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cestrela7 [59]
3 years ago
7

The existence of mitochondria within ________ cells is an example of co-evolution as the mitochondria has a different DNA sequen

ce than that of the nucleus in the host cell.
a.eukaryote
b.fungus
c.protist
Biology
2 answers:
Ivenika [448]3 years ago
7 0

I think it's A.

Hope it helps!

topjm [15]3 years ago
4 0

<em>The existence of mitochondria within </em><u><em>eukaryote</em></u><em> cells is an example of co-evolution as the mitochondria has a different DNA sequence than that of the nucleus in the host cell.</em>

<em>The answer is A. eukaryote</em>

<em>Hope this helps :)</em>

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They survived by moving to new areas.
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g Type I diabetes is caused by destruction of the beta cells of the pancreas, the cells that produce insulin. In most cases, the
vodka [1.7K]

Answer:

it is a failure of tolerance (self-tolerance) and specificity (recognition)

Explanation:

Type 1 diabetes is a chronic disease caused by the selective destruction of β cells that are involved in the production of insulin in the pancreas. Type 1 diabetes can be classified into two types: Type 1A diabetes (the immune form of the disease) and Type 1B diabetes (the non-immune form of the disease). Type 1A diabetes is considered an autoimmune disorder where immune responses against pathogens suffer a failure of tolerance to antigens in the β-cells of the pancreatic islets. Thus, Type 1A diabetes is characterized by the process of recognition of β-cell antigens (autoantigens) by the immune system. This disease is often caused by genetic factors associated with mutations in the human leukocyte antigen (HLA) class II region located on chromosome 6.

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3 years ago
In a population with two alleles, B and b, the allele frequency of B=0.59. What would be the frequency of the heterozygotes if t
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Answer:

0.483

Explanation:

The given population is in Hardy-Weinberg equilibrium. If the gene has two alleles, the sum total of the frequencies of these two alleles will be one.  

Therefore, the total of the frequency of allele B and frequency of allele b will be 1. f(B) + f(b)=1

If the frequency of allele "B" is 0.59, then the frequency of allele "b" will be=1-0.59= 0.41

The frequency of heterozygous genotype in the population= 2pq

p= frequency of the dominant allele

q= frequency of the recessive allele

So, 2pq= 2 x 0.59 x 0.41 = 0.483

6 0
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Glucose is broken down in the chemical processes of aerobic respiration.
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\bold{\huge{\purple{\underline{ Answer }}}}

<u>Aerobic </u><u>respiration </u>:- Aerobic respiration is process of respiration in which complete oxidation of glucose takes place in the presence of oxygen

<u>That </u><u>is</u><u>, </u>

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In aerobic respiration, Glucose with 6 carbon atoms broken into 2 pyruvic acid each with 3 carbon atom and produces water + CO2 + energy.

<h3><u>Whereas</u><u>, </u></h3>

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The end products of glucose in anaerobic respiration is

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5 0
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Here's one link:https://www.creative-biogene.com/support/Knockout-Cell-Lines-in-Antibody-Screening-and-Validation.html


8 0
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