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11 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.
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wrote...
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11 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
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padre, joybug200, jbaldwin
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wrote...
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11 years ago
The last answer should be 1/16 though !!
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tiffany.g96, Alfred-o
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wrote...
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11 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.
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wtehfunk, tkepper, padre, tyler5, jyk1213, Aironic, emvee, bcoletta, joybug200, cnlopez, LISAZ1984, ahong, janebercilla, danidoug808, alexbitz, mntex, emilyhynson, MRCHOUDH, immurani, buzzcraze123, sammy04, anitabanita, glen1590, adaizy, jacksp1, lsvan, constancelu, briannap1, jio23, lsueross, shagababe13, xEmptyx, thequeenofsoap, tiffany.g96, soniya patel, UnicornDemon, turtwig1998, elliemartines, shadowwolf592, Montanajon, Urm0m, magnussd, toirnas, devinparsons, ilcr, Wick, Toolxao, hallj6, sci-la, ctvygbh, bryacha1
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wrote...
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10 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.
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wrote...
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10 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.
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wrote...
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10 years ago
hermoine 21, tkepper, and dawn122 are correct. Thanks guys!
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wrote...
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10 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
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wrote...
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10 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
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Hard work beats talent when talent fails to work hard.” ― Kevin Durant
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10 years ago
Yeah hermione was correct, I didn't consider the phenotype being known for the father.
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9 years ago
This is the correct one
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eggbedif
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