Module 12 Hardy Weinberg Law Flashcards
Principle that states allele frequencies remain constant from generation to generation.
Hardy-Weinberg Principle
T/F, according to HW Principle, after the first generation of random mating, genotype frequencies do not remain constant.
F, After first generation of random mating, genotype frequencies also remain constant
Assumptions of Hardy-Weinberg Principle (7)
- Diploid Organism
- Reproduce Sexually
- Nonoverlapping Generations
- Random Mating
- Large Population Size
- Equal allele frequencies in the sexes
- No mutation, migration, or selection
Equation used in calculating genotypic frequencies in progeny
p^2 + 2pq + q^2
p: dominant allele freq.
q: recessive allele freq.
p^2= homozygous dom freq.
2pq= Heterozygote freq.
q^2= homozygous rec freq.
How do you calculate for frequency of mating types
Multiply p^2, 2pq, and q^2, depending on the parental genotypes.
At what value of p or q is max heterozygosity achieved?
When p=q=0.5
T/F a population cannot evolve if it meets the Hardy-Weinberg assumption
True
T/F a single generation of random mating produces the equilibrium frequencies of p2, 2pq, and q2
True
T/F the heterozygote frequency at equilibrium can be greater than 0.5
F, At HWE, Heterozygote frequency never exceeds 0.5
Albinism occurs with a frequency of about 1 in 20,000 in European populations. Assuming it to be due to a single autosomal recessive gene, and assuming the population to be in HWE, what proportion of people are carriers?
Given: freq(aa)= q^2 = 1/20,000
Required: freq (Aa)
Solution:
q^2=1/20000-> q=0.007
p=1-q=0.993
freq (Aa) = 2pq
2pq=2(0.007)(0.993)= 0.0139
freq(Aa)=0.0139
Chi-Square Analysis Problem
Jeffrey Mitton and his colleagues found 3 genotypes (R2R2, R2R3, and R3R3) at a locus encoding the enzyme peroxidase in ponderosa pine trees growing in Colorado. The observed number of these genotypes were 135 for R2R2, 44 for R2R3, and 11 for R3R3. Are the ponderosa trees in HWE at the peroxidase locus?
Answer with solution and minor explanation of steps in sheets:
bit.ly/Module12Chi-Square-Sample
In cases of di- or multi-hybrid genes, how do we solve for frequency of specific combinations? Let the loci be A and B with alleles A, a, and B, b respectively.
Expand the formula and apply product rule.
Locus A:
p^2+2pq+q^2 = 1
p^2=AA; 2pq=Aa; q^2=aa
Locus B:
r^2+2rs+s^2 = 1
r^2=BB; 2rs=Bb; s^2=bb
Multiply the above and add depending on what genotypes/phenotypes are being required.
Ex.
F(A_bb) = F(AAbb+Aabb)
F(A_bb) = (p^2 s^2)+ (2pqs^2)
Multihybrid gene example:
If the frequency of gene A is 0.4, gene b = 0.3, gene C = 0.2 and gene D = 0.5 and the genes are in different chromosomes, what is the frequency at genetic equilibrium of the AaBBccDd genotype?
Given: A=0.4, b=0.3, C =0.2, D=0.5
Required: Freq AaBBccDd
Solution:
Let AaBbCcDd=pqrstuvw
A=0.4-> a=0.6; b=0.3->B=0.7; C=0.2->c=0.8; D=0.5->d=0.5
AaBBccDd=(2pq)(r^2)(u^2)(2vw)
2(.4)(.6)(.49)(.64)*2(.5)(.5)
Freq AaBBccDd = 0.075264
For X-Linked genes, how do you get the p and q values for men and women
For women: Use HWE p^2+2pq+q^2
For men: Use allele freqs as there is only one X allele p+q=1
Why are recessive X-linked traits more common among males.
2 copies of recessive x needed in females, only one for males.
