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lesya [120]
3 years ago
6

(e) Predict the effect of the removal of the final electron acceptor on the rate of carbon fixation by the chloroplast. Justify

your prediction
Biology
1 answer:
Vinil7 [7]3 years ago
7 0

Answer:

The ATPs produced during Chemeiosmosis, in the chloroplast is needed during Calvin cycle to fix carbon for production of CHO, protein and fats  during conversion of 3C-glyceride phosphate(GP) to 2C- triose phosphate.(TP) .

The ATPs is needed for this reduction to take place in the presence of NADPH(which is oxdized to NADP).

Therefore, if the final electron acceptor is removed, there will be no electron transport chains which generated the PMF, for pumping H+ across into the thylakoid membrane of the chloroplast, to generate the electrochemical gradients, needed for  the energy for ATPase synthase enzymes,to synthesize ATPs production.

Therefore, the removal of the electron acceptor (oxygen) leads to stoppage of electron transport chain, and hence loss of ATP s, and therefore, no Carbon fixation as CHO, or protein, or fats.

Explanation:

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Black_prince [1.1K]

Answer:

D variations in individuals are not heritable

Explanation:

Genetic variation in individuals is what drives evolution. In a population of individuals in an environment, with the adequate resources, overpopulation can occur. However, resources are usually a limiting factor for population growth: there is not sufficient food/shelter etc. for everyone to survive.

Some individuals are better suited to success in the environment. This is because of random variation in their DNA that has arisen that give them a selective advantage. This allows them to thrive, take advantage of the limited resources over other members of the population, and be more likely to survive to reproductive age. This is natural selection.

Because this variation is genetic (in the DNA), they can then pass it on to their offspring. Over time, this leads to evolution over time.

8 0
3 years ago
A researcher synthesizes a new molecule called inhibitor X that inhibits the activity of phosphofructokinase (PFK), the enzyme t
Oduvanchick [21]

Answer : The dissociation constant of the PFK‑inhibitor complex is, 5 µM

Explanation :

The expression for reversible competitive inhibition when apparent Km affected by addition of the inhibitor is:

K_m_a=K_m[1+\frac{I}{K_i}]

where,

K_m_a = apparent value = 52 µM

K_m = Michaelis–Menten constant = 40 µM

I = inhibitor concentration = 1.5 µM

K_i = dissociation constant of the PFK‑inhibitor complex

Now put all the given values in the above formula, we get:

52\mu M=40\mu M[1+\frac{1.5\mu M}{K_i}]

K_i=5\mu M

Therefore, the dissociation constant of the PFK‑inhibitor complex is, 5 µM

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