Basic Genetics Problem

[unperturbed]

i encountered some basic genetics problem while reviewing the topic.
here are two questions i wasn't sure about.

Q1. Disease A is an autosomal recessive condition. Suppose that a husband and wife are both carriers (i.e. heterozygotes) for the disease. If they have three children, the probability of having at least 1 unaffected child is:
A. 1/2
B. 63/64
C. 1/8
D. 3/4

I put 3/4, since possbility having one Unaffected each time is 1/4. and it's asking for the at least... am i correct?

Q2.The occurrence of disease A is independent of a child’s sex. Assuming that female and male children are equally likely, what is the probability that the couple’s three children all have the disease, and are all boys?
A. 1/64
B. 1/32
C. 1/8
D. 1/512

I think it's D...but not sure...

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[MediumDef]

i encountered some basic genetics problem while reviewing the topic.
here are two questions i wasn't sure about.

Q1. Disease A is an autosomal recessive condition. Suppose that a husband and wife are both carriers (i.e. heterozygotes) for the disease. If they have three children, the probability of having at least 1 unaffected child is:
A. 1/2
B. 63/64
C. 1/8
D. 3/4

I put 3/4, since possbility having one Unaffected each time is 1/4. and it's asking for the at least... am i correct?

Q2.The occurrence of disease A is independent of a child's sex. Assuming that female and male children are equally likely, what is the probability that the couple's three children all have the disease, and are all boys?
A. 1/64
B. 1/32
C. 1/8
D. 1/512

I think it's D...but not sure...

Never mind I put something that would have led you needlessly astray

 

[holoclothes]

The first one is B. 63/64

You just read the question wrong. Probability of having an unaffected child is 3/4 and then the probability of having two more either with the disease or without is 100% each time so your chances of having at least one unaffected child increases from 3/4 to 15/16 to 63/64.

The second one is D and that is because the chance of having a boy is 1/2 each time and having a boy with the recessive disease is (1/2)(1/4) = 1/8 each time and having three boys all with the disease is (1/8)(1/8)(1/8) is 1/512

 

[TheBoondocks]

i encountered some basic genetics problem while reviewing the topic.
here are two questions i wasn't sure about.

Q1. Disease A is an autosomal recessive condition. Suppose that a husband and wife are both carriers (i.e. heterozygotes) for the disease. If they have three children, the probability of having at least 1 unaffected child is:
A. 1/2
B. 63/64
C. 1/8
D. 3/4

I put 3/4, since possbility having one Unaffected each time is 1/4. and it's asking for the at least... am i correct?

Q2.The occurrence of disease A is independent of a child's sex. Assuming that female and male children are equally likely, what is the probability that the couple's three children all have the disease, and are all boys?
A. 1/64
B. 1/32
C. 1/8
D. 1/512

I think it's D...but not sure...

You must learn to read such problems carefully. The key is atleast. You could have all unaffected 27/64, 2 affected 1 un, or 1 affected and 2 un. This is what the other poster is getting at.

 

[futuredoctor10]

Q2.The occurrence of disease A is independent of a child's sex. Assuming that female and male children are equally likely, what is the probability that the couple's three children all have the disease, and are all boys?
A. 1/64
B. 1/32
C. 1/8
D. 1/512

I think it's D...but not sure...

Yep I believe it is D also. Here's my work, if anyone gets something else let us know!

The probability to get the disease is 1/4 (Aa x Aa Punnett square: 3A_, 1aa).

The probability for all to get the disease [any gender]
P(A and B and C) = P(A) x P(B) x P(C) = 1/4 x 1/4 x 1/4

The probability for all to get the disease AND for all to be boys:
1/4 x 1/4 x 1/4 x 1/2 x 1/2 x 1/2 = 1/64 x 1/8 = 1/512

Q1. Disease A is an autosomal recessive condition. Suppose that a husband and wife are both carriers (i.e. heterozygotes) for the disease. If they have three children, the probability of having at least 1 unaffected child is:
A. 1/2
B. 63/64
C. 1/8
D. 3/4

I get 37/64 and I do not think B or D are right. Here's my work.

