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Rudik [331]
3 years ago
10

What would terminate translation?

Biology
1 answer:
gizmo_the_mogwai [7]3 years ago
5 0

Answer:the arrival of the ribosome to a stop condon, such as uga

Explanation:I’m taking the test right now

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Onion cell and elodea cell
NARA [144]
Both onion<span> and </span>elodea cells<span> had a </span>cell<span> wall and cytoplasm but the </span>onion<span> lacked chloroplasts. 1. </span>Onion<span> is a storage organ for the </span>onion<span> plant. It is underground so it can't do photosynthesis.</span>
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4 years ago
Two identical daughter cell result from mitosis and cytokinesis. In what ways are they " identical"?
Keith_Richards [23]
The two identical daughter cells resulting from mitosis and cytokinesis are identical in the following ways:1. Mitosis occurs when the nucleus of the cell divides into two identical nuclei, each with the same type and number of chromosomes. The cell's DNA is duplicated during this phase. Sometimes the cell's DNA isn't copied properly resulting in cancer-type cells. 2. Cytokinesis is when the cytoplasm divides into two identical daughter cells. Each cell is genetically identical and both are a similar size. 




6 0
3 years ago
Which of the following is not a common use for public land?
AleksandrR [38]

The answer is A. Logging

6 0
3 years ago
Read 2 more answers
Suppose a population is carrying a condition controlled by two alleles: P (dominant) and p (recessive). Only homozygous individu
Virty [35]

<span>The answer is B. 72.25 percent.

The Hardy-Weinberg principle can be used:</span>

<em>p² + 2pq + q² = 1 </em>and <em>p + q = 1</em>

where <em>p</em> and <em>q</em> are the frequencies of the alleles, and <em>p²</em>, <em>q²</em> and <em>2pq</em> are the frequencies of the genotypes.

<span>The <em>p</em> allele (<em>q</em>) is found in 15% of the population:
q = 15% = 15/100
Thus, q = </span><span>0.15

To calculate the <em>P</em> allele frequency (<em>p</em>), the formula <em>p + q = 1</em> can be used:
If p + q = 1, then p = 1 - q
p = 1 - 0.15
Thus, </span><span>p = 0.85

Knowing the frequency of the <em>P</em> allele (<em>p</em>), it is easy to determine the frequency of the <em>PP </em>genotype (<em>p²</em>):
p² = 0.85² = 0.7225

Expressed in percentage, p² = 72.25%.</span>

7 0
3 years ago
Three linked autosomal loci were studied in cucumbers: fruit color (dark vs. light), tendril size (long vs. reduced) and fruitin
Zina [86]

Answer:

Explanation:

1. Progeny observed in the experiment is calculated by adding all the progeny numbers observed in the :

TIME TO FRUIT                TENDRIL SIZE        FRUIT COLOR       NUMBER

 

 Normal                             normal                      dark                        370

 normal                              normal                      light                        32

 early                                  normal                      light                        971

 early                                   normal                     dark                         28

 early                                  reduced                    dark                          25

 early                                   reduced                   light                          483

 normal                                reduced                   light                          36

 normal                                reduced                    dark                         810

                                                                               TOTAL

                                                                                PROGENCY           <u>2755</u>

<u></u>

<u> </u>

2. The locus in the center is TIME TO FRUIT. Tendril size is at the beginning of the locus and the distance between the locus B and locus C is R2

<u> </u>

<u></u>

<u></u>

3 0
3 years ago
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