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lesya [120]
3 years ago
6

all of the triplet codes needed to produce a specific polypeptide chain are found in a(n)____________.

Biology
1 answer:
marshall27 [118]3 years ago
7 0

Answer:

Gene

Explanation:

Which is the basic physical unit of heredity.  Is a linear sequence of nucleotides along a segment of DNA that provides the coded instructions for synthesis of proteins.  

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Which of these best describes Earth’s core?
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Answer:  A.

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There core, or inner layer, is the most dense due to its magnetic pole, not to mention, the core is comprised of multiple molten materials, heated so hot, tht it practically exist between two states of matter simultaneously.

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The mRNA generated below was produced in the<br> of the cell.<br> 5' GCUACUAUGAACCUGCAAAUGAUUUCGU3'
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The mRNA generated below was produced in the  <em><u>nucleus </u></em>of the cell.

Messenger RNA or mRNA is a type of RNA that is an essential component of protein synthesis or gene expression. It is synthesized using the template that is the nucleotide sequence of DNA.

  • The synthesis of the mRNA s called transcription
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Thus, the process of making mRNA from DNA is called transcription, and it occurs in the nucleus.

Learn more about transcription:

brainly.com/question/11430054

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What prevents a lake or ocean from freezing solid
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A variety of opium poppy (Papaver somniferum L.) having lacerated leaves was crossed with a variety that has normal leaves. All
Ainat [17]

Answer:

The correct answer is-

F1 -  AaBb (lacerate)

F2 - A_B_; A_bb; and aaB_ (lacerate)

     - aabb (normal)

2. Two genes, with a dominant allele at either or both loci.

Explanation:

The given information gives THE following data:

Dominant: Lancerate leaves - AABB

Recessive: normal leaves - aabb

F1 has - all Lacerated leaves - AaBb

F2 by selfing F1:

         AB               Ab         aB         ab

AB    AABB        AABb      AaBB    AaBb

Ab     AABb       AAbb      AaBb     Aabb

aB     AaBB       AaBb       aaBB     aaBb

ab     AaBb       Aabb       aaBb      aabb

Here, 15/16 = lacerate which is 0.94 which is equal to the value of lacerate given in the question - 249/265 = 0.94

And normal 1/16 = 0.062 is almost same as 16/265 = 0.060

Thus, the genotypes of -

F1 -  AaBb (lacerate)

F2 - A_B_; A_bb; and aaB_ (lacerate)

     - aabb (normal)

5 0
3 years ago
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