Population Genetics

Question 1

What is a VNTR? Diseases associated with VNTRs?

Answer 1

A variable number tandem repeat (or VNTR) is a location in a genome where a short nucleotide sequence is organized as a tandem repeat. Fragile X-Syndrome.

Question 2

4 differences to measure genetic diversity:

Answer 2

Nucleotide state difference (sequencing)
Length difference (microsatellites)
Functional difference (ABO blood group)
Electrophoretic difference (allozyme)

Question 3

Hardy–Weinberg Equilibrium

Answer 3

p + q =1

p² + q² + 2pq = 1

Question 4

Calculate frequency of an allele?

Answer 4

AA Aa aa Calculate frequency of A
5 50 45

((5x2)+50)/200
=(10+60)/200
=70/200
=0.35

Question 5

General form of F statistics equation and what are F statistics?

Answer 5

F = 1 - (Ho/He)
Where Ho = Observed heterozygosity
and He = Expected heterozygosity
F statistics describe the statistically expected level of heterozygosity in a population.

Question 6

Fis

Answer 6

Departure from HW within a deme .
Used as a measure of inbreeding. 0 is no inbreeding.
Fis = 1 - (Ho/He)

Question 7

Fst

Answer 7

Departure from HW between demes.
Measures population structure, where 0 is unstructured and panmitic (randomly mating) and 1 is complete isolation and structure.
Fst = 1 - (Hs/Ht)
Fst = 1 - (seperated groups/ one group (i.e. panmitic))

Question 8

% of selfing equation

Answer 8

S= (2*Fis)/(1+Fis) = % of offspring arisen from selfing

Question 9

Inbreeding

Answer 9

Causes reduced fitness
Increased homozygosity and reduced heterozygosity
Can determine inbreeding by taking the shortest path (N) from one parent of an individual to the second.
F = (1/2)^N x (1+Fca) , where Fca is the inbreeding coefficient of the common ancestor. Fca is 0 if no inbreeding for common ancestor.

Question 10

Autozygous

Answer 10

Identical by descent

Question 11

Allozygous

Answer 11

Identical by state (i.e. not directly related, but same genetic info)

Question 12

Demes

Answer 12

Group of individuals that belong to the same taxonomic group

Question 13

Measuring gene flow:

Answer 13

1. Calculation of Nm from Fst
   - Nm is the average number of migrants per generations
   - Lower Fst (less structure) means higher gene flow
2. Private alleles
   - Alleles that only occur in one population
   - More private alleles = less gene flow
3. Direct estimates
   - Directly observing migrants moving between populations via marking after birth or the spread of rare alleles etc.

Question 14

General effects of gene flow:

Answer 14

Causes a reduction in genetic differences between populations.
Causes an increase of genetic variation within population.

Question 15

Know how to genotype a gel!

Answer 15

I.e. if bands appear in both A and a regions then heterozygote, if only in A or a then homozygote of the respective allele.

Question 16

Purging

Answer 16

Natural removing “unfit” individuals and deleterious alleles from a population e.g. in mice.
Leads to a reduction in inbreeding depressions in small populations.

Question 17

Heterosis

Answer 17

Heterozygote advantage, commonly used in plant breeding

Question 18

Calculate relative fitness:

Answer 18

Selective advantage = Fittest (W) /2nd Fittest (W)
Obs Exp Obs:Exp W(Relative Fitness) Sel. Adv.
AA 29 185 0.1567 0.15/1.15 = 0.13
AB 2993 2601 1.15 1.15/1.15 = 1 1.14
BB 9365 9525 0.983 0.98/1.15 = 0.85

Question 19

Muller’s Ratchet

Answer 19

A process in which absence of recombination, especially in an asexual population, results in accumulation of deleterious mutations, irreversibly.
These alleles are compensated for by beneficial ones.

Question 20

Populations have a balance between mutation and BLANK and mutation and BLANK

Answer 20

Balance between mutation and drift (production of alleles vs loss of alleles)
Balance between mutation and selection (production of alleles vs purging of alleles)
Note, positive mutations are rare, a small amount is sufficient to sustain a population.

