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solmaris [256]
3 years ago
8

Which cycle includes bacteria taking gas from the atmosphere and converting it into a material that plants can absorb through ro

ot systems?
A. The phosphate cycle
B. The water cycle
C. The nitrogen cycle
D. The oxygen and carbon dioxide cycle
Biology
2 answers:
Pachacha [2.7K]3 years ago
6 0

Answer:

C. The nitrogen cycle

Explanation:

In the nitrogen cycle, we have the presence of nitrogen-fixing bacteria, which are essential for this cycle to occur.

This is because, plants are not able to absorb atmospheric nitrogen. So these bacteria, present in the soil, capture nitrogen from the atmosphere and initiate a series of chemical reactions that confers nitrogen in a way that plants can absorb it through their roots.

Sauron [17]3 years ago
5 0
The nitrogen cycle would be the correct answer.
You might be interested in
An additional gene, gene W, was also examined. A test cross was made between true-breeding EEWW flies and EEWW flies. The result
Debora [2.8K]

This question is incorrect but here is the correct question below;

An additional gene,gene W was also examined. a test cross was made between true breeding EEWW flies and eeww flies. The resulting F₁ generation was then crossed with eeww flies. 100 offspring in the F₂ generation were examined and it was discovered that the E and W genes were not linked.

Which is the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked?

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

b) Linked: 25% Eeww, 50% eeWw; not linked:parental genotypes EeWw and eeww.

c) Linked genotypes (EeWw and eeww) and recombinant genotype ( Eeww & eeWw) in the F₂ generation are nearly the same irrespective of their linkage.

d) Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw.

Answer:

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

Explanation:

a test cross was made between true breeding EEWW flies and eeww flies

If EEWW self crossed, we have the following ( EW, EW, EW, EW)

Also, for eeww, we have ( ew, ew, ew, ew)

                   

                    EW                   EW                     EW                   EW

ew               EWew               EWew               EWew               EWew      

ew               EWew               EWew               EWew               EWew

ew               EWew               EWew               EWew               EWew

ew               EWew               EWew               EWew               EWew

All offspring are  (EWew)

The question goes further by saying "The resulting F₁ generation was then crossed with eeww flies".

And we are asked to find the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked

∴

To determine  the offsprings of the linked genes we need to go by the definition and understand what linked genes are: Linked genes are genes that are physically close together on the same chromosomes. Effect of recombinantion on linked genes, results in gene swaps which occur in chromosomes that are homologous.

Having said that; If  EWew × eeww

we have;                 EW   &   ew    ×    ew  &    ew

           EW               ew

ew       EeWw          eeww

ew       EeWw          eeww

offspring that

are linked in   ⇒     EeWw    EeWw     &      eeww      eeww

F₂   will be

\frac{1}{2} = 50% of EeWw of the total 100 offspring in the F₂ cross

\frac{1}{2} = 50% of eeww of the total 100 offspring in the F₂ cross

∴ Linked genes =  50% EeWw and 50% eeww.

For unlinked genes; If  EWew × eeww

if rearrangement occurs in EWew  and EWew self crossed, we have ( EW,Ew,eW,ew) as the traits needed for the unlinked gene F₂ crossing.

Also ewew will be (ew, ew, ew, ew).

                       EW                    Ew                    eW                    ew

ew                  EeWw               Eeww                eeWw                eeww

ew                  EeWw               Eeww                eeWw                eeww

ew                  EeWw              Eeww                 eeWw                eeww

ew                  EeWw              Eeww                 eeWw                eeww

We have the following results for the unlinked genes

\frac{1}{4} = EeWw  25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = Eeww   25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = eeWw   25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = eeww    25% of the total 100 offspring in the F₂ cross

∴ not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

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Answer:

A) Decreases cellular energy production

B) DCCD also affects K+ transport

Explanation:

A) Consequences are of DCCD on cellular energy production: <em>Decreases cellular energy production</em>

ATP-synthase pump is composed of two subunits: F1 catalytic subunit that synthesizes ATP, and F0 proton pumping subunits, that transport H+ through the membrane. F1 subunit might act independently of F0 to produce ATP, but this molecule can not be released without H+ gradient, which generates a movement necessary for ATP release from the catalytic center.

When any of the parts composing F0 react with DCCD, the subunit can not transport H+ through the membrane. DCCD inhibits the enzyme activity by blocking the protons´ flow.

As DCCD blocks the protons´ flow, and the protons´ flow is necessary to release the ATP molecule from the F1 subunit, no other ADP + Pi can enter to F1 subunit, and the production of ATP stops.

B) Other cellular effects of DCCD

There seem to be other effects of DCCD on cell activity, some of which are still under study. To name a few:

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  • When exposing the cell to high concentrations of DCCD for a long time, might occur an alteration in the electron transporting chain
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  • Inhibition of K+ transport, associated with the inhibition of H+ transport.

Concerning the effect of DCCD on the K+ transport, DCCD stops the extrusion of H+ and the consequent intrusion of K+.

DCCD strongly inhibits the simultaneous flow of H+ and K+. First, it inhibits H+ flow, acidification of the environment stops, but at this point, K+ keeps moving through the membrane. Once the H+ flow has ceased, the K+ flow slowly decreases until it finally stops moving. There is a lag time in the DCCD effect on K+ flow to the instantaneous effect on H+ flow.

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