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

A woman is a carrier for hemophilia A (X-linked recessive). Her husband is unaffected. What is the probability that their daughter will be a carrier but unaffected?

Answer: 50%

The mother's genotype is X^H X^h (carrier) and the father's is X^H Y. Each daughter receives one X from her father (X^H) and one X from her mother — 50% chance of X^H (normal) and 50% chance of X^h (carrier). So 50% of daughters will be carriers and unaffected; 50% will be completely normal. No daughter can be affected because they always get a normal X from dad.

In a cross between two plants heterozygous for two independently assorting genes (AaBb × AaBb), what fraction of offspring will be homozygous recessive for BOTH traits?

Answer: 1/16

For each gene independently, the probability of a homozygous recessive offspring (aa or bb) from a Aa × Aa cross is 1/4. Because the two genes assort independently, the probability of being homozygous recessive for BOTH is 1/4 × 1/4 = 1/16.

A man with type AB blood and a woman with type O blood have a child. Which blood type is IMPOSSIBLE for that child to have?

Answer: Type AB

The father (AB) can contribute either I^A or I^B alleles. The mother (OO) can only contribute i alleles. Possible offspring genotypes are I^A i (Type A) or I^B i (Type B). Type AB (I^A I^B) requires one child to receive both I^A and I^B alleles, which cannot happen here since the mother only donates i. Type O (ii) also cannot occur since the father cannot donate an i allele.

Two parents are both phenotypically normal but have a child with cystic fibrosis (autosomal recessive). What is the probability that their NEXT child will be a phenotypically normal CARRIER?

Answer: 1/2

Since both parents are phenotypically normal yet have an affected child (ff), both parents must be carriers (Ff × Ff). The Punnett square yields: 1/4 FF, 2/4 Ff, 1/4 ff. Of the 3/4 phenotypically normal offspring, 2/4 (= 2 out of 4 total) are carriers (Ff). The probability of the next child being a carrier is 2/4 = 1/2, regardless of any previous births.

In snapdragons, flower color shows incomplete dominance: RR = red, Rr = pink, rr = white. If a pink-flowered plant is crossed with a white-flowered plant, what percentage of offspring will be pink?

Answer: 50%

Pink × White = Rr × rr. The Punnett square gives: 1/2 Rr (pink) and 1/2 rr (white). Because there is incomplete dominance, there are no red offspring. Exactly 50% of offspring will be pink (Rr) and 50% will be white (rr).

A geneticist examines a pedigree and notices a trait that appears in every generation, affects both males and females equally, and an affected father passes it to all of his children. Which inheritance pattern BEST fits this data?

Answer: Autosomal dominant

The trait appearing in every generation and being transmitted from an affected father to ALL children (both sons and daughters) rules out X-linked recessive (affected fathers cannot pass their X to sons) and autosomal recessive (which typically skips generations). Mitochondrial inheritance is matrilineal — fathers never transmit mitochondria. Autosomal dominant (one copy sufficient) explains all observations: vertical transmission, equal sex ratios, and father-to-all-children inheritance if the father is homozygous dominant.