BMST - Basic Math and Science Genetics and Heredity Questions and Answers Flashcards

A woman is a carrier for an X-linked recessive disorder. Her husband is unaffected. What is the probability that their daughter will be a carrier?

Answer: 50%

The mother (X^A X^a) passes either X^A or X^a to each daughter, while the father (X^A Y) always passes X^A to daughters. So daughters are either X^A X^A (unaffected, non-carrier) or X^A X^a (carrier), each with 50% probability.

In a dihybrid cross between two plants heterozygous for both seed color (Yy) and seed texture (Rr), what fraction of offspring will be homozygous recessive for BOTH traits?

Answer: 1/16

Each trait independently segregates: probability of yy = 1/4, probability of rr = 1/4. By the product rule for independent events: 1/4 × 1/4 = 1/16.

A man with blood type AB has children with a woman of blood type O. Which of the following blood types is IMPOSSIBLE in their offspring?

Answer: AB

The father (I^A I^B) can contribute either I^A or I^B; the mother (ii) can only contribute i. Possible offspring genotypes are I^A i (type A) or I^B i (type B). AB (I^A I^B) is impossible because the mother has no I allele to pass along — wait, that's not right. AB requires both I^A and I^B, but the mother contributes only i, so offspring can be type A or type B, never AB.

Two true-breeding strains of mice are crossed: one with black fur (BB) and one with white fur (bb). The F1 offspring all have gray fur. When F1 mice are intercrossed, the F2 ratio is 1 black : 2 gray : 1 white. This pattern of inheritance is best described as:

Answer: Incomplete dominance

Incomplete dominance produces a blended intermediate phenotype in heterozygotes (gray), and the F2 ratio of 1:2:1 reflects the genotypic ratio (BB : Bb : bb) where each genotype produces a distinct phenotype. Codominance would produce both parental traits simultaneously, not a blend.

A geneticist discovers that two genes on the same chromosome show 32% recombination frequency in test crosses. What can be concluded?

Answer: The genes are linked and approximately 32 map units apart

Recombination frequency (RF) between linked genes can range from 0% (completely linked) to ~50% (effectively unlinked). An RF of 32% indicates the genes ARE linked and are approximately 32 centimorgans (cM) apart. RF only approaches 50% for very distantly linked loci; it does not mean they're on separate chromosomes.

In humans, phenylketonuria (PKU) is caused by a homozygous recessive allele. Two unaffected parents have an affected child. What is the probability that their NEXT child will be unaffected but heterozygous?

Answer: 1/2

Both parents must be carriers (Aa × Aa). The cross produces 1 AA : 2 Aa : 1 aa. The probability of being a carrier (Aa) is 2/4 = 1/2. The question asks for the probability of being heterozygous (carrier) in the NEXT child regardless of outcome, so it is simply 2/4 = 1/2. (Note: 2/3 applies only if you're told the next child is unaffected.)