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2 years ago

Clue 1:

Biology

1 answer:

tankabanditka [31]2 years ago

4 0

Answer:

The two problems that the transfer of information from DNA to protein must overcome are:

How to bring the information from the nuclear DNA to the place of protein synthesis?
How to convert this DNA information into amino acids and then into proteins?

Explanation:

The genetic information is found in the DNA and depends on a specific sequence of nitrogenous bases. This information is transcribed into the messenger RNA, whose base sequence is organized into triplets and codons, each of which encodes an amino acid, as well as establishing the pattern for starting and stopping the synthesis of a protein.

<h3>How to bring the information from the nuclear DNA to the place of protein synthesis? </h3>

The DNA must be transcribed into messenger RNA (mRNA), a process that occurs in the nucleus of the cell. mRNA leaves the nucleus and travels to the cytoplasm, where amino acid synthesis will take place.

<h3>How to convert this DNA information into amino acids and then into proteins?</h3>

Once in the cytoplasm mRNA binds to ribosomes, structures in charge of translating the sequence of nitrogenous bases RNA to synthesize amino acids. The set of ribosomes and rough endoplasmic reticulum are in charge of the assembly of amino acids to produce peptides and proteins.

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3 years ago

A trihybrid cross is a cross between two individuals who are heterozygous for three genes. For example: AaBbCc x AaBbCc. Assumin

Mariulka [41]

Answer:

27:9:9:9:3:3:3:1

Explanation:

The gametes of the parental genotypes would be

ABC, ABc, AbC, aBC, Abc, aBc, abC, abc

The cross between AaBbCc and AaBbCc will result into following offspring (image is also attached)

ABC ABc AbC aBC Abc aBc abC abc

ABC AABBCC AABBCc AABbCC AaBBCC AABbCc AaBBCc AaBbCC AaBbCc

ABc AABBCc AABBcc AABbCc AaBBCc AaBbcc AaBBcc AaBbCc AaBbcc

AbC AABbCC AABbCc AAbbCC AaBbCC AAbbCc AaBbCc AabbCC AabbCc

aBC AaBBCC AaBBCc AaBbCC aaBBCC AaBbCc aaBBCc aaBbCC aaBbCc

Abc AABbCc AABbcc AAbbCc AaBbCc AAbbcc AaBbcc AabbCc Aabbcc

aBc AaBBCc AaBBcc AaBbCc aaBBCc AaBbcc aaBBcc aaBbCc aaBbcc

abC AaBbCC AaBbCc AabbCC aaBbCC AabbCc aaBbCc aabbCC aabbCc

abc AaBbCc AaBbcc AabbCc aaBbCc Aabbcc aaBbcc aabbCc aabbcc

The genotype ratio is 27:9:9:9:3:3:3:1

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3 years ago

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Rudiy27

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2 years ago