Question 21

Lethal Equivalent

Answer 21

A combination of selective effects that on average have the same impact on the composition of the gene pool as one death; for example, two carriers at 50% risk of dying would be the lethal equivalent of one carrier at 100% risk;

Question 22

Mutation-Selection Equilibrium Equations

Answer 22

Δqmut = μp :Rate of p mutating to q
μ = mutation rate, p = 1-q, q = freq. of deleterious allele a
Δqsel = (-spq²) / (1-sq²) : rate of change due to selection
S = selection coefficient

Question 23

3 Types of selection

Answer 23

1. Directional
2. Stabilsing
3. Diversifying/Disruptive

Question 24

Equation when rate of mutation is equal to rate of selection:

Answer 24

q = √(μ/s)

Gives the mutation rate that offsets loss of alleles to keep it in equilibrium

Question 25

Linkage disequilibrium

Answer 25

The non-random association of alleles at two or more loci in a general population. e.g. Disequilibrium = A only paired with B and a only paired with b D = Pab - Pa*Pb Pab = Observed haplotype frequency ( Haplotype: set of genetic determinants located on a single chromosome that are usually inherited together.) With time, linkage disequilibrium (D) decays towards linkage equilibrium (0) Pa and Pb = Observed allele frequency

Question 26

Absolute Fitness

Answer 26

The change in a single genotype in abundance over one generation. n(t+1) = Wn(t) n = abundance of a genotype and time "t" and "t+1" W = Absolute fitness

Question 27

Examples of genetic markers:

Answer 27

SNP's, allozymes, Microsatellites, DNA sequencing | Genetic markers used to track inheritance.

Question 28

Homoplasy

Answer 28

A trait/sequence shared by a set of species that is not present in their common ancestor. I.e. obtained from different paths.

Question 29

4 effects of genetic drift on small populations

Answer 29

1. Allele frequency's fluctuate at random, some disappear, some fluctuate 2. Reduces genetic variation of the population 3. Frequency of harmful alleles can increase 4. Differences between populations can increase

Question 30

Expected loss of heterozygosity after "t" generations

Answer 30

ft = 1 - (1-(1/2N)^t Note: Initial frequency of a new mutant = 1/2N Loss of mutant = 1 - (1/2N)

Question 31

Effective Population size

Answer 31

Ne =(4𝑁𝑓*𝑁𝑚)/(𝑁𝑚+𝑁𝑓)

Question 32

Neutral Evolution and the Molecular Clock

Answer 32

``` t = d/2k d = neutral substitutions k = rate of accumulation of d t = generation time ``` Molecular clock = use of mutation rate to determine the time when two or more life forms diverged. 𝑘=2𝑁𝜇 x 1/2𝑁=𝜇

Question 33

Examples of Directional Selection

Answer 33

1. Size of black bears in Europe - Size decreased in interglacial periods - Size increased in glacial periods 2. Beak size of finches - Wet years -> More smaller seeds, rarely eat larger seeds - Dry years -> Less smaller seeds, eat more larger seeds - Beaks specialised to eat specific seed types 3. Sockeye salmon migration - To reproduce salmon migrate to the rivers in which they were born - Directional selection is causing salmon to arrive earlier (i.e have an earlier migration timing)

Question 34

Examples of Stabilising Selection

Answer 34

1. Plant height - Plant too small -> out-competed for sunlight - Plant too tall -> susceptible to wind damage - Stabilises plant size at a medium height 2. Bird clutch size - Lay too many eggs -> not enough resources -> death - Too little eggs -> minimal offspring production and survival - So the birds produce a moderate number of eggs to maximise offspring production and survival.

Question 35

Examples of Diversifying/Disruptive Selection

Answer 35

1. Rabbit fur - BB = Black fur, Bb = Grey fur, bb = white fur, incomplete dominance - Grey rabbits not camouflaged by either white or black environments and so are more susceptible to predation. This is an example of heterozygote disadvantage. 2. Peppered moths in London - BB = Black moth , Bb = Grey moth, bb = white moth incomplete dominance - Black moth camouflaged in industrial buildings - White moths camouflaged outside - Grey moths are seen in both environments - Heterozygote disadvantage