Module 13-15 Alteration of Genetic Equilibrium

Question 1

These are processes that cause a change in gene frequency magnitude and degree.

Answer 1

Systematic processes (Ex. Mutation, selection and migration)

Question 2

These are processes that change the magnitude but not the direction

Answer 2

Dispersive Processes (Genetic Drift and nonpanmictic/nonrandom mating)

Question 3

This is the ultimate source of genetic variation inclusive of all genetic changes.

Answer 3

Mutation

Question 4

T/F Mutation occurs at a generally very low rate per generation.

Answer 4

T, 10^-5 or 10^-6 per generation in most loci of most organisms

Question 5

Two types of mutation and differentiate

Answer 5

Non-recurrent: Rare, small chance of survival

Recurrent: Consequence of altering gene frequencies

Question 6

If p0 is the initial frequency of gene A and it mutates continuously to a at a rate, and the reverse mutation does not occur, A after n generations becomes pn. How is pn calculated?

Answer 6

pn=po*(1-u)^n

where:
pn = freq of A after n generations
po = initial freq of A
u = rate of mutation
n = number of generations

Question 7

Given initial A and a frequencies p and q, how will you calculate new frequencies with forward and backward mutations after 1 generation?

Answer 7

freq (A) = p1 = p + (vq) - (up)
freq (a) = q1 = q + (up) - (vq)

where:
u is the forwards rate
v is the reverse rate

Question 8

How do you calculate for p and q at mutational equilibrium?

Answer 8

At mutational equilibrium, up=vq therefore

p = v/(u+v) and q = u/(u+v)

Question 9

Using mutational equilibrium, how do you calculate the value of q after any specified generation time with both forward and backward mutation?

Answer 9

(u+v)n=ln((q-qe)/(qn-qe))

where:
q = initial q value
qe = q freq at mutational equilibrium
qn = q after n generations

Algebra go brrrr

Question 10

Refers to fluctuations in allele frequency that occur by chance, particularly in small populations, as a result of random sampling among gametes.

Answer 10

Random Genetic Drift

Question 11

Two Causes of Random Genetic Drift

Answer 11

Mendelian Segregation

Finite population size

Question 12

How to calculate variance in allelic frequency among populations under random genetic drift? What about sampling error?

Answer 12

(sp)^2 = (pq)/2N

where:
N = total population size
p and q = allelic freqs of p and q
(sp)^2 = Variance

To get sampling error/standard dev, get square root of Variance.

Question 13

The number of breeding individuals in an idealized population that would show the same amount of dispersion of allele frequencies under random genetic drift or the same amount of inbreeding as the population under consideration

Answer 13

Effective Population Size

Question 14

Causes of genetic drift (3)

Answer 14

Population size reduction by environmental factors

Founder Effect

Genetic Bottleneck

Question 15

Occurs when a population is established by a small number of individuals resulting in significantly less genetic diversity

Answer 15

Founder Effect

Question 16

Occurs when a population undergoes a drastic reduction in population size. Emphasis on drastic

Answer 16

Genetic Bottleneck

Question 17

Four Main Aspects of Random Genetic Drift

Answer 17

▪direction is unpredictable

▪magnitude depends on population size

▪long-term effect is to reduce genetic variation within a population

▪causes populations to diverge

Question 18

Composite of the forces that limit the reproductive success of a genotype

Answer 18

Selection

Question 19

Comparative ability of a genotype to withstand selection

Answer 19

Fitness

Question 20

The extent to which a genotype contributes to the offspring of the next generation relative to the other genotypes in a given environment

Answer 20

Adaptive Value (W)

Also known as fitness or Selective Value

Question 21

This is the percentage reduction in fitness.

Answer 21

Selection Coefficient (s)

Question 22

What is the relationship between Adaptive Value (W) and selection coefficient (s)?

Answer 22

W=1-s or s=1-W

Question 23

In calculating the frequency of genotypes before and after selection, assuming Hardy-Weinberg proportions before selection, what are the steps?

