## Related questions with answers

The height of the seed head in wheat at maturity is determined by several genes. In one variety, the head is just $9$ inches above the ground; in another, it is $33$ inches above the ground. Plants from the $9$-inch variety were crossed to plants from the $33$-inch variety. Among the F$_{1}$, the seed head was 21 inches above the ground. After self-fertilization, the F$_{1}$ plants produced an F$_{2}$ population in which $9$-inch and $33$-inch plants each appeared with a frequency of $1$/$256$.

(a) How many genes are involved in the determination of seed head height in these strains of wheat?

(b) How much does each allele of these genes contribute to seed head height?

(c) If a 21-inch F$_{1}$ plant were crossed to a $9$-inch plant, how often would you expect $18$-inch wheat to occur in the progeny?

Solution

Verifieda.For a monogenic phenotype,

The probability for a recessive phenotype in F2 = $\dfrac{1}{4}$

Similarly,

For a polygenic trait (determined by n number of genes),

The probability for a complete recessive phenotype in F2 = $\dfrac{1}{4}^{n}$

This formula is applicable for the parental cross: All homozygous dominant X All homozygous recessive

So, the given phenotype is determined by 4 genes

$4^n = 256$

n = 4

b.All recessive: aabbccdd = 9 inches

All dominant: AABBCCDD = 21 inches

Difference = 21 - 9 = 12

Number of genes = 4

Contribution by each gene = $\dfrac{12}{4}$ = 3 inches

Assuming that each gene contributes equally to the phenotype.

c. If a 21 inch plant is crossed to 9 inch plant,

The probability of getting 18 inch plant = $\dfrac{1}{4}$

AABBCCDD or AaBbCcDd = 21 inches

aabbccdd = 9 inches

For 18 inch phenotype, any of the three genes must be heterozygous or homozygous dominant.

(For ex: AaBbCcdd)

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