Answer:
5%
Explanation:
Frequency of white tigers = 4%
population of tigers = 200
number of heterozygotes = 10
<u>Determine the actual genotype frequency of Heterozygotes </u>
frequency of heterozygotes = Number of heterozygotes / population
= ( 10 / 200 ) * 100
= 0.05 * 100 = 5%
A Respiration by animals CO2 is being added.
B trees burning same
C tree leaves removed
D oil added
Answer:
a. True, b. False, c.True, d. True
Explanation:
a. Base excision repair is started by a DNA glycosylase that recognizes the changes and removes the altered base by cleavage of the glycosidic bond binding the base and the deoxyribose sugar together.
b. Nucleotide excision repair works by a cut-and patch mechanism that removes their heavy lesions, including pyrimidine dimers and nucleotides . Endonucleases are responsible for the lesion of the damaged strand.
c. Nucleotide excision repair is initiated by the proteins namely UvrA, UvrC, and UvrB in Escherichia coli.
-UvrD (helicase II) later removes the damaged strand
-DNA polymerase I (PolI) fills in the resulting gap.
d. DNA glycolases removes the damaged nitrogenous base.
-It leaves the sugar-phosphate backbone intact and thus creating an apurinic/apyrimidinic site, which is commonly referred to as an AP site.
e. Xeroderma pigmentosum complementation group A(XPA)
-This is an essential protein in the nucleotide excision repair pathway.
- It helps to make a pre-incision complex along with other proteins.
