In corn snakes, the wild-type color is brown. One autosomal recessive mutation causes the snake to be orange, and another causes the snake to be black. An orange snake was crossed to a black one, and the F1 offspring were all brow. Indicate what phenotypes and ratios you would expect in the F2 generation if orange pigment is the product of one pathway, black pigment is the product of another pathway, and brown is the effect of mixing the two pigments in the skin of the snake.

9 brown: 3 black : 3 orange : 1 unpigmented

Explanation:

Since we are told that orange is part of one pathway, and black is part of another, we can assume that they are alleles of different genes.

We know that an orange snake was crossed with a black snake to get all brown snakes. Therefore, this looks like a cross between two true breeding parents. The orange parent would be homozygous dominant at the orange allele (OO) and homozygous recessive at the black allele (bb). The black parent would be homozygous recessive at the orange allele (oo) and homozygous dominant at the black allele (BB)

OObb x ooBB

The F1 offspring will all be doubly heterozygous (OoBb). Since we are told they are all brown, we know that having one allele of each gives brown offspring.

The possible gametes produced by each F1 parent are:

OB, ob, Ob, oB

This produces 16 possible combinations of genotypes. See punnett square attached

We end up with a ratio of 9 brown: 3 black : 3 orange : 1?

The question mark is for the genotype oobb. How do we know what this phenotype is, since neither pathway has a dominant allele present?

From the information given, we assume this individual is unpigmented.

the answer is otpion c) understanding

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genetic variation will increase because of a new habitat and food source.

explanation: