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wrote...
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4 years ago
Individuals III-3 and III-4 are expecting their first child when they become aware that they both have a family history of this recessive condition. As their genetic counselor, you can calculate the probability that they are carriers and that their child will be affected with the condition.

Choices are: 1/12, 1/2, 3/4, 1/3, 1/16, 2/3, 1/6

The probability that III-3 is a carrier (Rr)=

The probability that III-4 is a carrier (Rr)=

The probability that IV-1 will be affected (rr)=
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wrote...
4 years ago
The probability that III-3 is a carrier (Rr)= 2/3

The probability that III-4 is a carrier (Rr)= 1/2

The probability that IV-1 will be affected (rr)= 1/3

If you want me to explain how to get the answers feel free to say it.
Dan101
wrote...
4 years ago
Can you please explain?
wrote...
4 years ago
Well now that I look at it, my answer was wrong.

for III-3 it's 1/2 because his mother (represented by a circle) is Rr and his father (represented by a square) whose genotype isn't stated is also Rr. This is deduced from the fact that one of their offspring (III-1) is positive for the condition (which means it's rr). This means that the father who isn't positive for the condition is carrying an r gene making him Rr. When you do a punnet square and cross both parents the probability that III-3 will be a carrier (Rr), is 1/2.

for III-4 it's 1/2, because when you use a punnet square and cross both parents. there's a 1/2
probability that III-4 is a carrier.

The probability that IV-1 will be affected (rr) is 1/8, because you multiply the probability that both parent's are carriers which is 1/2 * 1/2 = 1/4 then you multiply that by the probability that a cross between carriers will result in a homozygous recessive genotype (rr) which you can get by doing a punnet square of Rr x Rr which gives a 1/4 probability and 1/4*1/4 = 1/8

Heh, I had messed up pretty bad
Rose angel
wrote...
4 years ago
The last answer should be 1/16 though !!  Smiling Face with Halo
hermoine21
wrote...
4 years ago
The probability that III-3 is a carrier (Rr)= 2/3

The probability that III-4 is a carrier (Rr)= 1/2

The probability that IV-1 will be affected (rr)= 1/12

Note: The original answer of 2/3 for III-3 was correct because the probability that III-3 is a carrier is calculated based on the available options. A punnet square is not entirely correct because the person is showing the dominant phenotype. The punnet square shows 3 options that will give the dominant phenotype, 2 of which are Rr (carrier).

Multiplying 2/3 x 1/2 x 1/4 (likelihood that the child will be rr from punnet square) = 1/12.

I just did this problem on masteringgenetics and got it correct.
wrote...
3 years ago
The probability that III-3 is a carrier (Rr)= 2/3

The probability that III-4 is a carrier (Rr)= 1/2

The probability that IV-1 will be affected (rr)= 1/12

Note: The original answer of 2/3 for III-3 was correct because the probability that III-3 is a carrier is calculated based on the available options. A punnet square is not entirely correct because the person is showing the dominant phenotype. The punnet square shows 3 options that will give the dominant phenotype, 2 of which are Rr (carrier).

Multiplying 2/3 x 1/2 x 1/4 (likelihood that the child will be rr from punnet square) = 1/12.

I just did this problem on masteringgenetics and got it correct.


hermoine is correct, the others are not.
Dawn122
wrote...
3 years ago
The probability that III-3 is a carrier (Rr)= 2/3

The probability that III-4 is a carrier (Rr)= 1/2

The probability that IV-1 will be affected (rr)= 1/12

Note: The original answer of 2/3 for III-3 was correct because the probability that III-3 is a carrier is calculated based on the available options. A punnet square is not entirely correct because the person is showing the dominant phenotype. The punnet square shows 3 options that will give the dominant phenotype, 2 of which are Rr (carrier).

Multiplying 2/3 x 1/2 x 1/4 (likelihood that the child will be rr from punnet square) = 1/12.

I just did this problem on masteringgenetics and got it correct.


You can calculate probabilities in pedigrees by considering the requirements for a specific outcome. In this case, the outcome that individual IV-1 will be affected has four requirements:

1.Individual III-3 is a carrier (probability = 2/3).
2.Individual III-4 is a carrier (probability = 1/2).
3.Individual III-3 passes the r allele to his child (probability = 1/2, assuming III-3 is a carrier, which is accounted for in requirement 1).
4.Individual III-4 passes the r allele to her child (probability = 1/2, assuming III-4 is a carrier, which is accounted for in requirement 2).


Because requirements 1 AND 2 AND 3 AND 4 must be met for the outcome in question (individual IV-1 to be affected), you calculate the answer by applying the multiplication rule (as implied by “AND”):
The probability that IV-1 will be affected (rr) = 2/3 x 1/2 x 1/2 x 1/2 = 1/12.
wrote...
3 years ago
hermoine 21, tkepper, and dawn122 are correct. Thanks guys!
wrote...
3 years ago
The probability that III-3 is a carrier (Rr)= 2/3

The probability that III-4 is a carrier (Rr)= 1/2

The probability that IV-1 will be affected (rr)= 1/12

these are the actual answers
wrote...
3 years ago
I made up my mind the answer are:

The probability that III-3 is a carrier (Rr)= 2/3

The probability that III-4 is a carrier (Rr)= 1/2

The probability that IV-1 will be affected (rr)= 1/12

Hard work beats talent when talent fails to work hard.”
― Kevin Durant
wrote...
3 years ago
thanks!
wrote...
3 years ago
Yeah hermione was correct, I didn't consider the phenotype being known for the father.
wrote...
3 years ago
This is the correct one
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