n a study conducted by an undergraduate student, the obtained data showed that approximately 75% of the European Northern human population demonstrate the expression of a phenotype directly associated with the homozygous dominant genotype, while 25% display the mutant phenotype despite their genotypes being homozygous dominant. Using your knowledge propose two hypotheses for this observation.
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In a study conducted by an undergraduate student, the obtained data showed that approximately 75% of the European Northern human population demonstrate the expression of a
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- Propose a hypotheses for this observation: In a study conducted by an undergraduate student, he obtained data showing that approximately 75% of the European Northern human population demonstrate the expression of a phenotype directly associated with the homozygous dominant genotype, while 25% display the mutant phenotype despite their genotypes being homozygous dominant.Your task will ultimately be to generate a hypothesis and test this hypothesis using a chi square test. There are four possible phenotypes for these kernels:Purple and Smooth Purple and Shrunken Yellow and Smooth Yellow and Shrunken The purple colour results from the formation of a purple pigment in the skin of the kernel. The gene locus for the colour trait is designated “R”. The wrinkled phenotype arises from a decrease in the amount starch produced within the kernel. The gene locus for the texture trait is designated “S”. Create a Punnett square for this cross and present the expected phenotypic ratio. Present the results in a table as dihybrids and determine the observed phenotypic ratio of purple/smooth to purple/shrunken to yellow/smooth to yellow/shrunken. Consider the plant as a whole rather than the individual cobs. Generate a null hypothesis and perform a chi square test on the data. Report the equation used, the chi square value, the degrees of freedom, and the resulting…The following data for the genotypes at the alcohol dehydrogenase locus were observed from a sample of Drosophila melanogaster. Sample size 1000 FF 550 FF=0.55; FS=0.34; SS=0.11 O F-0.72; S=0.28 FF=0.078; FS=0.40; SS=0.52 FS What are the Hardy-Weinberg equilibrium genotype frequencies for each genotype based upon this sample? OFF=0.52; FS=0.40; SS=0.078 340 SS 110
- .. Nowadays, newborn babies are required to be tested for phynylketonuria (PKU), an autosomal recessive genetic disorder. If untreated, individual homozygous for PKU suffer mental retardation. In a recent year, 4 PKU babies were detected out of 126,000 tested. Assuming Hardy-Weinberg conditions, what is the frequency of the PKU gene in this population (Give your answer in decimal number with 3 decimal places)?PLEASE explain how to do this question step by tspe i am very confused! thank you! You have three independent mutant alleles in the Drosophila gene no legs (nlg): nlg1, nlg2, and nlg3. You assay the phenotype of Drosophila, which is an insect, that are heterozygous for the three allele (heterozygous for a wild-type and a mutant allele), and that are homozygous for the three alleles. Allele nlg1 nlg2 nlg3 heterozygous Wild-type Four pairs of legs Wild-type homozygous Stumpy legs Five pairs of legs No legs nlg1 is a ___X___ allele, nlg2 is a ___Y___ allele, and nlg3 is a ___Z___ allele. X Y Z A Gain of function Null Hypomorphic B Hypomorphic Gain of function Null C Null Gain of function Hypomorphic D Null Hypomorphic Gain of function E Hypomorphic Null Gain of function Referencing the table above, select the correct answer: 1 : A 2 : B 3 : C 4 : D 5 : E Correct answer is BSteven Frank and Laurence Hurst argued that a cytoplasmically inherited mutation in humans that has severe effects in males but no effect in females will not be eliminated from a population by natural selection because only females pass on mtDNA (S. A. Frank and L. D. Hurst. 1996. Nature 383:224). Using this argument, explain why males with Leber hereditary optic neuropathy are more severely affected than females.
- The mutation rate of the genome of SARS-COV2 is estimated to be between 0.8-2.38 x 103. Let's use the mutation rate of 0.0008 of a particular site converting allele A2 to A1 (careful with this one). Let's imagine that the initial allele frequencies are: A1 = 0.8 = p %3D A2 = 0.2 = q What is the allele frequency of allele A1 in the next generation? Please use five decimal points (Using fewer decimals will result in no points for this question).A certain form of congenital glaucoma is caused by an autosomalrecessive allele. Assume that the mutation rate is 10-5 and that peoplewith this condition produce, on the average, only about 80% of theoffspring produced by people who do not have glaucoma.a. At equilibrium between mutation and selection, what will the frequencyof the gene for congenital glaucoma be?b. What will the frequency of the disease be in a randomly matingpopulation that is at equilibrium?Star eye A peculiar eye condition known as "star” is manifested as a dominant gene in Drosophila. Its recessive allele R* produces the normal eye of wild type. The expression of R can be suppressed by the dominant allele of another locus, Ru-R. Ru-R*, as the recessive allele of the said locus, has no inhibitory effect on R*. When a normal-eyed male of genotype Ru-R Ru-R RR is crossed to a homozygous wild-type female of genotype Ru-R* Ru-R* R*R*, what phenotypic ratio is expected in the F2?
- O Mutation-selection balance Suppose that one allele A₁ mutates to another allele A2 at some rate, μ. Suppose as well that A₁ is dominant over A2 such that A₁A1 and A₁A2 both have the same fitness, but that individuals that are homozygous recessive (A2A2) for the mutant allele A2 are less fit than the dominant genotype by some amount s, the selection coefficient. In this case, A2 mutant alleles come into the population at rate µ, and are removed from the population only when the show up in homozygous genotypes. The gory mathematical proof can be found in Box 7.8, which tells us that: O ■ Example: Suppose A₁ mutates to A2 at rate 0.005, but A₂A2 homozygous recessives are 50% less fit (s = 0.5) than either A₁A1 or A₁A2. What are the expected equilibrial abundances of A₁ and A2? μ = 0.005 S = 0.5 p* = 1-sqrt(µ/s) = 1-sqrt(0.005/0.5) = 0.9 = sqrt(µ/s) = sqrt(0.005/0.5) = 0.1 ● * p = = 1-sqrt(µ/s) q* = sqrt(μ/s) ● Question: Suppose A₁ mutates to A2 at rate 0.01, but A₂A2 homozygous…Based on this information (picture) A. What is the probability that a randomly sampled individual from the population has two copies of the a allele (that is, that it has an aa genotype)? B. What is the probability that both members of a randomly sampled married couple (man and woman) are aa at the asparagus-smelling gene? C. What is the probability that both members of a randomly sampled married couple (man and woman) are heterozygotes at this locus (meaning that each person has one allele A and one allele a)? D. Consider the type of couple described in (c). What is the probability that the first child of such a couple also has one A allele and one a allele (is a heterozygote)? Remember that the child must receive exactly one allele from each parent.In a complementation cross between Drosophila wiltype(brown eyes). you get a 9:3:3:1 ratio with white eyes being the lowest ratio. what would account for the viabilty of this white mutant strain compared to the brown eyes wild type. additionally why would we see a more robust female population than male population.