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Calculating probabilities in pedigrees
jessiijessii
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Calculating probabilities in pedigrees
Individuals III3 and III4 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 III3 is a carrier (Rr)=
The probability that III4 is a carrier (Rr)=
The probability that IV1 will be affected (rr)=
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sanflash
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The probability that III3 is a carrier (Rr)=
2/3
The probability that III4 is a carrier (Rr)=
1/2
The probability that IV1 will be affected (rr)=
1/3
If you want me to explain how to get the answers feel free to say it.
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Dan101
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#2
2 years ago
Can you please explain?
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sanflash
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Well now that I look at it, my answer was wrong.
for
III3
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 (III1) 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 III3 will be a carrier (Rr), is 1/2.
for
III4
it's 1/2, because when you use a punnet square and cross both parents. there's a 1/2
probability that III4 is a carrier.
The probability that
IV1
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
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Rose angel
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#4
2 years ago
The last answer should be 1/16 though !!
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hermoine21
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#5
2 years ago
The probability that III3 is a carrier (Rr)= 2/3
The probability that III4 is a carrier (Rr)= 1/2
The probability that IV1 will be affected (rr)= 1/12
Note: The original answer of 2/3 for III3 was correct because the probability that III3 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.
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Quote from: hermoine21 (2 years ago)
The probability that III3 is a carrier (Rr)= 2/3
The probability that III4 is a carrier (Rr)= 1/2
The probability that IV1 will be affected (rr)= 1/12
Note: The original answer of 2/3 for III3 was correct because the probability that III3 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.
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Dawn122
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Quote from: hermoine21 (2 years ago)
The probability that III3 is a carrier (Rr)= 2/3
The probability that III4 is a carrier (Rr)= 1/2
The probability that IV1 will be affected (rr)= 1/12
Note: The original answer of 2/3 for III3 was correct because the probability that III3 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 IV1 will be affected has four requirements:
1.Individual III3 is a carrier (probability = 2/3).
2.Individual III4 is a carrier (probability = 1/2).
3.Individual III3 passes the r allele to his child (probability = 1/2, assuming III3 is a carrier, which is accounted for in requirement 1).
4.Individual III4 passes the r allele to her child (probability = 1/2, assuming III4 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 IV1 to be affected), you calculate the answer by applying the multiplication rule (as implied by “AND”):
The probability that IV1 will be affected (rr) = 2/3 x 1/2 x 1/2 x 1/2 = 1/12.
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hermoine 21, tkepper, and dawn122 are correct. Thanks guys!
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The probability that III3 is a carrier (Rr)= 2/3
The probability that III4 is a carrier (Rr)= 1/2
The probability that IV1 will be affected (rr)= 1/12
these are the actual answers
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I made up my mind the answer are:
The probability that III3 is a carrier (Rr)= 2/3
The probability that III4 is a carrier (Rr)= 1/2
The probability that IV1 will be affected (rr)= 1/12
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thanks!
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sanflash
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11 months ago
Yeah hermione was correct, I didn't consider the phenotype being known for the father.
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This is the correct one
Attached file
bio.docx
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