

ratio applies only to the physical phenotypes.

Step-by-Step Solution

Determine the parental genotypes and gametes produced during meiosis according to Mendel's First Law.

Both parents are heterozygous (

Ss

). Each parent produces gametes with either the

S

allele (

50\%

) or the

s

allele (

50\%

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 monohybrid cross (

Ss \times Ss

Offspring genotypes are

1\ SS

2\ Ss

, and

1\ ss

Combining the gametes gives

25\%\ SS

50\%\ Ss

, and

25\%\ ss

Compare the resulting genotypic ratio with the ratio stated in the prompt.

The genotypic ratio is

1:2:1

, whereas the phenotypic ratio (smooth to wrinkled) is

3:1

The prompt incorrectly asserts that

3:1

is the genotypic ratio, confusing genotypic outcome with phenotypic outcome.

Key Concept

Mendel's Law of Segregation and Monohybrid Cross Ratios

Estimated Time:1m 0s

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