You are studying the evolution of tusk length in a populatio

You are studying the evolution of tusk length in a population of African elephants. You make the tusk-length measurements on 4 different families for whom the adults were captured in nature and brought to a zoo where their offspring were born and reared. These values are in the table below. What is the heritability of tusk length in these animals?

Family

Mother

Father

Son

Daughter

1

2m

6m

2m

2m

2

9m

11m

6m

4m

3

4m

8m

4m

2m

4

6m

10m

5m

3m

In this same population of elephants, the offspring of 4 other breeding pairs were orphaned when poachers killed their parents. The parents’ tusks were recovered and the offspring were ‘adopted’ by other families in the same population. You take measurements of (dead) parents and offspring after they are mature, and calculate that h² for tusk length is 0.85.

Assume both studies have measured animals that are representative of the population. Give a reasonable explanation for why the h² values calculated in the two studies differ even though they are taken from the same population.

Family	Mother	Father	Son	Daughter
1	2m	6m	2m	2m
2	9m	11m	6m	4m
3	4m	8m	4m	2m
4	6m	10m	5m	3m

Solution

Heritability depends upon two factors genetic variation (V_G) and total phenotypic variation (V_P). First, we have to calculate variance s to find the genetic and phenotypic variations.

Mother

Father

Son

Daughter

2

6

2

2

9

11

6

4

4

8

4

2

6

10

5

3

Xi=

21

35

17

11

Mean X =Xi/n

5.25

8.75

4.25

2.75

[Xi-X]²

[21 – 5.25]² =15.75² =90.56

689.02

162.56

68.06

Variance = (Xi-X)²/ n-1

S²=30.18

S= 5.5

S= 15.15

7.36

4.76

S is standard deviation

Mean X =Xi/n = where Xi is individual tusk length value and n is total individuals

To count Variance we calculate s²= (Xi-X)²/ (n-1)

Genetic variation (V_G) = V_P – Ve, where Ve is environmental variance

To calculate environmental variance Ve = (standard deviation of mother + father)/2 =[ 5.5 + 15.15]/2 =10.32

Phenotypic variance= (Variance of son + Daughter)/2 = [7.36+4.76]/2= 6.06

Vg= Vp-Ve = 6.06-10.32 = -4.26

H² = Vg/Vp = 4.26/6.06= 0.7 [ignore minus mark]

Therefore, the values of Heritability differ. Individuals within a population have different genetic variations even they live in same environment.

	Mother	Father	Son	Daughter
	2	6	2	2
	9	11	6	4
	4	8	4	2
	6	10	5	3
Xi=	21	35	17	11
Mean X =Xi/n	5.25	8.75	4.25	2.75
[Xi-X]2	[21 – 5.25]2 =15.752 =90.56	689.02	162.56	68.06
Variance = (Xi-X)2/ n-1	S2=30.18 S= 5.5	S= 15.15	7.36	4.76