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
The nitrogenous bases in DNA are adenine, guanine, thymine, and cytosine.
The nitrogenous bases in RNA are adenine, guanine, uracil, and cytosine.
So the base in RNA that is different than in DNA is uracil.
Hope this helps.
Ecosystem. it is defined as a biological community of interacting organism and their physical environment.
Mitotic spindles are microtubule-based structures that separate chromosomes during mitosis. So the cell would lack structure.
Answer:
Option-D, a segment of DNA
Explanation:
A gene represents the particular segment of DNA which contains the instruction for the trait of an organism. The nucleotide sequence present in the segment provides the instruction in the form of codons.
The gene is present in the form of its alternative form called alleles which are located on the chromosomes at locus. The alleles represent the variants of the gene and therefore controls the two variety of a single trait.
Thus, Option-D is correct.
