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arsen [322]
3 years ago
15

In which organelles does the chemical process of photosynthesis occur?

Biology
2 answers:
ludmilkaskok [199]3 years ago
8 0
That would be the chloroplast. 
NemiM [27]3 years ago
5 0
Chloroplasts is the answer
hope this helps :)
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Assume there are 10,000 Joules (J) of energy available in trophic levelll in the figure According to the conventional model of e
geniusboy [140]

Answer:

To complete this calculation, we divide the amount from the higher trophic level by the amount from the lower trophic level and multiply by one hundred. That is, we divide the smaller number by the bigger one (and multiply by one hundred)

Explanation:

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A mutation creates a dominant negative allele of a particular gene. The normal allele of the gene encodes a protein that forms a
Wewaii [24]

Answer:

87.5% of the trimers in the cell will be inactive.

Explanation:

A monomer is a single subunit. Look at an easy comparison. If a monomer is a ball. For the protein to be functional, it needs to be a trimer .A trimer = 3 monomers joined / bound together with bonds. Therefore, if a monomer is a ball, we need 3 balls, we attach it together and that is the primer . if one ball becomes missing, that protein won't work .

In chemistry, a trimer is regarded as three molecules or ions of the same substance (that is, same type) which are associated together .

Double negative : Each gene has 2 alleles. So, for a positive protein, both allele needs to be same, let's say + means on. So for a x protein , if it's x+/x+ it means the protein is homozygous and active. If for some mutation, it's x+/x- , it's a heterozygous protein. Double negative means when it is x-/x- and this means the protein has lost is function . It's not functional at all.

The heterozygous cell will form one or more subunits has the mutant structure when the gene is expressed. From the question if there is a mutation in dominant negative allele, it will cause the inactivation of one trimer resulting in 87.5% inactive trimers for a heterozygous individual.

There are 87.5 % percent of the trimers present in the cell which will be inactive for a heterozygous individual, If at least one of the subunits has the mutant structure.

The probability for each individual subunit to be a mutant is 1/2

Gene codes for a protein that forms trimer so the percentage of the trimers present in the heterozygous individual is third power of 1/2 =1/8

1/8=0.125 will be active.

1- 0.125 will be inactive.

=0.875 will be inactive.

0.875 × 100= 87.5%

Therefore 87.5% will be inactive.

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Answer:

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Explanation:

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