Reductions of chromosome numbered
The correct answer should be something like: "Gene therapy involves the use of bacteria or viruses to deliver new or modified genes to cells". That is what can be concluded based on the paragraph you provided.
Answer:
c. Two sister chromatids did not separate into the proper daughter cells during anaphase.
Explanation:
The observed cell is undergoing mitosis which does not include separation of homologous chromosomes. During anaphase of mitosis, two sister chromatids of each chromosome separate from each other. They move to opposite poles. This results in equal distribution of two complete sets of chromosomes to each daughter cell.
However, the failure of two sister chromatids of a chromosome during anaphase would lead to the formation of two abnormal daughter cells. One of the daughter cells would have one extra chromosome (2n+1) while the other would lack one chromosome from the diploid set (2n-1).
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:
The cell structure has specific functions that are essential to carry out life's processes. These components include the cell wall, cell membrane, cytoplasm, nucleus, and cell organelles.
Explanation: