Chapter 18

Assume 2 alleles, A1 and A2 in a diploid pop.: freq. of genotype A1A1 = P = p^2 freq. of genotype A1A2 = H = 2pq freq. of genotype A2A2 = Q = q^2 Frequency of alleles A1 = p = P + (0.5)H Freq. of alleles A2 = q - Q + (0.5)H

* memorize *

Mutation of CF

- Deleterious Mutations in the CFTR gene cause CF. CFTR transports Cl- in and out of the cells, helps to control the movement of water in tissues, which is necessary for the production of thin, freely flowing mucus.

Mutation CCR5 delta 32 in humans

- beneficial mutation Prevents CCR5 from appearing on their cells, making them resistant to HIV.

stabilizing selection example

B-hemoglobin in humans

Equilibrium disturbed by random genetic drift

Increases variance among populations

Genotype: AA. AS. SS Freq.: p^2. 2pq. q^2 s: selection coefficient for AA t: selection coefficient for SS Relative fitness: 1-s = 0.89 1-t = 0.2 Frequency of A (p) and S (q) reaches equilibrium when ....

1 - s = 0.89 so, s=0.11 1 - t = 0.2 so, t=0.8 p= t / (s+t) -> 0.8 / (0.11 + 0.8) = 0.879 Then, q = 0.121 Freq.: AA = p^2 = 0.879^2 = 0.773 or 7.33% SS = q^2 = 0.121^2 = 0.015 = 1.5% Meaning: 1.5% of the population have sickle cell anemia while 21.1% carry the risk allele and 77.3% are immune

HWE assumptions

1. No Selection; genotypes survive at equal rates and contribute gametes to the next generation equally 2. No Mutation : the alleles we are accounting for stayed the same, none disappeared or were created anew by mutation 3. No Migration : genotypes do enter/exit the population 4. No Chance Events (drift) : the size of the population is infinite 5. Individuals choose mates at random : violating this assumption affects genotype frequency, not allele frequency - diploid, sexually-reproducing with non-overlapping generation NO EVOLUTIONARY FORCES

Relative fitness (W)

A genotype's ability to survive and reproduce relative to other genotypes in the population - also measures differential reproduction

Distribution of malaria overlaps with that os sickle cell anemia because

Carrier (AS) is resistant to malaria, thus preserving the S allele

Equilibrium disturbed by selection

Causes variable effects, depending on the form

Equilibrium disturbed by non-random mating

Causes variable effects, depending on the type

cystic fibrosis (CF)

Chromosome 7 mutations for protein called cystic fibrosis transmembrane conductance regulator (CFTR) Expressed in lining of lungs and intestines

directional selection

Continuously remove individuals from one end of the phenotypic distribution - selection against one extreme

Artificial selection practiced in plants is what type of selection

Directional

Three ways selection can act

Directional Stabilizing Disruptive

Null Hypothesis (H0)

General assumption of any statistical test that there are no significant deviations between the measured results and the predicted ones

Natural selection for bacterial antibiotic resistance

Resistance of Mycobacterium tuberculosis is caused by mutations. Tx involves combination therapy

Cystic fibrosis is an autosomal recessive disease. The frequency of the disease among European Americans is approximately 1 in 2000. Assume the locus is in HWE, calculate the frequency of the disease allele and the frequency of homozygous healthy individuals.

Genotype: aa Assume frequency of a is q and A is p. aa = 1 /2000 = 0.0005 q = sq.rt. 0.0005 = 0.022 (frequency of the diseased allele) P = 1 - 0.22 = 0.978 Frequency of homozygous healthy individuals: F(AA) = p^2 = 0.978^2 = 0.956

In cats, all-white color is dominant over colors other than all white. In a population of 100 cats, 19 are all white. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of the all-white allele in this population?

Genotypes: AA, Aa 19 are all white Other : 100 - 19 = 81 Assuming a = q and A = p Other colors : 81/100 = 0.81 Sq.rt. Of 0.81 = 0.9 P = 1 - 0.9 = 0.1

Dimples are a dominant phenotype, determined by one gene with 2 alleles (D,d). Among 400 American genetics students, 96 had dimples. What are the frequencies of the 2 alleles in this population, assuming the HWE?

Genotypes: DD, Dd 96 students with dimples 304 students without dimples Assuming d = q and D = p W/o dimples: 304/400 = 0.76 Sq.rt. Of 0.76 = 0.872 P = 1 - 0.872 = 0.128

DNA typing is used to compare evidence DNA (E) left at a crime scene to 2 suspects (S1 and S2). S1 is excluded by the evidence, but S2 remains included. What is the frequency of S2's genotype if the allelic frequencies in the population are f(A1) = p = 0.1, f(A2) = q = 0.2 and f(A3) = r = 0.7, and the population is at HWE?

HWE with 3 alleles: (p + q + r)^2 + 2pq + 2pr + 2qr S2's genotype is A1A2 or 2pq 2 x 0.1 x 0.2 = 0.04

How CF is passed on

If both parents are carriers, three outcomes can result. 25% not carrier; 50% carrier; 25% affected Must be homozygous for recessive allele

Equilibrium disturbed by migration

Introduces new alleles

Equilibrium disturbed by mutation

Introduces new alleles

Consider the following M-N blood group allele frequencies: M = p = 0.6 N = q = 0.4 Complete the phenotypic frequencies: f(MM) f(MN) f(NN)

MM = P = p^2 = (0.6)^2 = 0.36 MN = H = 2pq = 2 x 0.6 x 0.4 = 0.48 NN = Q = q^2 = (0.4)^2 = 0.16

Only way new alleles are created for a species

Mutation

Calculating allele frequency

Number of copies of a particular allele present in a sample divided by total number of alleles

Calculating Genotype Frequency

Number of individuals possessing the genotype divided by total number of individuals in sample

Stabilizing selection is also called

Overdominance/ heterozygote advantage

Hardy-Weinberg Law

Population is large, randomly mating, not affected by mutation, migration, or natural selection Allele frequencies do not change The genotype frequencies stabilize

Critical values

Probabilities

Suppose that the avg. number of viable offspring produced by 3 genotypes is: A1A1. A1A2. A2A2 Mean number of offspring: 10. 5. 2

Relative fitness: W11 = 10/10 = 1.0 W12 = 5/10 = 0.5. W22 = 2/10 - 0.2 s = (1-W). s11 = 0. s12 = 0.5. s22 = 0.8

selection coefficient

Relative intensity of selection against a genotype

Mutation in CFTR gene prevents intestinal infection by

Salmonella typhi

Effects of new mutation

Slow and gradual

absolute fitness

The average reproductive rate of individuals with he same genotype

If the X^2 value is greater than the critical value...

Then the null hypothesis has been rejected and a a significant deviation from predicted values was observed

X^2 statistic used in genetics

To illustrate if there are deviations from the expected outcomes of the alleles in a population

A particular human population has 500 MM, 300MN, and 700NN individuals. Calculate allergic frequencies and determine if the population is in Hardy-Weinberg equilibrium.

Total number of individuals: 1500 Genotype freq.: MM: 500/1500 = 0.333 MN: 300/1500 = 0.2 NN: 700/1500 = 0.467 Allele freq.: M: 0.333 + 0.2 / 2 = 0.433 N: 0.467 + 0.2 / 2 = 0.567 Expected number: p^2 + 2pq + q^2 MM: 0.433^2 x 1500 = 281 MN: 0.433 x 0.567 x 2 x 1500 = 737 NN: 0.567^2 x 1500 = 482 Chi-square test: (O - E)^2 / E MM: (500 - 281) ^2 / 281 = 170 MN: (300 - 737)^2 / 737 = 259 NN: (700 - 482)^2 / 482 = 99 + ——————————————————- 528 = chi-square value P < 0.05 = population is NOT in HWE

The following vole genotypes were found where Te and Tf represent different alleles. TeTe : 407 TeTf : 170 TfTf : 17 -> calculate the genotype and allele frequencies for this population

Total number of individuals: 594 Genotype freq.: TeTe : 407 / 594 = 0.685 TeTf : 70 / 594 = 0.286 TfTf : 17 / 594 = 0.029 = 1.0 Allele freq.: Te : 0.685 + 0.286 /2 = 0.828 Tf : 0.286 + 0.029 /2 = 0.172 = 1.0

disruptive selection

natural selection in which individuals at the upper and lower ends of the curve have higher fitness than individuals near the middle of the curve - selection against the mean

Consider an autosomal locus (A), with 2 alleles A1 and A2. If p = frequency of A1 allele and q = frequency of A2 allele and if there are only 2 alleles in this population then p+q = ?

p + q = 1 and q = 1 - p

In a Chi-square analysis, what condition causes one to reject (fail to accept) the null hypothesis?

p < 0.05

Hardy-Weinberg equilibrium

p^2 + 2pq + q^2 = 1.0

stabilizing selection

removes extreme variants from the population and preserves intermediate types - selection against both extremes

selection coefficient equation

s = 1 - w