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evablogger [386]
3 years ago
14

What kind of relationship exists between the birds and the cattle

Biology
1 answer:
Katena32 [7]3 years ago
7 0
The answer to this is, they both are commensalism.
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Which statement explains how producers are dependent upon consumers for their survival?
MrRa [10]

Answer:

It explains what the consumers do for the survival of the producers

Explanation:

6 0
3 years ago
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In the Calvin cycle the conversion of energy poor carbon dioxide into energy rich glucose
rodikova [14]

Answer:

D)NADPH is made

A)ATP is used

Explanation:

Sorry to ask the two questions, I asked why the two options are in the process:Reduction. In the second stage, ATP and NADPH are used to convert the 3-PGA molecules into three-carbon sugar molecules, glyceraldehyde-3-phosphate ( G3P ). At this stage, it gets its name because NADPH donates, or reduces , electrons to a three-carbon intermediate to form G3P.[Ocultar detalhes]

The reduction stage of the Calvin cycle, which needs ATP and NADPH, converts 3-PGA (produced in the fixation stage) into a three-carbon sugar. This process takes place in two main stages:

Simplified diagram of the reduction step of the Calvin cycle showing the carbon atoms, but not the complete molecular structures. A 3-PGA molecule first receives a second phosphate group from ATP (generating ADP). Then, the doubly phosphorylated molecule receives electrons from NADPH and is reduced to form glyceraldehyde-3-phosphate. This reaction generates NADP + and also releases an inorganic phosphate.

Simplified diagram of the reduction step of the Calvin cycle showing the carbon atoms, but not the complete molecular structures. A 3-PGA molecule first receives a second phosphate group from ATP (generating ADP). Then, the doubly phosphorylated molecule receives electrons from NADPH and is reduced to form glyceraldehyde-3-phosphate. This reaction generates NADP + and also releases an inorganic phosphate.

First, each 3-PGA molecule receives a phosphate group from ATP, becoming a doubly phosphorylated molecule called 1,3-bisphosphoglyceride (and leaving an ADP as a by-product).

Second, 1,3-bisphosphoglycerate molecules are reduced (gain electrons). Each molecule receives two electrons from NADPH and loses one of its phosphate groups, becoming a three-carbon sugar called glyceraldehyde-3-phosphate (G3P) . This step produces NADP^+

+

start superscript, plus, end superscript and phosphate (\text P_iP

i

start text, P, end text, start subscript, i, end subscript) as by-products.

The chemical structures and real reactions are:

Reactions of the Calvin cycle reduction step, showing the molecular structures of the molecules involved.

Reactions of the Calvin cycle reduction step, showing the molecular structures of the molecules involved.

The ATP and NADPH used in these steps are products of the photo-dependent reactions (first stage of photosynthesis). That is, the chemical energy of ATP and the reducing potential of NADPH, both produced with the use of light energy, keep the Calvin cycle running. Conversely, the Calvin cycle regenerates ADP and NADP^+

+

start superscript, plus, end superscript, providing the necessary substrates for photo-dependent reactions.

7 0
2 years ago
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Select all that apply.
Brums [2.3K]
A couple of the methods of molecular biotechnology are cloning and recombinant DNA.
3 0
3 years ago
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The body's reaction to long-lasting stress is associated with ____.
tatyana61 [14]
Depression is the answer
4 0
3 years ago
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In a population of heffalumps, there is a single gene that controls the presence of hair. The H allele, which is a complete domi
maksim [4K]
<h2>Frequency of allele </h2>

Explanation:

Hardy Weinberg Equilibrium is used to calculate the allelic as well as genotypic frequency

Allelic frequency of dominant and recessive allele is represented by p and q respectively whereas genotypic frequency of dominant genotype is represented by p2 and q2 respectively

Given:

H allele (p) = hairy heffalump (dominant)

h allele (q) = hairless heffalump (recessive)

36% of heffalump population is hairless represents the % of recessive genotype, hh (q2) =36%

Calculation of frequency of the h allele (q) :

Frequency of genotype hh (q2) will be: 36/100=0.36 or 0.6*0.6

Frequency of h allele (q) will be 0.6

3 0
3 years ago
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