Answer:
The correct answer would be C) 75% large-toothed and 25% small-toothed.
It can be explained with the help of the law of dominance which states that the dominant allele expresses itself completely over the recessive allele in the heterozygous condition.
So, the genotype TT, as well as Tt, will result in the production of dominant trait or character in the offspring.
Thus, three (1 TT and 2 Tt) out of four offspring would have large teeth and only one offspring (tt) would have small teeth.
Hence, we can conclude that 75% of the offspring will have large teeth and only 25% will have small teeth.
(8.314 J/molK)(310K)ln(3E5) = E-EcatE-Ecat
<span>=32504.22 J/mol = 32.504 KJ/mol (pay attention to any requirement on sig fig)
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Answer:
When the patch occupancy rate (c) equals the patch extinction rate (e), patch occupancy (P) is 0
Explanation:
According to Levin's model (1969):
<em>dP/dt = c - e</em>
where P represents the proportion of occupied patches.
<em>c</em><em> </em>and <em>e </em>are the local immigration and extinction probabilities per patch.
Thus, the rate of change of P, written as dP/dt, tells you whether P will increase, decrease or stay the same:
- if dP/dt >0, then P is increasing with time
- if dP/dt <0, then P is decreasing with time
- if dP/dt = 0, then P is remaining the same with time.
The rate dP/dt is calculated by the difference between colonization or occupancy rate (<em>c</em>) and extinction rate (<em>e</em>).
c is then calculated as the number of successful colonizations of unoccupied patches as a proportion of all available patches, while e is the proportion of patches becoming empty. Notice that P can range between 0 and 1.
As a result, if the patch occupancy rate (c) equals the patch extinction rate (e), then patch occupancy P equals to 0.
The carry the deoxygenated blood from the heart to the lungs and Pulmonary veins do the opposite!
