chapter 2 problem 18 Galactosemia is a recessive human disease that is treatable
ID: 4451 • Letter: C
Question
chapter 2 problem 18Galactosemia is a recessive human disease that is treatable byrestricting lactose and glucose in the diet. Susan Smithers and herhusband are both heterzygous for hte galactosemia gene.
D. If the couple has four children, what is the probability that atleast one child will have galactosemia?
since its a monohybrid cross I used the probabilities of
(1/4)*(3/4)*(3/4)*(3/4)=27/256
1:4 chance of having galactosemia and 3:4 chance of not
what am I doing wrong? Ive been told the answer is 175/256
F. If the couple has three children, What is the probability thattwo of the children will have galactosemia and one will not,regardless of order?
since its a monohybrid cross I used the probabilities of
(1/4)*(1/4)*(3/4)=(3/64)
1:4 chance of having galactosemia and 3:4 chance of not
What am im doing wrong? ive been told the answer is 9/64
Explanation / Answer
Let A be the normal allele, a be the allele of Galactosemia Aa x Aa will produce AA, Aa, Aa and aa Thus, the probability of their child having galactosemia (aa) is1/4 (D) ***Note the word AT LEAST (So you need to sum up the probabilityfrom one children to four children. On the other words, you can getthe probablity by subtracting the probablity of no child havinggalactosemia from 1) P(At least one child will have galactosemia) = 1 - P(No child will have galactosemia) = 1 - 4C0(0.25)0(0.75)4 = 1 - 81/256 = 175/256 (F) P(2 children have, 1 do not have galactosemia) = 3C2 (0.25)2(0.75)1 = 9/64 I think your problem is you did not consider the many combinationsituations (As in binomial), so this cause your answer differ fromthe correct answer. Hope this helps!
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