In maize (*Zea mays*), the allele for starchy endosperm (Su) is completely dominant over t… — JAMB UTME

In maize (*Zea mays*), the allele for starchy endosperm ( $Su$ ) is completely dominant over the allele for sugary endosperm ( $su$ ). Based on Mendel's First Law (Law of Segregation), which expected progeny outcome on the right correctly matches each monohybrid parental cross scenario on the left?

Cross between two heterozygous starchy plants ( $Su\,su \times Su\,su$ ), evaluated for the genotypic ratio among starchy progeny only $1\ Su\,Su : 2\ Su\,su$ ( $1 : 2$ ratio)
Test cross of a heterozygous starchy plant ( $Su\,su \times su\,su$ ), evaluated for the ratio of dominant allele ( $Su$ ) carriers to non-carriers $1 : 1$ ratio ( $50\%$ carriers to $50\%$ non-carriers)
Self-pollination of an $F_1$ plant ( $Su\,su$ ), evaluated for the total percentage of pure-breeding (homozygous) progeny $50\%$ of total progeny
Cross between a heterozygous starchy plant ( $Su\,su$ ) and a homozygous starchy plant ( $Su\,Su$ ), evaluated for the percentage of sugary ( $su\,su$ ) progeny $0\%$ of total progeny

Answer

Cross between two heterozygous starchy plants matches

1\ Su\,Su : 2\ Su\,su

; Test cross of a heterozygous starchy plant matches

1 : 1

ratio (

50\%

carriers to

50\%

non-carriers); Self-pollination of an

F_1

plant matches

50\%

of total progeny; Cross between a heterozygous starchy plant and homozygous starchy plant matches

0\%

of total progeny.

Each monohybrid parental cross is correctly matched by calculating the specific genotypic or phenotypic probabilities based on allele segregation during meiosis according to Mendel's First Law.

Step-by-Step Solution

Analyze the cross between two heterozygous starchy plants (

Su\,su \times Su\,su

Genotypes produced are

1/4\ Su\,Su

1/2\ Su\,su

, and

1/4\ su\,su

Mendel's Law of Segregation states alleles segregate into gametes with equal probability (

1/2\ Su

1/2\ su

). Starchy plants comprise

Su\,Su

and

Su\,su

. Among starchy progeny, the genotypic ratio is

1\ Su\,Su : 2\ Su\,su

Analyze the test cross of a heterozygous starchy plant (

Su\,su \times su\,su

Genotypes produced are

1/2\ Su\,su

and

1/2\ su\,su

Carriers of the dominant allele

Su

are

Su\,su

(

50\%

), while non-carriers are

su\,su

(

50\%

), yielding a ratio of

1 : 1

Analyze self-pollination of an

F_1

plant (

Su\,su \times Su\,su

) for pure-breeding individuals.

Pure-breeding individuals (

Su\,Su

and

su\,su

) constitute

50\%

of offspring.

Pure-breeding offspring are homozygous.

25\%\ Su\,Su + 25\%\ su\,su = 50\%

of total progeny.

Analyze the cross

Su\,su \times Su\,Su

for sugary (

su\,su

) progeny.

Proportion of

su\,su

offspring is

0\%

The homozygous dominant parent contributes a dominant

Su

allele to all offspring, preventing the expression of the recessive

su\,su

genotype.

Key Concept

Mendel's First Law and Monohybrid Cross Proportions

Estimated Time:2m 30s

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