The \(A\) and \(B\) antigens in humans may be found in water-soluble form in secretions, including saliva, of some individuals \((\text {Se/Se and Se/se})\) but not in others (se/se). The population thus contains "secretors" and "nonsecretors." (a) Determine the proportion of various phenotypes (blood type and ability to secrete) in matings between individuals that are blood type \(A B\) and type \(O,\) both of whom are \(S e / s e\) (b) How will the results of such matings change if both parents are heterozygous for the gene controlling the synthesis of the H substance \((H h) ?\)

For the first mating between blood type AB (IAIB) and blood type O (ii) parents, both of whom are Selse (they are secretors, so their genotype is Se/se). Parent 1: IAIB (AB blood type) and Se/se (secretor) Parent 2: ii (O blood type) and Se/se (secretor)

To determine the genotypes of offspring, we'll perform two separate Punnett square analyses: one for the blood type genes and one for the secretor genes. Punnett square for blood type genes: | IA | IB --------------- i | IAi| IBi --------------- i | IAi| IBi Punnett square for secretor genes: | Se | se -------------- Se | SeSe| Sese -------------- se | Sese| sesese

Using the Punnett squares, we can determine the frequency of each phenotype among the offspring. Blood types: - IAi (A blood type): 50% - IBi (B blood type): 50% Secretor status: - SeSe (non-secretor): 25% - Sese (secretor): 50% - sesese (non-secretor): 25%

Using the frequencies obtained in step 3, we can calculate proportions of phenotypes in the offspring. Multiply the frequency of each blood type by the frequency of each secretor status to get the proportions: - A blood type (50%) and secretor (50%): 0.50*0.50 = 25% - A blood type (50%) and non-secretor (50%): 0.50*0.50 = 25% - B blood type (50%) and secretor (50%): 0.50*0.50 = 25% - B blood type (50%) and non-secretor (50%): 0.50*0.50 = 25% (a) Thus, in matings between blood type AB and blood type O individuals who are Selse, the proportion of various phenotypes in the offspring would be 25% A secretor, 25% A non-secretor, 25% B secretor, and 25% B non-secretor.

In the second part of the exercise, we have to determine the results of matings between heterozygous H gene parents. Since both parents are heterozygous (Hh) for the gene controlling synthesis of the H substance, there will be a chance that the offspring will receive two recessive genes (hh) from both parents. A child born with the hh genotype will not be able to produce the H substance, which is needed for A, B, or O blood types. This lack of H substance leads to a rare blood type, called the Bombay blood type (Oh). To evaluate this possibility, perform a Punnett square analysis for the H gene: Punnett square for the H gene: | H | h ----------- H | HH | Hh ----------- h | Hh | hh Among these offspring, 25% would have the HH genotype, 50% would have Hh genotype, and 25% would have the hh genotype. We can also multiply the blood type frequencies with H substance frequencies: - AB blood type: 0 - A blood type: 0.50 * (0.25 + 0.50) = 0.375 - B blood type: 0.50 * (0.25 + 0.50) = 0.375 - O blood type: 0.25 * (0.25 + 0.50) = 0.1875 - Oh blood type (Bombay): 0.25 * 0.25 = 0.0625 (b) If both parents are heterozygous for the H gene, the results of such matings would change as follows: 37.5% A blood type, 37.5% B blood type, 18.75% O blood type, and 6.25% Oh blood type (Bombay phenotype).