Lecture 17

Question 1

Species abundance spectrum:

Answer 1

• Extinct, critically endangered, endangered, threatened, abundant, pests (too many!)

Question 2

Extinction vortex:

Answer 2

• Species that go extinct have a common downward spiral
• They have a small population (due to human influence, natural disasters, colonisation events etc) which leads to a smaller population leading to extinction eventually

Question 3

Small populations lead to:

Answer 3

• Inbreeding

- Genetic drift

Question 4

Inbreeding and genetic drift leads to:

Answer 4

• A loss of genetic variability

Question 5

A loss of genetic variability leads to:

Answer 5

• A reduction in individual fitness and population adaptability

Question 6

Reduction in individual fitness and population adaptability:

Answer 6

• Lower reproduction

- High mortality

Question 7

Inbreeding:

Answer 7

• mating between relatives
• Homozygosity increase (deleterious alleles are expressed)
• Heterozygosity decreases (heterozygote advantage decreases)

Question 8

Inbreeding coefficient:

Answer 8

• F
• The probability that an individual receives 2 alleles at a given locus, that are identical by descent
• Copies of a single allele from a common ancestor
• A level of inbreeding

Question 9

Inbreeding depression:

Answer 9

• Reduction of fitness due to inbreeding

- Measured through deformaties, fertility, survival, body weights and litter size

Question 10

Alleles identical by descent:

Answer 10

• Two copies of an allele are descended from the same copy in a common ancestor, so they are identical by descent and identical by state.

Question 11

Alleles identical by state:

Answer 11

• Two copies of an allele that are the same in structure and function, but are descended from two different copies in ancestors, so they are identical by state by not identical by descent.

Question 12

Inbreeding coefficient:

Answer 12

• A: a1a1
• B: a3a4
• C: a1a3
• D: a1a4
• I: a1a1, because A-> C = 1/2,
  C -> I = 1/2 A -> D = 1/2
  D -> I = 1/2
  P of I = a1a1 = 1/2 x 1/2 x 1/2 x 1/2 = 1/16 = 1/4

Question 13

Generalised equation to calculate F:

Answer 13

1. List all common ancestors
2. List all pathways through each common ancestor
3. Count the number of ancestors (any individual can be present in a pathway only once, but can be present in more than one pathway)

Question 14

p sum of i = 1 (1/2) ni (1+ Fai)

Answer 14

• p = the number of possible pathways
• n1 = the number of individuals in the ith pathway (excluding the inbred individual)
• Fai = the inbreeding coefficient of the common ancestor in the ith pathway

Question 15

Different ways of look at n:

Answer 15

• Counting gametic transmission (loops)

- Counting the number of ancestors (include the inbred individual)

Question 16

What about a sex-linked gene?

Answer 16

1. F = 1 for any male (but only one X-chr)
2. Paths connecting 2 consecutive males are excluded (no X-chr transmission)
3. Only females in a (valid) path will be counted