BIOL 202 Lecture Notes - Lecture 10: Coefficient Of Relationship, Genotype Frequency, Distant Relatives

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redclam140

24 May 2021

School

McGill University

Department

Biology

Course

BIOL 202

Professor

Western, Dankort, Nilson, Hipfner and Schoen

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Document Summary

Cystic fibrosis is a recessive disease: aa = diseased individual, aa = unaffected individual, aa = carrier individual. The population is in hardy-weinberg proportions: p2 + 2pq + q2 ratios applies. A, b and c are 3 separate genes with 2 alleles each: version 1 (a1, b1 and c1, version 2 (a2, b2 and c2) A is biallelic (a1/a2) at the chosen locus: frequency of a is given by 2pq. B and c are monoallelic (b2/b2 and c2/c2) at the chosen locus: frequencies of b and c are given by p2 and/or q2. Assumptions: genes a, b and c are unlinked frequencies are independent of each other. Affects genotype frequencies and not allele frequencies. Is an important factor in genetic variation distribution in a population. If inbreeding is absent and hardy-weinberg conditions hold: f (a/a) = p2 f (a/a) = 2pq f (a/a) = q2. P (a passing pink allele to b) = 1/2.

Related Questions

Define each of the Hardy-Weinberg expected frequencies (using p,q,r) for autosomal locus, with three traits, where A and B are co-dominant to C, and A is more common: Frequency f(AA) f(BB) f(CC) f(AB) f(AC) f(BC)

H-W Term (q^2)

Phenotypic Characteristics (A)

blackfly450

II think this is a pretty basic Hardy-Weinberg Problem, I justdon;t know how to solve it yet. Thank you for your help!

Assume a population of 800 individuals with the followinggenotypes:452 AA, 296 Aa, and 52 aa. This is the 1stgeneration.

a. Calculate the frequencies for A___ and a___. Do this fromgenotype frequencies (do NOT assume Hardy Weinberg Equilibrium)

b. Use the allele frequencies from part a to calculate theexpected genotype frequencies for AA___ Aa___ and aa__

c. Based on your allele frequencies above, how many of eachgenotype would you expect if the population of 800 people were inHardy-Weinberg Equilibrium? AA____ Aa____aa____

d. Suppose that the original population (452 AA; 296 Aa; 52 aa)experienced an epidemic of an infectious disease that kills 75% ofeach homzygous class ( AA and aa). and 25% of the heterozygotes.How many of each genotype would survive?

e. The results of part e are still the first generation, butpost-epidemic survivors. What are the new allele frequencies?

f. Assume the 1st generation post-epidemic survivors materandomly, and there is no selection, no migration, no mutation, nogenetic drift. What frequency of each genotype is expected in the2nd generation?

g. If the 2nd generation contains 350 indivduals how many ofeach genotype do you expect?

cerisekangaroo944

Suppose that red flower color (A/A or A/a) is dominant to white flower color (a/a) in petunias. The common allele (A) encodes the red pigment for petal colour. The rare allele (a) encodes a nonsense mutation in the pigment gene, so no pigment is produced (such that petals are not coloured and appear white). Among the petunias in a greenhouse population, 480 are A/A and 90 are A/a. There are no white-flowered petunias observed. Is this population of petunias in genetic equilibrium? Show all of your steps and calculations clearly (including use of a Ï 2 test) and write a brief (2-3 sentence) conclusion about whether this population of petunias is in HardyâWeinberg equilibrium. Make sure the expected genotype frequencies equal to 1

You have selected a petunia plant from the greenhouse with red flowers and want to determine whether the plant is A/A or A/a.

a) What kind of experiment (what kind of cross) could you perform to determine the genotype of the selected petunia plant? Explain briefly.

b) How will the cross in (a) help you determine the genotype of your red-flowered petunia? That is, explain how will the results from this cross differ if the red-flowered petunia is A/A versus A/a?

c) You end up choosing two red plants that are both of the genotype A/a and you cross them. If the cross produces six plants, what is the probability that four of them will be white and two will be red?

MY ANSWERS SO FAR

Allele frequency of p= 1/2 (2pq)+ p²

P= 1/2 (0.1578) + 0.8421

=0.921

Allele frequency of q = 1/2 (2pq) + q²

⁼ 1/2 (0.1578) + 0.0

= 0.0789

Expected genotype of the population:

AA(p2)= 0.848 aa (q2)= 0.006

(2pq)Aa =0.0105

I'm confused as to why this doesn't total upto 1

goldcrab34

BIOL 202 Lecture Notes - Lecture 10: Coefficient Of Relationship, Genotype Frequency, Distant Relatives

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