1. a) Our expectation was that Allele A1 recurrence would increment after Some time since it is the most qualified. Yes, my expectation was right. b) The last recurrence expanded at a much slower rate than some time recently. 2. A1 is the prevailing allele. Since A1 is shown in A1 A2 and A1A1 genotype. These two genotype delineates one and only phenotype, which is constantly present in both of these genotypes. Additionally, this phenotype is preferential in natural selection. 3. The kind of natural selection that was happening in the cases above was directional in light of the fact that the diagram was leveling out asymptotically at 1, demonstrating that variety will be moving in one course. 4. Yes, in the long run after some time …show more content…
10. Heterozygote advantage of interest is said to keep up hereditary variety since it contains both dominant and recessive alleles. The allele recurrence for heterozygotes will increment and they will be defended from malaria and sickle cell sickness. The homozygotes will fall to either malaria or sickle cell illness. HWE questions a) • No Natural Selection • There is no migration • There is no genetic drift • There is no mutation b) 1. 4/10,000 = 0.0004, this is q^2 2. 0.0004 = 0.02. 3. p = 1 – 0.02 = 0.98. p^2 = 0.9604 2pq = 0.0392 q^2 = 0.0004 Accepting Hardy-Weinberg for this situation is not right on the grounds that there natural selection acting against the allele that causes the illness, given that kids with the ailment will die before they pass it on to their kids. c) • y frequency q = 0.8 • p = 1 – 0.8 = 0.2 • p^2 = 0.04, 2pq = 0.32, q^2 = 0.64 d) i. 336 + 64 = 400 animals in total, q^2 = 64/400 = .016, this is the homozygous recessive. ii. Since q^2 is 0.16, q = 0.4. p+q=1 so p is equal to 0.6 the dominant allele. iii. 2pq = (2)(0.6)(0.4) = 0.48 *(400) = 192, the heterozygous iv. p = 0.6, q =
p2 + 2pq + q2 = 1 ; where ‘p2’ represents the homozygous dominant genotype, ‘2pq’ represents the heterozygous genotype, and ‘q2’ represents the homozygous recessive genotype
Figure 4. This graph depicts the average allele frequency of male cichlid fish when a mutation that goes from recessive to dominant arises within the population at a rate of 0.001. The average value over five trials is shown to be 0.318.
The results of the allele frequency changes in the five different trials are all in close range of one another. The trials 1 and 5 being the same at .85 and were the highest, trial 4 was the lowest at .70, and the two middle results were trial 2 at .75 and trial 3 at .80. In a population size of 1000, the difference in the green and albino alligator’s pigments will inevitably affect the fitness. The green alligator is the dominant big “A” allele, while the
Suppose the feather color of a bird is controlled by two alleles, D and d. The D allele results in dark feathers, while the d allele results in lighter feathers.
Calculate the ratios of the genotypes and phenotypes of the offspring in the F1 generation.
The two recessive alleles are both on the same chromosome. Genes A and B completely follow Mendel’s principles of inheritance; genes B and C are physically connected together and never are separated from each other at any time during any cell division cycle or fertilization event. Draw below the gamete genotypes that this individual could produce.
was used to find the frequencies of the dominant and recessive alleles. After noting and calculating the results of the first fifty people at the mall, I tested a second experiment, deducing that attached ear lobes seemed to be more prevalent in younger females. My new hypothesis was females under age thirty had a higher frequency of attached ear lobes, so I then observed the next fifty females that appeared under the age of thirty and recorded the results. The Hardy Weinberg formula was again used in the calculations.
Directional Selection occurs when an extreme phenotype at one end of a population distribution is favored over all other
i) (one point) What proportion of gametes produced by the F1 have the dominant allele for both loci? What proportion of the gametes produced by line B have the d
The purpose of the Population Genetics Lab was to use the Hardy Weinberg theory of genetic equilibrium and examine the relationship between evolution and allele frequencies. Additionally this lab allowed us to examine how microevolution factors such as natural selection, heterozygote advantage, and genetic drift affected the genotype and allele frequencies. Finally the population genetics lab permitted us to become masters of finding allele frequencies and establishing a better knowledge on evolution in populations and how we can calculate if that occurred using the Hardy-Weinberg equations. This experiment is relevant because
12) In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency of allele a is 0.2. What is the frequency of individuals with Aa genotype?
trait by other genes or the environment) is the rule rather than the exception” (pp. 23).
B. The incidence of sickle-cell anemia is an excellent example of a "balanced polymorphism" because "It illustrates balanced polymorphism because carriers are resistant to malaria,
This table helps show all the possible genotypes from one set of parents. The table shows that the genotypes purple and starchy are dominant, and the genotypes yellow and sweet are recessive.(stallsmith)
The pairs of alternative traits examined segregated among the progeny of a particular cross, some individuals exhibiting one traits, some the other