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
Answer:
cool
Explanation:
just cool that's good for you.
Grasses are the producers. Gazelles, warthogs, and wildebeests are primary consumers. Lions and cheetahs would be the secondary consumers.
This is called Biomagnification - concentration of toxins in an organism as a result of its ingesting other plants or animals in which the toxins are more widely disbursed.
Answer:
C. His mother carried at least one allele for that trait
Explanation:
-Men have a single allele of each gene on the X chromosome, inherited from their mother, and a single allele of each gene on the Y chromosome, from their father.
-A sex-linked trait is a trait that is controlled by a gene or an allele located on the sex chromosome.
-X-linked traits are maternally inherited from carrier mothers or from an affected father. Each son born to a carrier mother has a 50% probability of inheriting the X-chromosome carrying the mutant allele.
