Answer:
5 is the best conclusion draw for Africa
Option B: Under anaerobic conditions, cells generate ATP through anaerobic glycolysis and creatine phosphate.
Our body produces ATP through oxidative phosphorylation. ATP is used in various functions and gets hydrolyzed into ADP and inorganic phosphate. But during intense exercises like sprinting, our body becomes unable to produce sufficient ATP through oxidative phosphorylation.
In this condition, creatine phosphate is used to regenerate ATP molecules for a short time. Creatine phosphate, when short of oxygen, transfers high-energy phosphate to ADP. ADP then gets transformed into ATP and produces creatine out of the reaction.
Another mechanism to produce ATP when short of oxygen is through anaerobic glycolysis. In this method, glucose is converted to lactate. This is a faster mechanism that produces 2 molecules of ATP per glucose molecule. The energy produced through oxidative phosphorylation is 100 times slower than anaerobic glycolysis.
To know more about ATP production, refer to the following link:
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Horizontal or transverse or axial cut is if you cutting the object with the horizontal plane. The cut would look like ( -- ).
Vertical or longitudinal is cutting in the vertical plane. The cut would look like ( | )
<span>There is also coronal cuts which were a cut with direction from front to back. </span>
<span>. In light-dependent </span>reactions high energy electrons<span> help transform ADP and NADP into ATP and NADPH. These are then sent to power light-independent </span>reactions<span> that go onto create sugar. </span>
Answer:
linkage with approximately 33 map units between the two gene loci
Explanation:
If two genes are not linked, number of recombinants and parental offspring will be equal. Here it is clearly visible that recombinants are less than parental offspring hence the genes are linked. Given, the offspring are in following numbers:
AaBb = 106 = Parental
aabb = 94 = Parental
Aabb = 48 = Recombinant
aaBb = 52 = Recombinant
Recombination frequency = (Number of recombinants/ Total progeny) * 100 = (100/300) * 100 = 33.33 %
1% recombination frequency= 1 map unit of distance between the two gene loci. So here the distance between the two gene loci is approximately 33 map units.
Hence, these results are consistent with linkage with approximately 33 map units between the two gene loci
