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The frequency of aa is 36%, which means that q2 = 0.36. If q2 = 0.36, then q = 0.6.As q is the frequency of the a allele, the frequency is 60%.
You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the frequency of the "A" allele.
The frequency of aa is 36%, which means that q2 = 0.36. If q2 = 0.36, then q = 0.6.Since q = 0.6, and p + q = 1, then p = 0.4.The frequency of A is equal to p, so the answer is 40%
You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the frequency of the genotypes "AA" and "Aa.
The frequency of aa is 36%, which means that q2 = 0.36. If q2 = 0.36, then q = 0.6.Since q = 0.6, and p + q = 1, then p = 0.4.The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, calculate the frequency of the recessive allele.
The homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04).The square root (q) is 0.2 (20%).
Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the percentage of butterflies in the population that are heterozygous.
Since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 = 0.4. To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1 - 0.63 = 0.37. The percentage of butterflies in the population that are heterozygous would be 2pq so the answer is 2 (0.37) (0.63) = 0.47.
A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population.
35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 - q then 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate the frequency of the remaining genotypes in the population (AA and Aa individuals). AA = p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa = q2 = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they should equal 1.
After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island?
There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = 0.05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% of the F1 population will be born with cystic fibrosis.
Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the frequency of the recessive allele in the population.
q2 is 1/2,500 or 0.0004. Therefore, q is the square root, or 0.02. The frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%).
In a given population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population.
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To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB. The frequency of A equals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number of individuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 - p, then q = 1 - 0.792 or 0.208.