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It depends. It could be 2 am and 14:00 pm if it's on a mechanical clock
Answer:
980,100
Explanation:
let the recessive condition (aa) be represented as q² which is the genotypic frequency = 100/1,000,000 = 0.0001
allelic frequency of a = q = √q² = √0.0001 = 0.01. thus q = 0.01
From the formula p + q = 1 where p is the allelic frequency of A.
since q = 0.01
p = 1 - 0.01 = 0.99. The allelic frequency of A (p) = 0.99
p² = 0.99² = 0.9801. Genotypic frequency of AA= 0.9801
= 0.9801 x 1,000,000 = 980,100 individuals with AA (homozygous normal)
For 2pq genotypic frequency for hetrozygous (Aa).
Using the formula p² + 2pq + q² = 1. Since p² = 0.9801 and q² = 0.0001
2pq = 1 - (p² + q²)
= 1 - (0.9801 + 0.0001)
= 1 - (0.9802)
= 0.0198 = 0.0198 x 1,000,000 = 19,800 individuals with Aa
Answer:
A.Glycogenesis: Glycogen synthase
B. Glucogenesis: Fructose 1,6 biphosphate phosphatase
C. Urea cycle : Carbamoyl phosphate synthetase
D.Fatty acid synthesis: Acetyl CoA carboxylase
E.Glycolysis : Phosphofructokinase 1
F. Pentose phosphate pathway: Glucose-6-phosphate dehydrogenase
Explanation:
A. Glycogen synthase converts glucose into glycogen during glycogenesis.
B. Fructose 1,6 biphosphate phosphatase catalyzes condensation of dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate during glucogenesis.
C. Carbamoyl phosphate synthetase I catalyses production of arbamoyl phosphate during urea cycle.
D. Carboxylase controls fatty acid metabolism.
E. The phosphofructokinase 1 is an important enzyme that regulate formation of two-phosphate sugar molecules during glycolysis.
F. Glucose-6-phosphate dehydrogenase participates in the pentose phosphate pathway. This pathway gives reducing energy to cells.