Answer 23

1. Get frequency before selection p^2, 2pq, and q^2
2. Multiply by their respective selection coefficients to get their weighted contributions
3. Sum all weighted contributions to get the weighted total
4. Divide the weighted contributions by the weighted total to get the frequency after selection

Question 24

Calculation of Change in Frequency (Delta q) of Gene a when |s|= 1 (homozygous recessive is lethal) for Homozygous Recessive Genotype

Answer 24

Delta q = -q^2/(1+q)

Applicable when aa is lethal

Question 25

Calculation of Change in Frequency of a (Delta q) for a Gene with Varying Degrees of Deleterious Effects (0<s<1)

Answer 25

Delta q = ((-sq^2)(1-q))/(1-(q^2 * s))

Applicable when 0<s<1

For summarized formulas, just check slide 11 from module 14

Question 26

Calculation of Change in Frequency of A (Delta p) for a Gene with Varying Degrees of Deleterious Effects on the Dominant Allele. Let t be the selection coefficient for p

Answer 26

Delta p = (-tp(1-p)^2)/(1-tp(2-p))

For summarized formulas, just check slide 11 from module 14

Question 27

Calculation in Change in Frequency of Gene a (Delta q) in One Generation of Selection in which the Heterozygotes have Adaptive Value of 1 and the Homozygous Selected Against by t and s, respectively.

Answer 27

Delta q = (pq(pt-qs))/(1-(tp^2)-(sq^2)

For summarized formulas, just check slide 11 from module 14

Question 28

How to calculate for change in q (Delta q) with complete dominance (selection against a)?

Answer 28

Delta q = ((-sq^2)(1-q))/(1-sq^2)

Slide 15 on Module 14

Question 29

How to calculate for change in q (Delta q) with complete dominance (selection favoring a)?

Answer 29

Delta q = ((sq^2)(1-q))/((1-s) (1-q^2))

Slide 16 on Module 14

Question 30

Let A1 and A2 donate two alleles for 1 gene. How do you calculate for change in q (Delta q) with no dominance (selection against A2).

Answer 30

Delta q = (0.5*sq(1-q))/(1-sq)
Slide 17 on Module 14

Question 31

Let A1 and A2 donate two alleles for 1 gene. How do you calculate for change in q (Delta q) with no dominance (selection favoring A2).

Answer 31

Delta q = (0.5*sq(1-q))/(1-s+qs)
Slide 17 on Module 14

Question 32

How do you calculate for the allele frequency of q after n amount of generations assuming s=1 and there is no mutation?

Answer 32

qt = q/(1+tq)

Where:
q = initial q frequency
t = number of generations to change q to qt
qt = you, jk, its the final q allele frequency

Note: Only use when s=1 (Selection is against homozygous recessive, no mutation)

Question 33

At equilibrium the change due to mutation will be equal and opposite to the change due to selection (Delta q mutation = - Delta q selection)

At mutation/selection equilibrium, q can be calculated how?

Answer 33

qe=sqrt(u/s)

Where:
u = mutation rate
s = selection coefficient.

Question 34

What are the three types of selection and what do they select against/for

Answer 34

Stabilizing Selection - Selects against extremes, selects for intermediate phenotype

Directional Selection - Selects for a particular genotype

Disruptive Selection - Selects against the intermediate phenotype, selects for the extremes.

Question 35

What are the implications of stabilizing selection

Answer 35

Increased fitness of intermediate phenotype favors adaptation to existing environmental conditions but reduces phenotypic and genotypic diversity

Question 36

What are the implications of Disruptive selection

Answer 36

Increased fitness of extreme phenotypes results in higher phenotypic and genetic diversity resulting in a population with a bimodel distribution

Question 37

Can be thought of as two special kinds of natural selection, competition and preference

Answer 37

Sexual Selection

Question 38

Offers explanations for the evolution of large human brains and behaviors such as humor, music, and poetry that do not have obvious survival value

Answer 38

“mating-mind” hypothesis

Question 39

Technical term for altruistic behavior that has been shown, or is theoretically supposed, to be explained by kin selection or self sacrifice methods

Answer 39

Kin Altruism

Nepotism or favor towards own kin is needed for kin-selection to occur

Question 40

This is the mathematical explanation of kin selection. What is the name and formula?

