Answer:
<em>C) Eat locally-produced and organic food; cur meat and dairy consumption.</em>
Explanation:
Eating organic food will be of no help because humans do not usually use forest trees as food resources. And hence, it will be of no help to stop the carbon dioxide emissions. Hence, option C will be of no help.
The consumption of green products will lead to lesser production of carbon dioxide and hence can be helpful. Using alternate energy sources which are not carbon based will also help to reduce the amount of carbon dioxide. If lands are restored then, these trees will not change over time. Hence, option A, B and D are helpful for no substantial change in the tree species over time.
Answer:
G1 checkpoint
Explanation:
The cell cycle consists of the series of activities that occurs from the replication of DNA to the actual division of the cell in an organism's cell. However, the cell has been internally programmed to ensure that all conditions are in place/favorable before allowing progression into the next stage of the cell cycle. This internally controlled mechanism ia called CHECKPOINTS. This checkpoint occurs at three main stages of the cell cycle;
after G1 phase, G2 phase and during M phase.
In the G1 checkpoint, the cell ensures that the cell is in the right shape to proceed into the Synthesis phase of the cell cycle where it will double it's DNA. The checkpoint checks majorly for damage or change to the DNA before replicating it. If any error is found, that cell is halted and prevented from proceeding to the S-phase of the cell cycle.
This is the case in the question, the cell has been halted and prevented from entering the S-phase. Hence, its DNA is still the same as it started. However, the DNA of the cells surrounding it has been doubled as they have undergone DNA replication during S-phase.
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:
Need more details. Like the picture.
Explanation:
Replicate every 40 minutes. calculate the total number of nucleotides in the bacterial chromosome of
