Answer:
A. VG = 80
B. Broad sense heritability = 0.80
C. Narrow sense heritability = 0.30
D. Average yield = 430 Units
Explanation:
A. Given that
VA = 30
VD = 50
VI = assumed not available
Therefore
Total genetic variance (VG) = VA + VD
= 30 + 50
= 80
VG = 80
B. Given that
VP = 100
VG = 80
Broad sense heritability, H2 = VG/VP
= 80/100
= 0.80
C. Given that
VA = 30
VP = 100
Narrow sense heritability, h2 = VA/VP
=30/100
= 0.30
D. The difference in selection = 500 - 400
= 100
Recall,
Selection response is heritability multiplied by selection differential.
That is
R = h2S
Selection differential = 100
Heritability h2 = 0.30
Selection response = 0.30 × 100
= 30units
Therefore, expected average yield = 400 + 30
= 430 Units
Enterocytes is the answer.
QUESTION ONE
Here are the answers:
1. The purpose of zoo described in question 1 is EDUCATIONAL: it provides knowledge about animals' lives.
2. CONSERVATION: It protects the animals from becoming extinct.
3. RESEARCH: It is a source of animal specimens for research studies.
4.ENTERTAINMENT: people go to zoos to relax and to have fun.
QUESTION 2
The statement is TRUE.
Zoo has a lot to offer in term of educating the public and that includes the topic of biodiversity. Biodiversity refers to the variety of living organisms in the world and the relationships that exist among them. Those animals have the way they relate with other organisms. Thus, the topic of biodiversity can be treated using zoo.
