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balu736 [363]
2 years ago
5

11) In C3 photosynthesis, the reactions that require ATP take place in A) the light reactions alone. B) the Calvin cycle alone.

C) both the light reactions and the Calvin cycle. D) neither the light reactions nor the Calvin cycle. E) the chloroplast, but is not part of photosynthesis.
Biology
1 answer:
valentinak56 [21]2 years ago
5 0

Answer:

In C3 photosynthesis, the reactions that require ATP take place in

B) the Calvin cycle alone.

Explanation:

In C3 PLANTS PHOTOSYNTHESIS OCCUR IN TWO STEPS-

1. LIGHT stage , WHICH TAKES PLACE IN LIGHT. In this stage water is split and ATP is released.

2. DARK stage, which takes place in dark. This the stage where calvin cycle takes place which require ATP, i.e energy. In this process sugar is produced with help of carbon dioxide. In this carbon dioxide is accepted by 5- carbon molecule ribulose-1, 5-biphosphate, which result into two molecules of 3- phosphoglycerate. this three carbon compound. hence is called as C3 pathway. hence C3 plants.

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19. If a scientist needs to use a large dataset to explore how corn grows in different soils, how would this data be classified?
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<u>ANSWER:</u>

The correct option is B- Qualitative.

<u>EXPLANATION:</u>

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<u>Answer:</u>

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Madge” was found wandering the streets of New Jersey. She was brought to the attention of a licensed clinical social worker beca
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A probe is launched from Earth and lands on Mars. The gravitational acceleration on Mars is 3.7 m/s2.
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In the fruit fly, Drosophila melanogaster, a spineless (no wing bristles) female fly is mated to a male that is claret (dark eye
Korolek [52]

Explanation:

We have three genes in the same chromosome and are trying to determine their order and relative distance to each other.

<u>The genes and possible alleles are:</u>

  • claret (c/c+)
  • spineless (s/s+)
  • hairless (h/h+)

All mutations are recessive: two copies of the mutant allele are needed for the fly to show that trait.

<u>Parental cross</u>

  • claret hairless male:\frac{s^+\  c\ h}{s^+\  c \ h}
  • spineless female: \frac{s\  c^+\ h^+}{s\  c^+ \ h^+}

Each parent can produce 1 type of gamete only, so the F1 will be homogeneous:

<u>F1</u>

<u>\frac{\ s\  \ \  c^+\ \ h^+}{s^+\   \ c \ \ \ h}</u>

During meiosis, the F1 females can produce 8 types of gametes: 2 parentals and 6 recombinants (two of them, the result of a double crossing over).

If they are test crossed to homozygous recessive males (which can only produce a \frac{s  \ c\   h}{} gamete), the following phenotypes are obtained (I just write the alleles they inherited from the female fly, as the ones that came from the male are the same for all of them):

  • 321 spineless (s c+ h+) ----> Parental
  • 309 claret, hairless (s+ c h) ----> Parental
  • 130 claret, spineless (s c h+) ----> Recombinant
  • 140 hairless  (s+ c+ h) ----> Recombinant
  • 32 hairless, claret, spineless  (s c h) ----> Recombinant
  • 38 WT  (s+ c+ h+) ----> Recombinant
  • 18 claret (s+ c h+) ----> Double Recombinant
  • 12 hairless, spineless (s c+ h) ----> Double Recombinant

The phenotypes observed in the highest frequency are always the parentals, and the ones in the lowest frequency are always the double recombinants.

<u>To determine the order of the genes:</u>

  1. we have to write down the genotype of the F1 female three times, changing the order of the genes each time.
  2. Then, we hypothesize what the double recombinant gametes would look like.
  3. When the theoretical double recombinants we obtain are the same as the ones observed in the F2, we know that <em>that </em>is the correct order of the genes.

In this problem, only if the middle gene is h+/h the double crossing over gives us the observed double recombinant gametes, therefore <u>hairless</u> is the middle gene.

\frac{s\ h^+\ c^+}{s^+\  h\ c}

<u> Double recombinants:</u>

  • s h c+ ----> spineless hairless
  • s+  h+ c  ----> claret

<u>To determine the distance between the genes:</u>

Genetic distance (m.u.) = Recombination Frequency x 100

  • Distance between the spineless and hairless genes:

Distance \ [s-h]= \frac{number\ of\  recombinants \ [s-h]}{Total number of individuals}  * 100\\\\\\Distance \ [s-h]= \frac{32+38+12+18}{1000}  * 100\\\\Distance \ [s-h]= 10\  map\ units

  • Distance between the hairless and claret genes:

Distance \ [h-c]= \frac{130+140+12+18}{1000}  * 100\\\\Distance \ [h-c]= 30\  map\ units

<h3><u>The gene map for these genes is:</u></h3>

spineless -----------------hairless ---------------------------claret

                   10 m.u.                            30 m.u.

3 0
2 years ago
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