Please help me with this task. An inbred strain of plants has a mean height of 4

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Please help me with this task.

An inbred strain of plants has a mean height of 48 cm. A second strain of the same species has a mean height of 57 cm. When plants from the two strains are crossed together, all the plants in the F1 generation is 52.5 cm tall. On the other hand, crossing two F1 plants provides a wide range of different heights in the F2 generation: Most of the plants are the same height as the F1 generation plants, while fewest plants are in the most extreme height categories (respectively 30 cm and 75 cm). In F2 approximately 1 out of 1000 plants are 30 cm tall. Simultaneously about 1 out of 1000 F2 plants are 75cm tall.

a) What mode of inheritance is occuring here?

b) How many loci controls the trait?

c) How many gene pairs are involved?

d) How much does each allele contribute to plant height in cm?

e) Indicate one possible set of genotypes for the original P1 parents and the F1 plants that could account for these results.

f) Indicate possible genotypes that could account for F2 plants that are 50.5 cm and 43.5 cm high.

Explanation / Answer

a) Polygenic mode of inheritance is occuring here. The phenotypes of plant height values obtained by crossing F1 show continuous variation because of being "quantitaive" in nature. Some feaures are controlled by many genes, each of which contributes some amount to the overall phenotype. When a large numbers of genes are involved, it is hard to distinguish the effect of each individual gene according to Mendelian rules. This is known as polygenic inheritance.

c) The formula used to calculate number of gene pairs (n) :

Ratio of F2 progeny expressing either extreme phenotype = (1/4)n

Here, 1 out of 1000 plants show extreme phenotype i.e. 30 cm or 75 cm.

(1/4)n = 1/1000

Solving for n, we get n = 5 (approx.)

Therefore, 5 gene pairs are involved.

d) Assuming each gene has one contributing allele and one no contributing allele, total no. of contributing alleles for 5 gene pairs will be 10 (5*2 = 10).

The difference between heights of extreme phenotype i.e. 45 (75 - 30 = 45) must be due to the presence of contributing alleles. The tallest plant has 10 contributing alleles, so each dominant allele contributes an average of 4.5 cm to the height of the plant.

e) Height of original P1 parent with height 48 cm is a result of 4 contributing alleles. (48 - 30 = 18 ; 18/4.5 = 4)

Possible genotype will be : AaBbCcDdee

Height of original P1 parent with height 57cm is a result of 6 contributing alleles. (57 - 30 = 27 ; 27/4.5 = 6)

Possible genotype will be : AaBbCcDDEe

Height of original F1 parent with height 52.5 cm is a result of 5contributing alleles. (52.5-30 = 22.5 ; 22.5/4.5 = 5)

Possible genotype will be : AaBbCcDdEe

f) Possible Genotype of F2 plant that is 50.5 cm - AaBbCcDdEe

Genotype of F2 plant that is 43.5 cm has 3 contributing alleles. Its genotype will be - AaBbCcddee

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