Answer:
With 6 we should multiply gross photosynthesis (as we were measuring gross photosynthesis for 10 min).
Explanation:
According to this question:-
Net photosynthesis- 3mm/20min- 1.5mm/10min
Cellular respiration- 1mm/15min- 0.66mm/10min
Gross photosynthesis = Net photosynthesis + Cellular respiration = 1.5+0.66 = 2.16 (mm/10min)
Rate of photosynthesis = (gross photosynthesis)6 = 2.16 *6 = 12.96 mm/hr
As we gage the photosynthesis rate in hr.
Answer: Survival rate up to start of April = 76.67%
Explanation: size of colony = 150mice.
Size of colony in February =125mice.
Size of colony in march = 115 mice.
1. Survivorship rate in February
= Size of colony / size in February *100.
= 122 / 150 * 100
= 83.33% to the nearest hundredth
2. Survivorship rate in march
= Size of colony / size in march *100.
= 115/150 * 100
= 76.67% to the nearest hundredth.
Answer:
a. DNA polymerase proofreading: consequence of its absence is the DNA mutation
b. Mismatch repair enzymes
: consequence of its absence impedes homologous recombination resulting in the final mutation
c. Nucleotide excision repair enzymes
: the absence of nucleotide cleavage repair enzymes would impede the functioning of damaged DNA repair mechanisms
Explanation:
a. DNA polymerases are the enzymes that form the DNA in cells. During DNA replication (copying), most DNA polymerases can "check their work" with each base they add. This process is called review. If the polymerase detects that you have added a wrong nucleotide (incorrectly paired), remove it and replace it immediately, before continuing with DNA synthesis
b. In homologous recombination, the information from the homologous chromosome that matches that of the damaged one (or from a sister chromatid if the DNA has been copied) is used to repair the fragmentation. In this process the two homologous chromosomes are approached and the undamaged region of the homologue or the chromatide is used as a template to replace the damaged region of the broken chromosome. Homologous recombination is "cleaner" than the union of non-homologous ends and does not usually cause 11 mutations
c. Excision repair: damage to one or a few DNA bases is usually fixed by removing (excising) and replacing the damaged region. In repair by base cleavage, only the damaged base is removed. In nucleotide excision repair, as in the mating repair we saw earlier, a nucleotide section is removed
Inserting foreign genes into fertilized goats eggs
