In maize (*Zea mays*), purple endosperm color (P) is dominant over yellow endosperm color … — JAMB UTME

In maize (*Zea mays*), purple endosperm color ( $P$ ) is dominant over yellow endosperm color ( $p$ ), and starchy kernel texture ( $S$ ) is dominant over waxy kernel texture ( $s$ ). A plant heterozygous for both traits ( $PpSs$ ) is allowed to self-pollinate, producing 480 kernels in the $F_2$ generation. How many of these kernels are expected to display the purple endosperm and waxy kernel texture phenotype?

90Cevap
B
270
C
30
D
120

Cevap

90 kernels are expected to show the purple endosperm and waxy kernel texture phenotype.

In a dihybrid cross between two heterozygous individuals (

PpSs \times PpSs

), the offspring phenotypes segregate in a classical 9:3:3:1 ratio. The phenotype representing one dominant trait and one recessive trait (purple endosperm and waxy texture,

P\_ss

) occurs with a frequency of

\frac{3}{16}

. Out of 480 kernels,

\frac{3}{16} \times 480 = 90

kernels will manifest this specific phenotype.

Adım Adım Çözüm

Determine parental cross and gametes

The self-pollinated parent is

PpSs

. Gametes formed are

PS

Ps

pS

, and

ps

in equal proportions.

Mendel's Law of Independent Assortment states that alleles of different genes segregate independently during gamete formation.

Calculate the expected phenotypic fraction for purple and waxy kernels

Probability of purple (

P\_

) =

\frac{3}{4}

; Probability of waxy (

ss

) =

\frac{1}{4}

. Combined probability =

\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}

Because the two genes assort independently, the joint probability is the product of their individual probabilities.

Calculate expected count from total offspring

Expected count =

\frac{3}{16} \times 480 = 90

Multiply the expected phenotypic frequency by the total kernel population size.

Anahtar Kavram

Mendel's Law of Independent Assortment and Dihybrid

F_2

Phenotypic Calculations

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