Mid-term

Question 1

4 Primary genetic factors in the extinction of small populations are

Answer 1

1) loss of genetic diversity
2) inbreeding depression
3) mutational meltdown
4) hybridisation

Question 2

Loss of genetic diversity

Answer 2

Loss of diversity (i.e. heterozygosity) due to drift is greatest in small populations

But loss of heterozygosity is also affected by population growth rate and the presence of overlapping generations

Question 3

Loss of heterozygosity each generation equation

Answer 3

Ht+1 = Ht[1 - (1/2Ne)]

Question 4

Loss of diversity is common in threatened populations

Answer 4

Threatened species have significantly lower genetic diveristy (Ht) than non-threatened species (Hnt), across all taxa

e.g. Kakapo

Question 5

Loss of alleles

Answer 5

• although moderate population size reduction may not have a large effect on heterozygosity, it will still have a large effect on allelic diversity because of genetic drift will rapidly eliminate low frequency alleles
• long term response to selection depends on allelic diversity
• short-term persistence may be affected if loss of alleles at major histocompatibility complex loci, due to increased susceptibility to disease

Question 6

Inbreeding Depression

Answer 6

• inbreeding depression is a decrease in fitness due to mating between related individuals
• results from the presence of partially recessive deleterious mutations maintained by selection - mutation balance
• mating between relatives increases homozygosity and therefore the deleterious effects become fully expressed and fitness of inbred individuals decreases
• increased homozygosity may also reduce the opportunity for expression of over-dominance which will also lead to reduced fitness

Question 7

Mutational meltdown

Answer 7

• mutational meltdown is the decline in reproductive rate and downward spiral towards extinction due to chance fixation of new mildly deleterious mutations in small populations i.e. it is the accumulation of mildly deleterious mutations due to genetic drift
• two phases: 1) fitness drops but populations are still able to more than replace themselves 2) fitness is less than required for replacement and the population declines towards extinction. Mutational meltdown strictly refers to the second phase

Question 8

Hybridisation

Answer 8

Abthropogenic hybridisation (i.e. resulting from human activities) can result in extinction where hybridisation is extensive and 'pure' individuals no longer exist 
- complete admixture leads to extinction

Question 9

Two common measures of genetic diversity

Answer 9

1. Allelic diversity
   - average number of alleles (A = total alleles / # loci)
   - effective number of alleles (ne = 1/∑pi2)
2. Expected heterozygosity ( expected under Hardy-Weinberg equilibrium). Expected heterozygosity reported rather than observed because it is less sensitive to sample size.
   He =1-∑pi2

Question 10

Hardy-Weinberg equilibrium

Answer 10

Uses in conservation genetics:
1. Null model to test real populations against. Describes genotype frequencies when there is no evolution (i.e. no mutation, no migration, no selection, no genetic drift) and is randomly mating with no population subdivision.
2. Estimation of recessive allele frequencies.
3. Estimation of frequency of carriers (heterozygotes).
Hardy-Weinberg Equilibrium (HWE) for single locus with only 2 alleles:
p2 +2pq+q2 =1.

Question 11

HWE & Sex-linked loci

Answer 11

• Homogametic sex (e.g. XX in mammals) expected genotype frequencies = same as for autosomal loci (p2 + 2pq + q2 for 2 alleles).
• Heterogametic sex (e.g. XY in mammals) expected genotype frequencies for locus on X = p + q (i.e. the allele frequencies).

Question 12

HWE & Polyploids

Answer 12

Polyploids – more than 2 copies of each chromosome (common in plants).
e.g. tetraploid – 4 copies of each chromosome (4n).

Question 13

Deviations from HWE

Answer 13

Usually seen as either:

• Deficit of heterozygotes
  1. Non-random mating (inbreeding / assortative mating)
  2. Selection against heterozygotes
  3. Population subdivision
  4. Null alleles
• Excess of heterozygotes
  1. Selection for heterozygotes
  2. Non-random mating (disassortative mating)

Question 14

Effective inbreeding coefficient

Answer 14

Effective inbreeding coefficient (Fe) can be calculated from deviation from HWE:
Fe = 1 – (Hobs/Hexp) Inbreeding can be compared:
1. Within a population at one time (i.e. to assess non-random mating).
2. Between 2 populations (compare founder pop to source): Fe = 1 – (Hobs founder/Hobs source)
3. Over time (compare population today to same population several generations ago). Fe = 1 – (Hobs time 2/Hobs time 1)