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What is the likelihood than a given gene in individual 9 would be homozygous by decent?

A) 1/2
B) 1/8
C) 1/64
D) 1/16

User Taudris
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Final answer:

Using the sum and product rules of probability, the chance that an offspring will express the dominant phenotype at all four loci in a tetrahybrid cross is 27/64.

Step-by-step explanation:

The likelihood of an offspring having a homozygous dominant or heterozygous genotype for a given gene, and thus expressing the dominant phenotype, can be determined using the sum and product rules of probability. For each gene, the probability of being homozygous dominant (AA) is 1/4, and the probability of being heterozygous (Aa) is 1/2. Using the sum rule, the probability that a gene will be either homozygous dominant or heterozygous is 1/4 + 1/2 = 3/4.

When considering multiple genes, we use the product rule to calculate the combined probability. For a tetrahybrid cross with genes A, B, C, and D, the probability that an offspring will express the dominant phenotype at all four loci is 3/4 × 3/4 × 3/4 × 3/4, which equals 81/256 or simplifies to 27/64. This calculation applies to a mating between individuals with multiple heterozygous alleles and accounts for the likelihood of dominant expression at each locus.

User Statox
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