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The frequency that aa is 36%, which way that q2 = 0.36. If q2 = 0.36, climate q = 0.6.As q is the frequency the the a allele, the frequency is 60%.
You have sampled a population in which you know that the percent of the homozygous recessive genotype (aa) is 36%. Making use of that 36%, calculate the frequency that the "A" allele.
The frequency the aa is 36%, which way that q2 = 0.36. If q2 = 0.36, then q = 0.6.Since q = 0.6, and p + q = 1, then ns = 0.4.The frequency that A is same to p, therefore the prize is 40%
You have sampled a population in i m sorry you understand that the percent of the homozygous recessive genotype (aa) is 36%. Making use of that 36%, calculate the frequency the the genotypes "AA" and also "Aa.
The frequency the aa is 36%, which means that q2 = 0.36. If q2 = 0.36, climate q = 0.6.Since q = 0.6, and p + q = 1, then ns = 0.4.The frequency the AA is same to p2, and also the frequency the 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 room 100 student in a class. Ninety-six did fine in the food whereas 4 blew that totally and also received a grade of F. In the extremely unlikely occasion that this traits are genetic rather than environmental, if this traits show off dominant and recessive alleles, and if the four (4%) stand for the frequency that the homozygous recessive condition, calculate the frequency of the recessive allele.
The homozygous recessive because that this gene (q2) represents 4% (i.e. = 0.04).The square source (q) is 0.2 (20%).
Within a populace of butterflies, the color brown (B) is leading over the color white (b). And, 40% of all butterflies are white. Provided this basic information, i m sorry is something the is really likely to be on one exam, calculate the percent of butterflies in the population that are heterozygous.
Since white is recessive (i.e. Bb), and also 40% that the butterflies space white, then bb = q2 = 0.4. To identify q, which is the frequency the the recessive allele in the population, simply take the square root of q2 which functions out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, climate p must be 1 - 0.63 = 0.37. The percentage of butterflies in the population that are heterozygous would be 2pq therefore the answer is 2 (0.37) (0.63) = 0.47.
A very huge population the randomly-mating activities mice includes 35% white mice. White coloring is resulted in by the dual recessive genotype, "aa". Calculation allelic and genotypic frequencies for this population.
35% room white mice, i m sorry = 0.35 and represents the frequency the the aa genotype (or q2). The square root of 0.35 is 0.59, which equates to q. Because p = 1 - q climate 1 - 0.59 = 0.41. Now that we know the frequency of every allele, we can calculate the frequency that the continuing to be genotypes in the populace (AA and Aa individuals). AA = p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and also as before aa = q2 = 0.59 x 0.59 = 0.35. If you add up every these genotype frequencies, they should equal 1.
After graduation, you and 19 of her closest friend (lets to speak 10 males and also 10 females) charter a aircraft to go on a round-the-world tour. Unfortunately, you all crash floor (safely) ~ above a deserted island. No one finds you and also you begin a brand-new population entirely isolated native the remainder of the world. Two of her friends carry (i.e. Room heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency that this allele does not adjust as the population grows, what will be the incidence the cystic fibrosis on her island?
There room 40 full alleles in the 20 world of i m sorry 2 alleles room for cystic fibrous. So, 2/40 = 0.05 (5%) the the alleles are for cystic fibrosis. The represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% that the F1 population will it is in born v cystic fibrosis.
Cystic fibrosis is a recessive problem that affects around 1 in 2,500 babies in the Caucasian populace of the united States. You re welcome calculate the frequency that 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 also "B" alleles are existing in the ABO system; there space no people with form "O" blood or through O alleles in this certain population. If 200 people have kind A blood, 75 have form AB blood, and also 25 have form B blood, what are the alleleic frequencies of this population.
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To calculation the allele frequencies because that A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with form AB blood room heterozygous AB, and also individuals with form B blood space homozygous BB. The frequency that A equals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number of individuals). Hence 2 x (200) + (75) separated by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is just 1 - p, then q = 1 - 0.792 or 0.208.