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Lubov Fominskaja [6]
3 years ago
14

The joint comsec monitoring activity provides opsec assistance by:

Biology
1 answer:
marishachu [46]3 years ago
8 0
The answer to this question is: Monitoring of COMMAND SECURITY (COMSEC) processes and procedures.Communications security (COMSEC<span>) is the steps that could be carried oyt in the prevention of unauthorized access to telecommunications traffic The procedures will help people to understand basic first-aid to prevent their personal gadget from being hacked by a third party.</span>
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B) limestone

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Which statements accurately describe how fructose metabolism in the liver differs from glucose metabolism?
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The fructose 1-phosphate pathway can deplete intracellular phosphate/ ATP.

Explanation:

Fructose 1-phosphate is a derivative of fructose. For understanding in better way fructose metabolism has three enzymes. Fructose- bisphosphate aldolase B, fructokinase and Adenosine triphosphate. These all are present in liver and kidney of human as well rat. In liver rapidly fructose is change to fructose 1 through fructokinase.

After it is converted into trioses dihydroxyacetone phosphate as well as glyceraldehyde through aldolase.  With glucose metabolism Fructose get synergistic effect

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What are two events that can take place for Secondary succession to occur?
Nadya [2.5K]
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Disturbances such as forest thinning, floods, fire and wind can all lead to secondary succession. Examples of secondary succession are the gradual replacement of old fields by forest or the vegetation recovery and change following wildfire occurrence.
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3 years ago
Glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as an X-linked recessive allele in humans. A woman whose father
prohojiy [21]

Answer:

(a) 1/2; (b) no

Explanation:

Glucose-6-phosphate dehydrogenase deficiency (G6PD) is an X-linked recessive disorder and the woman's father was diseased so it means that woman is a carrier of the allele but has normal phenotype. It means that she will have XXᵇ genotype.

In contrast to this, her husband is diseased so his genotype will be XᵇY.

The Punnett square diagram related to the cross is attached.

(a) Proportion of their sons expected to be G6PD is 1/2:

They both may give birth to 4 progeny with genotypes XXᵇ, XᵇXᵇ, XY and XᵇY. It means they both may have 2 sons out of which one with genotype XᵇY will be diseased while the one with genotype XY will be healthy. So the proportion of their sons having G6PD is 1/2 or 50%.

(b) If the husband were G6PD deficient, the answer will not change.

The reason behind this is that this disease is caused by an allele located in X chromosome. But father contributes only Y chromosome to his son not X chromosome. The X chromosome will affect the genotype of his daughter not son that is why answer will not change. It means they will still have 1/2 of their sons diseased.  

7 0
3 years ago
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