Question: Two varieties of maize, one 11 cm, and other 47 cm high were crossed,

Question

Question: Two varieties of maize, one 11 cm, and other 47 cm high were crossed, and the F1 hybrids were crossed to generate F2. In the F2 there were a total 13,923 plants with a continuous variation in heights between the two extremes and only 3 as large as 47 cm high, and 5 of 11cm high.

a. How many genes are involved in determining height in this plant?
By using 1/4^n, n=6, = 1/4096, there are 6 gene pairs are involved in determining height. <-Is it correct?

b. What is the contribution of each allele to the phenotype in cm?
(47 - 11) / 6 = 6. <-Is it correct?

c. What would be the size and genotype of the F1 from a cross between a true-breeding 11cm plant and a true-breeding 47 cm plant?

d. The F1 from c. is then crossed to give an F2. What proportion would you expect in F2 of either one of the extreme phenotypes?

e. In a cross between a 29 cm plant and a 20 cm plant, what would be the genotype giving the smallest number of phenotypes? And specify the phenotypes observed.

Explanation / Answer

a. This example depicts continuous variation,wherein more genes on different parts of genome are responsible for the variable phenotype.Often this is called as polygenic inheritance.

The formula used to calculate number of genes (n) that have an additive effect on the phenotype is  

(1/4)^n = proportion of F2 individuals showing extreme phenotype (8 is the number of F2 extremes here).

The (2n+1) rule also determines number of gene pairs involved since, 2n+1 = number of phenotype categories.

b The contribution of each allele also depends upon the concept of dominance (incomplete here), co dominance.Plus the environmental factors also play a role.The formula should be 47-11/2 = 18cm.

c. Genotype of F1 from the cross would be T(47cm)t(11cm), heterozygous at this locus. Phenotype based on b would be 36cm.

d.The proportion of either of the extreme phenotypes should be 1/2.

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