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Gnesinka [82]
3 years ago
5

Does the distribution of bases in sea urchin DNA and salmon DNA follow Chargaff’s rules? a) Yes, because the %A approximately eq

uals the %T and the %G approximately equals the %C in both species. b) No, because %A + %T does not equal %G + %C in both species. c) No, because %A is higher than %T, and %G is higher than %C in both species. d) Yes, because the %A + %T is greater than the %G + %C in both species.
Biology
1 answer:
MArishka [77]3 years ago
3 0

Answer:

A. Yes, because the %A approximately equals the %T and the %G approximately equals the %C in both species.

Explanation:

According to Chargaff's rule, in all cellular DNAs, the number of adenosine residues (A) is equal to the number of thymidine residues (T). And the number of guanosine residues (G) is equal to the number of cytidine residues (C). Therefore, the sum of the purine residues equals the sum of the pyrimidine residues (A+ G= C+ T). It is based on the fact that a purine base always pairs with a pyrimidine base in a double helix DNA.

Chargaff’s rule is followed in all the double-helical DNA molecules irrespective of the species. In DNAs of sea urchin and salmon, the percentage of adenine is equal to that of the thymine and the percentage of guanine is equal to that of the cytosine. Therefore, Chargaff's rule is followed.

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Need this ASAP!!HELP!!
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A student does an experiment for a science fair to study whether temperature affects the timing of a cricket’s chirps. The stu
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The frequency of chirps increases as the temperature increases. The correct option is B. <em>The chirps occur closer together as the temperature increases.  </em>

-------------------------------

<h2><u>Available data:</u></h2>

<h3>Data recorded by the student </h3>

<em><u>Day      Temperature   Average time between chirps (sec)</u></em>

1            21                             2. 5

2           22                            2. 6

3           23                            2. 2

4           24                            2. 3

5           25                            2. 0

6           26                             1. 8

7           27                              1. 9

8           28                             1. 8

9          29                             1. 4

10          30                             1. 2

11           31                              1. 5

<u>12          32                             1. 1</u>

In this experiment the student recorded the time in seconds <u>between </u><u>chirps</u>.

We can see that <em>as the </em><em>temperature increases</em><em>, in general, the</em><em> time</em><em> between</em><em> chirps decreases</em>.

  • At 21ºC the time between chirps is 2.5 seconds
  • At 26ºC the time between chirps is 1.8 seconds
  • At 32ºC the time between chirps is 1.1 seconds

According to this information, if the average time between chirps decreases with the temperature increase, we can assume that<em> the </em><em>frequency</em><em> of chirps </em><em>increases</em><em> as the temperature gets higher. </em>

<h3>Options,</h3>

<em>A) The higher the temperature, the fewer chirps there will be in 10 seconds</em>.

Wrong. The number of chirps in 10 seconds will depend, not only on the frequency but also on how long the chirps last. Since the student did not record the time of chirps, we can not make this conclusion.  

<em>B) The chirps occur closer together as the temperature increases</em>.

True. The chirps increase in frequency as the temperature increases, so they occur closer together.

<em>C) The chirps become farther apart as the temperature increases</em>.

Wrong. We can see how the time between chirps decreases as temperature increases, meaning that they are closer not farther.

D) There is no relationship between temperature and the time between chirps.

Wrong. If we make a graph we will see the tendency and the relationship between chirps and temperature.

Graphs usually explain the relationship between variables. In this example, the relationship would be inverse.

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Explanation:

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5'... C TTA CAG TTT ATT GAT ACG GAG AAG G... 3'  NO STOP CODON

5'... CT TAC AGT TTA TTG ATA CGG AGA AGG... 3'  NO STOP CODON

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2)Then you find the complementary sequence (which you already have)

3'... GAA TGT CAA ATA ACT ATG CCT CTT CC... 5'

3'... GA ATG TCA AAT AAC TAT GCC TCT TCC... 5'

3'... G AAT GTC AAA TAA CTA TGC CTC TTC C... 5'

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4)Now you need to find start (AGT in terms of DNA) and stop codons(<u>TAA</u>, <u>TAG,</u> <u>TGA</u>, also in terms of DNA), in 5->3 and de reverse complementary.

<em />

<em>As you can see, you have for open reading frames that lack of stop codon, unfortunately, they also lack of AGT, the start codon.</em>

I added a table where you can find thestart and stop codons as well as the proteins that they translate to in terms of RNA.

I hope you find this information interesting and useful, good luck!

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