4. Mendel worked with pea plant varieties with contrasting traits. Suppose that

Question

4. Mendel worked with pea plant varieties with contrasting traits. Suppose that you are

studying pea plants with either round peas (R) or wrinkled peas (r), yellow peas (Y) or green

peas (y), purple flowers (P) or white flowers (p), tall plants (T) or short plants (t). Assume that

you are starting with a cross where one plant (1) has the genotype R/r; y/y; P/p; T/t and the

second plant (2) is R/r; Y/Y; P/p; t/t. If you were to cross the two plants, what proportion of the

progeny will:

a.have the genotype r/r; Y/y; p/p; t/t?

b.have a phenotype that is wrinkled, yellow, white and short?

c.have a dominant phenotype for all four traits?

d. be pure breeding for the seed shape phenotype?

e.be pure breeding for round peas and purple flowers?

f. be pure breeding for all four traits?

g. have the genotype R/r; Y/y; p/p; T/t?

Explanation / Answer

We can easily solve this problem with the help of probability. Here, all the questions are asked about the F1 Generation. The genotype for the two plants in consideration are:

RryyPpTt and RrYYPptt

When these two plants are crossed if we consider no linkage,

a.) The proportion of the F1 Progeny with genotype rrYypptt

1/4 × 1 × 1/4 × 1/2= 1/32 = 3.125 %

This means 3.125% of the progeny will have the said genotype.

b.) The proportion of the progeny that will have a phenotype that is wrinkled, yellow, white and short, would have the genotype of : rrYypptt

Following the rules of probability, the proportion of the progeny that will have genotype rrYypptt:

1/4×1×1/4×1/2 = 1/32.

This means 3.125% of the progeny will have the said genotype.

c.) To have dominant phenotype for all four traits, one individual should have either of the two genotypes,

RrYyPpTt or RRYyPPTt. Homozygous dominant genotype for Y and T are not possible because the parental genotypes do not support such combination.

Then, following simple probability rule, we have:

1/4×1×1/4×1/2 = 1/32

1/2×1×1/2×1/2= 1/8

Adding them will give us their combined probability, which is: 1/32+ 1/8 = 0.15625

So, 15.625% of the progeny will have a dominant phenotype for all for traits

d.) To be pure breeding for the seed shape phenotype, we need to consider only the genotype combination of seed shape. Pure breeding means either homozygous dominant or homozygous recessive for the seed shape.

The probability for the above condition will be

1/4 + 1/4= 1/2 ( Since there is a 25% probability for the progeny to be homozygous dominant or recessive for seed shape)

hence, 50% of the progeny will be pure breeding for seed shape.

e.) If we consider both round peas and purple flower, the probability of them being pure breeding for both phenotypes is:

1/4×1/4 = 1/16

hence, 6.25% of the progeny will be pure breeding for round peas and purple flower.

f.) Pure breeding for all four traits is not possible in the progeny as one of the parents being homozygous dominant for pea colour and the other being homozygous recessive for pea colour, all the individuals of the progeny will be heterozygous for the phenotype pea colour.

g.) The proportion of the Progeny with genotype RrYyppTt

1/2×1×1/4×1/2 = 1/16

This means 6.25% of the progeny will have the said genotype.

If there is any question or confusion regarding this answer, please free to ask me in the comments section.

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