Lecture 17 Flashcards Preview

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Flashcards in Lecture 17 Deck (16):
1

Species abundance spectrum:

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

2

Extinction vortex:

- 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

3

Small populations lead to:

- Inbreeding
- Genetic drift

4

Inbreeding and genetic drift leads to:

- A loss of genetic variability

5

A loss of genetic variability leads to:

- A reduction in individual fitness and population adaptability

6

Reduction in individual fitness and population adaptability:

- Lower reproduction
- High mortality

7

Inbreeding:

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

8

Inbreeding coefficient:

- 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

9

Inbreeding depression:

- Reduction of fitness due to inbreeding
- Measured through deformaties, fertility, survival, body weights and litter size

10

Alleles identical by descent:

- 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.

11

Alleles identical by state:

- 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.

12

Inbreeding coefficient:

- 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

13

Generalised equation to calculate F:

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)

14

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

- 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

15

Different ways of look at n:

- Counting gametic transmission (loops)
- Counting the number of ancestors (include the inbred individual)

16

What about a sex-linked gene?

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