2. Suppose that you and 19 of your classmates (giving a final population of 10 males and 10 females) are on a cruise, and your s
hip sinks near a deserted island. You and all of your friends make it to shore and start a new population isolated from the rest of the world. Two of your friends carry the recessive allele (i.e., are heterozygous) for phenylketonuria. If the frequency of this allele does not change as the population on your island increases, what will be the incidence of phenylketonuria on your island?
20 people start the new population. So there are 20 genes or 40 alleles for the recessive disorder phenylketonuria. 2 out of 40 alleles are recessive for the condition hence frequency of the allele = 2/40 = 0.05
Frequency of the allele does not change when the population increases so it is in Hardy-Weinberg equilibrium. According to it, if q is the frequency of recessive allele, q² = frequency of the recessive condition
Here, q = 0.05 So,
q² = (0.05)² = 0.0025
In percentage, it is 100 * 0.0025 = 0.25%
Hence, incidence of phenylketonuria in the new population is 0.25%
First move the x from the right side to the left by subtracting. 3x-6+2x-x=-2. combine like terms. 4x-6=-2. add six to both sides. 4x=4. divide 4 on both sides. x=1. hope this helped!
There would still be a 50% chance for the offspring to be Heterozygous (Hybrid), but then there would be another 50% chance for the offspring to be Homozygous recessive (True-breeding short plants)
