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2 years ago

A bacterium engulfed by a white blood cell through phagocytosis will be digested by enzymes contained in _____.

2 answers:

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A bacterium engulfed by a white blood cell through phagocytosis will be digested by enzymes contained in "Lysosomes"

Hope this helps!

ycow [4]2 years ago

7 0

Answer: Lysosome

Explanation:

This is an double membrane organelle found in the cytoplasm of the eukaryotic cell which contains variety of digestive enzymes.

Lysosomes contain enzymes such as DNAses, proteases, RNAses, cellulase and other digestive enzymes which act on the specific biomolecules and degrade the.

The lumen of lysosomes is acidic in nature thus, the foreign substance which gets ingested in the cell is easily digested by the enzymes.

In the given question since, the engulfed bacteria is digested by the enzymes which could have been provided by the lysosomes there fore lysosomes is the correct answer.

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The enzyme telomerase is part protein and part RNA. What would be the most likely effect of a large deletion in the gene that en

Artist 52 [7]

Answer: option A, B and D

Explanation:

The telomeres found at the terminal ends of eukaryotic chromosomes are long repetitive sequences that protects the DNA from being identify as a damage molecule to encourage fusion with other chromosomes. Since replication does not extend down to these 3' ends, the cell recruits the telomerase enzyme which is a ribonucleoprotein that helps to add telomeric repeats using its own RNA component as template to extend this end. Thus if mutation like a large deletion occurs in the gene encoding for this enzyme, the telomerase would not be able to associate correctly with the telomere, the catalytic center also encoding for its RNA polymerase would be altered and the enzymes will not be able to synthesize new telomeric repeats sequences

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2 years ago

OverLord2011 [107]

Answer:

1.hypothesis.

hope it helps!

5 0

2 years ago

Read 2 more answers

Suppose that of a cohort of 150 mice in a mouse colony born in February, 125 are still at the end of March and 115 are still ali

kogti [31]

Answer: Survival rate up to start of April = 76.67%

Explanation: size of colony = 150mice.

Size of colony in February =125mice.

Size of colony in march = 115 mice.

1. Survivorship rate in February

= Size of colony / size in February *100.

= 122 / 150 * 100

= 83.33% to the nearest hundredth

2. Survivorship rate in march

= Size of colony / size in march *100.

= 115/150 * 100

= 76.67% to the nearest hundredth.

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2 years ago

Suppose a human blood cell containing a 0.9 percent solute concentration were put into a container of 0 percent solute solution

Vladimir [108]

Answer and explanation:

If a human blood cell with a 0.9% solute concentration were to be put into a container of 0% solute solution, the cell would get BIGGER.

The cell contains a more concentrated solution than the solution in the container. The difference in concentration would produce an osmotic gradient that would cause water from the container to get inside the cell to even the concentrations - this is going to make the cell much bigger because the entering water would bloat the cell.

In this example, the solution in the container is hypotonic in relation to the cell, while the solution inside the cell is hypertonic in relation to the solution in the container. This is why the water will be moving from outside of the cell to the inside of the cell.

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1 year ago

Which abiotic factor is most likely in a desert?

finlep [7]

Abiotic means is not or never was living.

b. the wind is not a living thing

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2 years ago