Biology and Organic Chemistry Essay Example | Topics and Well Written Essays - 1000 words

biota and Organic Chemistry - Es grade ExampleAlso, allow q2 to check homozygous recessive allele individuals, which agency that q = the squarely root of the number that equals q2. Finally, we can gather the answer for 2pq, given all of the above information.When spy nation consisting of 1,000 squirrels, there are 2 expressions of coat colors, blushing(a) and black. We have observed that 292 squirrels were homozygous dominant, 440 squirrels were heterozygous and 268 were homozygous recessive.We pull up stakes say that R will represent the allele for dominant, red fur. Lets then say that r will stand for the recessive allele which when expressed in a homozygous pair, shows up as black fur coats on the squirrels. If we know we have 292 squirrels that are homozygous dominant, this tells us that 29.2% of the squirrels were RR (homozygous dominant) and red surface. If we have 440 squirrels which are heterozygous, that means that 44% of the total number of squirrels were Rr and had red coats. If there were268 squirrels which were homozygous recessive then 26.8% of the total number of squirrels were rr and had black coats. These figures were gathered by dividing the number of squirrels that possessed the same genotype (rr, RR or Rr) by the number of the total squirrel universe. This number reflects the actual number of squirrels possessing the same genotype into a percentage of the population of squirrels as a whole.To figure out the allelic frequency, we need to loo... We can safely assume from the data above that p2 = .292 and stands for the percent of homozygous dominant squirrels. When we take the square root of .292, we get.50. So, 50 is the frequency of dominant alleles. Also, we need to find out the frequency of recessive alleles. Essentially, we will want to know that q2 = the percentage of recessive squirrels. The square root of that number is equal to the frequency of recessive alleles. So, q2 = .268. When we take the square root of that we get.52. So then, the frequency of recessive alleles is equal to .52. 2. Was the population from question 1 in Hardy Weinbergs equilibrium Explain using the ki square test.The chi square model is a test that can be used on for each one folk recessive and dominant. We are able then to examine what we expect to see and compare that to what is expressed. We can define if the population is a Hardy Weinberg equilibrium or not. The way in which we can do this is by using the following formula (the sum of) (O-E)2/E. Let us say that O equal the numbers that we have observed while E stands for the numbers we expect to see in each group. We believe based on standard frequencies that we will see 750 red coated squirrels and 250 black coated squirrels. We have actually observed 732 red coated squirrels and 268 black ones. When we use these figures in the formula of the chi square model, we end up with 1.73 with one degree of freedom. There are only 2 assertable phenotypic categories here, red or b lack. 3. Ten years ago, we could see similar results on the same population of squirrels, but only phenotypes were recorded.3a. Assuming Hardy-Weinberg equilibrium, calculate the allelic