Answer:
Each FADH2 yields about 1.5 ATP via oxidative phosphorylation.
Explanation:
Most of the ATP molecules are produced by oxidative phosphorylation, not by substrate-level phosphorylation. During glycolysis, 2 ATP molecules per glucose are produced by substrate-level phosphorylation. Similarly, Kreb's cycle also yields 2 ATP per glucose by substrate-level phosphorylation.
For each pair of electrons transferred to O2 from FADH2 via electron transport chain, 4 and 2 protons are pumped from matrix towards the intermembrane space by complex III and complex IV respectively. It generates the proton concentration gradient required to drive the synthesis of 1.5 ATP molecules. Since oxidation of FADH2 is coupled to the phosphorylation of ADP to form ATP, the process is called oxidative phosphorylation.
Answer:
The mRNA interacts with a specialized complex called a ribosome, which "reads" the sequence of mRNA bases. Each sequence of three bases, called a codon, usually codes for one particular amino acid. (Amino acids are the building blocks of proteins.)
Explanation:
Answer:
One offspring would have red leaves.
Explanation:
Given
Yellow leaves are dominant and red leaves are recessive
Let us assume that the trait of Yellow color of a leaf be represented by "Y"
and the trait of red color of a leaf be represented by "y"
When two parent of genotype "Yy" are crossed with each other, the following offsprings are produced as shown in the punnet square below
Y y
Y YY Yy
y Yy yy
Total four offpsrings are produced out of which only one has red leave as red color is a recessive trait and it will appear only when both the allele in a cell are of red color.
Thus, one offspring would have red leaves.
Answer:
True
Explanation:
The linkage relation of two genes can be determined on the basis of calculating recombination frequency.
If the recombination frequency value is greater than 50% then the genes are not linked and if the value is less than 50% then the genes are linked.
In the given question, the crossover gametes formed will be Ab = 100
and aB= 106.
Therefore,
Since the recombinant frequency value is larger than 50% therefore the genes do not show linkage.
Thus, True is the correct answer.