Answer 40

Hamilton’s Rule

rB>C

Where:
r = Genetic relatedness of recipient to actor
B = Additional reproductive benefit gained by recipient
C = Reproductive cost to actor

Question 41

Also known as gene flow, this is the influx of genes from other populations leading to decreased genetic divergence and increased genetic variation between and within populations

Answer 41

Migration

Question 42

In a population of 600 natives with a q frequency of q_o=0.2, 400 immigrants with a q frequency of q_m=0.6 were assimilated. What is the frequency of q in the mixed population (q1) and the change in frequency?

Answer 42

Given:
400 immigrants, 600 natives
q_m = 0.6, q_o = 0.2

Required: q1 and Delta q

Solution:
Get m, the proportion of immigrants in the mixed.
Use the formula
q1 = mq_m+(1-m)q_o

m=400/(600+400)=0.4
q1 = 0.4(0.6)+(1-0.4)0.2
q1= 0.36

Delta q = q1-q_o = 0.36-0.2 = 0.16

Question 43

The value of q after n generations of migration is provided by the formula

Answer 43

q_n - q_m = (1-m)^n (q_o - q_m)

Algebra go brrr

Question 44

Percentage of alleles contributed by the migrants (or proportion of gene flow/admixture) can be calculated by this formula:

Answer 44

m = (q_o-q1)/(q_o-q_m)

Question 45

Matings between closely related individuals that leads to an increase in the proportion of homozygotes and a decrease in the proportion of heterozygotes

Answer 45

Inbreeding

Question 46

This is the avoidance of mating between related individuals

Answer 46

Outcrossing

Question 47

Result from increased homozygosity for heterozygous alleles and is associated with reduced fitness and lower survival rates among the offspring

Answer 47

Inbreeding Depression

Question 48

Measures the extent of inbreeding occurring in a population and the probability that two alleles of a given gene in an individual are derived from a common ancestral allele or are identical by descent.

Answer 48

Inbreeding Coefficient (F)

Question 49

if 2 alleles in an inbred individual are identical by descent, the genotype at the locus is said to be (a) otherwise it is said to be (b).

Answer 49

(a) Autozygous
(b) Allozygous

Question 50

When inbreeding occurs, the heterozygote frequency is _______ at each generation

Answer 50

halved

Question 51

At HWE, how do you solve for the inbreeding coefficient F?

Answer 51

F= 1- (Ho/He)
which can be further expanded to
F=1-(Ho/2pq) = (2pq-Ho)/2pq

Question 52

Genotype frequencies when inbreeding occurs can be calculated by

Answer 52

AA:p^2 (1-F) + pF or p^2 +pqF
Aa:2pq(1-F)
aa:q^2 (1-F) + qF or q^2 +pqF

Question 53

Xeroderma pigmentosum (XP) is an often-fatal skin cancer resulting from a recessive mutant allele that affects DNA repair. In the United States, the frequency of homozygous- recessive affected people is approx. 1 in 250,000.

What is the expected frequency of XP among the offspring of first-cousin matings?

Answer 53

Given: q^2 = 1/250000, First Cousins

Required: f(aa)

Solution:
First Cousins: F=1/16
q^2 = 0.000004->q=0.002
p=1-0.002=0.998

f(aa) = q^2 +pqF = 0.000004 + (0.998)(0.002)(1/16)

f(aa) = 0.000129

Question 54

Xeroderma pigmentosum (XP) is an often-fatal skin cancer resulting from a recessive mutant allele that affects DNA repair. In the United States, the frequency of homozygous- recessive affected people is approximately 1 in 250,000. It was found that frequency of XP among offspring of first-cousin matings was f(aa) = 0.000129

What is the ratio of XP among the offspring of first- cousin matings to that among the offspring of nonrelatives?

Answer 54

Given:
Inbred f(aa) = 0.000129
Outbred f(aa) = 0.000004

Required: inbred/outbred

Solution:
0.000129/0.000004 = 32

Question 55

Mating pattern in which similar phenotypes mate with one another more frequently than what is expected.

Answer 55

Assortative Mating (Homogamy)

A_ x A_ and aa x aa

Question 56

Mating Pattern where in dissimilar phenotypes mate with one another more frequently than what is to be expected

Answer 56

Disassortative Mating (Heterogamy)

A_ x aa