Aa x Aa Punnett yields 3 A_ to every 1 aa.
3/4 or 75% to be A_ and unaffected, 1/4 chance to be aa and affected.

The probability that at least one is affected is equal to 1 minus the probability that no one is affected. This is using a little statistics- complement events!

P(no one affected) = 3/4 x 3/4 x 3/4 = 27/64
P(at least one is affected) = 1 - P(no one affected) = 1 - 27/64 = 37/64

Thoughts? I don't know why 37/64 is not an answer choice; this method makes sense to me...

 

[TheBoondocks]

Yep I believe it is D also. Here's my work, if anyone gets something else let us know!

The probability to get the disease is 1/4 (Aa x Aa Punnett square: 3A_, 1aa).

The probability for all to get the disease [any gender]
P(A and B and C) = P(A) x P(B) x P(C) = 1/4 x 1/4 x 1/4

The probability for all to get the disease AND for all to be boys:
1/4 x 1/4 x 1/4 x 1/2 x 1/2 x 1/2 = 1/64 x 1/8 = 1/512

I get 37/64 and I do not think B or D are right. Here's my work.

Aa x Aa Punnett yields 3 A_ to every 1 aa.
3/4 or 75% to be A_ and unaffected, 1/4 chance to be aa and affected.

The probability that at least one is affected is equal to 1 minus the probability that no one is affected. This is using a little statistics- complement events!

P(no one affected) = 3/4 x 3/4 x 3/4 = 27/64
P(at least one is affected) = 1 - P(no one affected) = 1 - 27/64 = 37/64

Thoughts? I don't know why 37/64 is not an answer choice; this method makes sense to me...

That can't be applied here.
all unaffected (3/4)^3=27/64
first unaffected, second unaffected, third one affected= 9/64
first unaffected, second affected, third unaffected 9/64
first unaffected, next two affected 3/64
first affected, next two unaffected 9/64
first affected, second non affected, third affected 3/64
first affected, second affected, third unaffected 3/64
finally, all three affected 1/64

The sum comes to 63/64. I have a hunch this is hw and not MCAt. The second question seems mcat, but not the first. I looked at it and guessed 63/64 because atleast means one or more, plus, you must add up different ways. The most important thing is that it isn't necessary to calculate 63/64. You just need to realize that the other 3 answers don't apply here.

 

[futuredoctor10]

Thanks for that explanation Boondocks... that's what you and others were indicating (to combine all the possibilities), but it only makes sense now...

I agree the first question does not seem very MCAT-like!

 

[csau]

Hello,

The first question can be done easily through considering the complement. For example, the set of combinations where all children are affected would be the complement of at least 1 unaffected child.

This way, you can calculate the probably of all children being affected, then calculate for its complement.

The solution to this method would be:

(1) Probability of one child being affected: 1/4
(2) Probability of all three children being affected: (1/4)^3 = 1/64
(3) Probability of the complement of (2): 1 - 1/64 = 63/64

In addition, this solution partially answers the second question. To factor in the probability of all three being boys, you just multiply by the corresponding probabilities.

Solution to the second question:

(1) Probability of all three children being affected: 1/64 (from previous)
(2) Probability of being male: 1/2
(3) Probability of all three children being male: (1/2)^3 = 1/8
(4) Probability of (1) AND (3): 1/64 * 1/8 = 1/512



The solution for question one is similar to futuredoctor10's solution. However, futuredoctor10 calculated for at least one affected child and the question asks for at least one unaffected child.

 

[futuredoctor10]

Thanks csau!!! I knew complement events had to work, but just answered the wrong question.

So P(one unaffected) = 1 - (P all affected) = 1 - (1/4 x 1/4 x 1/4) = 1 - (1/64) = 63/64