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sandmanfan01 sandmanfan01
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9 years ago Edited: 9 years ago, sandmanfan01
An alternative to using the expanded binomial equation and Pascal's triangle in determining probabilities of phenotypes in a subsequent generation when the parents' genotypes are known is to use the following equation:

(n!/s!t!) (a^s*b^t)

where n is the number of offspring, s is the number of offspring in one phenotypic category, t is the number of offspring in the second phenotypic category, a is the probability of the occurance of the first phenotype, and b is the probability of the occurance of the second phenotype. Using this equation, determine the probability of a family of 10 offspring having exactly 2 children afflicted with sickle cell anemia (an autosomal recessive disorder) when both parents are heterozygous for the sickle-cell allele.

I don't know how to answer this.
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Staff Member
Educator
9 years ago
n = 10

s = 3

t = 7

a = .25 (A mendelian cross of two heterozygous parents will produce the genotypes SS, Ss, Ss, and ss. Because only autosomal recessive individuals (ss) are affected by the disease, one-fourth (0.25) children are afflicted with sickle cell anemia.)

b = 1 - .25 = .75



P = (10! / (3! X 7!)) (.25^3)(.75 ^ 7)

P = 0.25
Mastering in Nutritional Biology
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Answer accepted by topic starter
sandmanfan01 Authorsandmanfan01
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Posts: 2
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9 years ago Edited: 9 years ago, sandmanfan01
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Staff Member
Educator
9 years ago
Thanks for the update
Mastering in Nutritional Biology
Tralalalala Slight Smile
wrote...
3 years ago
thank you
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