10) You and your spouse are Carriers for an Autosomal Recessive disease. If you have 6 kids, what is the probability that 3 of your children are unaffected (normal) and 3 are affected with this disease. Report your answer with 3 decimals or as a percentage.
11) You and your spouse are carriers for an Autosomal Recessive disease i.e., your Genotype is "Aa". If you have 6 kids, what is the probability that only 1 of your kids will be affected by this disease. Report your answer with 3 decimals or as a percentage.
12) Foxes can have 3 different coat colors: Black (AA), Silver (Aa), and White (aa). If you cross 2 Silver foxes and produce 6 offspring, what is the probability that all 6 will be Silver. Report your answer with 3 decimals or as a percentage. For full credit, show your work in the following manner: (1/4)^6 = 0.0002 or 0.02%.
10) To calculate the probability of having 3 unaffected and 3 affected children, we need to use the binomial probability formula.
The probability of having an unaffected child is 1/4 (since both parents are carriers, the chance of having an unaffected child is 1/2 * 1/2 = 1/4).
The probability of having an affected child is 1/4 (since both parents are carriers, the chance of having an affected child is 1/2 * 1/2 = 1/4).
Using the binomial probability formula, the probability is:
P(x = 3) = (6C3) * (1/4)^3 * (3/4)^3
P(x = 3) = (6! / (3! * (6-3)!)) * (1/4)^3 * (3/4)^3
P(x = 3) = (6*5*4) / (3*2*1) * (1/4)^3 * (3/4)^3
P(x = 3) = 20 * (1/64) * (27/64)
P(x = 3) = 540 / 16384
P(x = 3) ≈ 0.033 or 3.3%
Therefore, the probability that 3 of your children are unaffected and 3 are affected with this disease is approximately 0.033 or 3.3%.
11) Since both parents are carriers with the genotype "Aa", the probability of having an affected child is 1/4 (since there is a 1/2 chance of transmitting the recessive allele from each parent).
Using the binomial probability formula, the probability of having only 1 affected child is:
P(x = 1) = (6C1) * (1/4)^1 * (3/4)^5
P(x = 1) = (6! / (1! * (6-1)!)) * (1/4)^1 * (3/4)^5
P(x = 1) = 6 * (1/4) * (243/1024)
P(x = 1) = 1458 / 4096
P(x = 1) ≈ 0.356 or 35.6%
Therefore, the probability that only 1 of your kids will be affected by this disease is approximately 0.356 or 35.6%.
12) Since both parents are Silver foxes (Aa), the probability of having a Silver offspring is 1/2 (since there is a 1/2 chance of transmitting the dominant allele from each parent).
Using the binomial probability formula, the probability that all 6 offspring will be Silver is:
P(x = 6) = (6C6) * (1/2)^6 * (1/2)^0
P(x = 6) = (6! / (6! * (6-6)!)) * (1/2)^6 * (1/2)^0
P(x = 6) = 1 * (1/64) * 1
P(x = 6) = 1 / 64
P(x = 6) ≈ 0.016 or 1.6%
Therefore, the probability that all 6 offspring will be Silver is approximately 0.016 or 1.6%.
10) To determine the probability that 3 of your children are unaffected and 3 are affected with the autosomal recessive disease, we can use the binomial probability formula.
The probability of a child being unaffected is 1/4 (since it requires both parents to pass on the normal allele), and the probability of a child being affected is 1/4 (since it requires both parents to pass on the disease-causing allele).
Using the binomial probability formula, which is given by:
P(X=k) = (n C k) * p^k * q^(n-k)
Where:
P(X=k) = probability of getting k successes (unaffected children)
n = total number of trials (number of children = 6)
k = number of successes (unaffected children = 3)
p = probability of a success (unaffected child = 1/4)
q = probability of a failure (affected child = 1/4)
Applying this formula, we can calculate the probability as follows:
P(X=3) = (6 C 3) * (1/4)^3 * (3/4)^(6-3)
= (6 C 3) * (1/4)^3 * (3/4)^3
= (20) * (1/64) * (27/64)
= 540/16384
≈ 0.033 or 3.3%
Therefore, the probability that 3 of your children are unaffected (normal) and 3 are affected with the disease is approximately 0.033 or 3.3%.
11) Similarly, to determine the probability that only 1 of your kids is affected, we can use the same binomial probability formula.
P(X=1) = (6 C 1) * (1/4)^1 * (3/4)^(6-1)
= (6 C 1) * (1/4)^1 * (3/4)^5
= (6) * (1/4) * (243/1024)
= 1458/4096
≈ 0.356 or 35.6%
Therefore, the probability that only 1 of your kids will be affected by the disease is approximately 0.356 or 35.6%.
12) To determine the probability that all 6 offspring will be Silver, we can use the probability of independent events. Since both parents are Silver (Aa x Aa), each offspring has a 1/4 chance of being Silver.
The probability of all 6 offspring being Silver is calculated as:
P(all 6 Silver) = (1/4)^6
= 1/4096
≈ 0.000244 or 0.0244%
Therefore, the probability that all 6 offspring will be Silver is approximately 0.000244 or 0.0244%.