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Answer :
Final answer:
Inheritance of nose length in snuffleups is determined by four genes, each with two alleles, because extreme nose lengths occur with a frequency of 1 in 256, indicating a (1/2)^8 probability.
Explanation:
The question involves understanding the inheritance pattern of nose length in a hypothetic species known as snuffleups, based on a Mendelian genetics problem. When a 19-inch-nosed snuffleup (presumably homozygous for long nose alleles) is crossed with a 3-inch-nosed snuffleup (presumably homozygous for short nose alleles), the F1 generation all have 11-inch noses, suggesting a blending inheritance model or multiple gene involvement. Crossing two F1 individuals results in nose lengths that vary from 3 inches to 19 inches in the F2 generation, with extreme lengths being very rare (occurring 1 in 256), potentially indicating a Mendelian inheritance pattern with multiple genes involved.
If we consider the extreme phenotypes being a result of either homozygous dominant or homozygous recessive genotypes and these phenotypes occur in a ratio of 1 in 256, we can infer that four genes are involved in determining the trait, each with two alleles. This is because (1/2)^4 x (1/2)^4 = 1/256. Thus, four genes, each with two alleles, are involved in the inheritance of the nose length trait of the snuffles. We assume each gene has one dominant and one recessive allele, and the extreme nose lengths are seen when all four genes are either all dominant or all recessive alleles.
The correct answer is:
e. 4 alleles
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