A parent that is heterozygous for both traits (AaBb) can create how many different allele possibilities.

1 answer:

8 0

Distribute the A to the B and the b, then the a to the B and the b. You should get four different possible combinations: AB Ab aB ab. These would then be used to set up one side of a punnett square, and the other parent's genotype would be set up the same, then put on the other side of the punnett square.

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<h2>Part 1.</h2><h2>Answer:</h2>

Only the traits or allele for the green trait will be needed to make a green cucumber.

<h3>Explanation:</h3>

Here in the given example there are two allele responsible for the color of cucumber. One is Y which is dominant allele and codes for the yellow color.

While y is recessive allele and it will be code for the green color.

In case of heterozygous in which the genotype will be Yy but the phenotype will be always yellow because Y is dominant.

In case of genotype yy the phenotype will be green.

So for green cucumber there will be two y in the genotype.

<h2>Part 2.</h2><h2>Answer:</h2>

For yellow cucumber, there is need of only one dominant allele which codes for yellow.

<h3>Explanation:</h3>

According to the given punnet squre, There will be 4 genotypes from the breeding of a yellow cucumber and green cucumber.

The yellow cucumber is heterozygous or homozygous and its genotypes can be Yy and YY relatively.

But for green cucumber the genotype can only be homozygous which is yy.

In the punnet squre the 2 phenotype are yellow and green.

Hence for the yellow cucumber there is only need of one Y allele in genotype.

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