Remember for a di hybrid cross with two completely heterozygous parents, the phenotypic ratio will always be 9:3:3:1
9 will always display both dominant traits, black and rough hair. 3 will display black hair and smooth hair and 3 will display not black hair and rough hair and 1 will display both receive traits
The correct answer is:
A. begin with the breakdown of glucose in glycolysis.
Explanation:
They both begin with a sequence of reactions known as glycolysis, which breaks glucose particles into smaller pyruvate molecules. They are also related in that through both processes, ATP is generated for the cell to use. Glycolysis is the metabolic pathway that transforms glucose C6H12O6, into pyruvate, CH3COCOO− + H+. The free energy delivered in this process is applied to form the high-energy molecules ATP and NADH .
The physiological process occurring in the muscle cells that account for the gradual onset of muscle fatigue is called anaerobic the body is breaking down oxygen after than it can get in.
Answer:
<h2>Recombination rate or recombination frequency between locus A and locus C is either 2% or 50%.
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Explanation:
As given,
Recombination frequency between locus A and B is 35%, means that the distance between A and B is 35 cM.
Recombination frequency between locus B and C is 33% , means the distance between B and C is 33 cM.
Therefore, if the locus C is present between A and B, then the recombination frequency between A and C would be 2%
A-----C ----------B
A---C= 2%
A----B= 35%
B---C= 33%
So, here the recombination frequency between A and C is 2%, means their distance is 2 cM.
If the locus C is as, A---35---B--33----C ; A-----B------C
A-----B= 35%
B------C= 33%
A-----C= 33+35= 68%
As rule, maximum frequency can not exceed 50%, as here the Recombination frequency between A and C = 50%.