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Ludmilka [50]
3 years ago
10

What organelles is responsible for sorting proteins before they are sent to the body of the organism

Biology
2 answers:
wariber [46]3 years ago
7 0

Answer:

Ribosomes

Explanation:

Ribosomes are small round bodies that are found free in the cytoplasm or attached to the <u>endoplasmic reticulum</u>(a system of membrane lined sacs that forms channels throughout the cytoplasm).

Ribosomes are the sites for protein synthesis. So they sort the type and quantity of protein (enzymes) produced in the cytoplasm by the help of the Nucleus.

Peace.

Readme [11.4K]3 years ago
7 0

Answer:Its the Golgi Apparatus

Explanation:

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Answer: If a red blood cell is place in hypertonic solution this will make water to move out of the cell(solute of greater concentration), the cell will shrink,it contented may become concentrated and may die.

Explanation:

Hypertonic solution is a solution that have higher concentration of solutes compared to hypotonic and isotonic solution. Hypertonic solution contains more dissolved particles like salt and electrolytes. When a cell is placed in it, the high concentration of the solutes will make water to leave the cell and the cell will shrink.

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3 years ago
7. Pyrethrums are chemicals used to kill insects. In bed bugs, the mutant-type allele, r, confers resistance, but only in homozy
natita [175]

Answer:

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

Explanation:

It is given that 99% of the bugs were killed after the spray of  pyrethrum. This suggests that 1% of the bugs that were not killed must be homozygous for the mutant type allele "r"

Thus, the frequency of homozygous "rr" species i.e q^2 = 0.01

From this we can evaluate the frequency of mutant "r" allele.

Thus, q = \sqrt{0.01} \\

q = 0.1

As per Hardy-Weinberg first equilibrium equation, p + q = 1

Substituting the value of q in above equation, we get

p = 1 - q\\p = 1 - 0.1\\p = 0.9

Thus, the frequency of homozygous normal bug is equal to

p^2 = 0.9^2 = 0.81

As per Hardy-Weinberg second equilibrium equation-

p^2 + q^2 + 2pq = 1\\

Substituting all the available values we get -

0.81 + 0.01+ 2pq = 1\\2pq = 0.18

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

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Answer:

1

Explanation:

they are always allowed to

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Explanation:

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Explanation:

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