2. In cucumbers, heart-shaped leaves (h) are recessive to normal leaves (H) and fruit spines (n) are recessive to no fruit spines (N). The genes encoding leaf shape and presence of fruit spines are located on the same chromosome 25 m. u. apart. A plant with heart-shaped leaves and fruit spines is crossed with a plant homozygous for normal leaves and no fruit spines. The F1 are testcrossed with plant with heart-shaped leaves and fruit spines. Of 1000 offspring, what would be the expected phenotypes and in what numbers would they be expected

The phenotypes and the numbers of each phenotypes are:

1. Normal leaves and no fruit spines (250 cucumbers)

2. Normal leaves and fruit spines (250 cucumbers)

3. Heart-shaped leaves and fruit spine (250 cucumbers)

4. Heart-shaped leaves and no fruit spine (250  cucumbers)

Explanation:

First let us state out the traits:

heart-shaped leaves (h) = recessive

normal leaves (H) = dominant

fruit spine (n) = recessive

no fruit spine (N) = dominant.

For first cross between the parents;

parents genotypes = P₁ = hhnn (heart-shaped leaves and fruit spines)

P₂ = HHNN (normal leaves and no fruit spine)

Next, let us do the parent crossing t get the F1 generation, this is done in the diagram attached. Putting into application the law of independent assortment, which states that for a dihybrid cross, the alleles are not linked to one another, hence can be expressed independent of the other. After the parent crossing, the genotype and phenotype of the F1 generation are;

F1 = HhNn (all offspring have Normal leaves and no fruit spine)

Next, the F1 are testcrossed with plant of heart-shaped leaves and fruit spines which has genotype hhnn

so the cross is between the plants; HhNn vs hhnn, as follows;

                           hn                   hn                hn               hn

HN                    HhNn               HhNn         HhNn         HhNn

Hn                    Hhnn                Hhnn           Hhnn         Hhnn

hn                     hhnn                hhnn            hhnn         hhnn

hN                    hhNn                hhNn           hhNn         hhNn

From the cross above, out of 16 offspring, the phenotypes are;

1. HhNn = Normal leaves and no fruit spine = 4 plants

2. Hhnn = Normal leaves and fruit spine = 4 plants

3. hhnn = Heart-shaped leaves and fruit spine = 4 plants

4. hhNn = Heart shaped leaves and no fruit spine = 4 plants

Therefore, the ration of the four phenotypes are 4:4:4:4 = 1 : 1 : 1 : 1

Therefore the four phenotypes are shared equally among the ofsprings

Hence in 1000 offsprings, the phenotypic distribution of the four phenotypes will be;

1000 ÷ 4 = 250 offsprings for each phenotype, shown as;

1. HhNn = Normal leaves and no fruit spine = 250 plants

2. Hhnn = Normal leaves and fruit spine = 250 plants

3. hhnn = Heart-shaped leaves and fruit spine = 250 plants

4. hhNn = Heart shaped leaves and no fruit spine = 250 plants.

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explanation:

i was told this before by a teacher, so yeah

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