BIOLOGY COLLEGE

ATP is produced during the light reactions of photosynthesis. This ATP is used to_____________in the Calvin Cycle.A. split water into oxygen has and hydrogen
B. produce NADPH
C. reduce carbon dioxide into glucose
D. None of these are correct

Answers

Answer 1
Answer:

Answer:

C

Explanation:

The energy source is ATP, and the reducing agent is NADPH, which adds high-energy electrons to produce sugar.

Related Questions

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Scientists construct an experimental bacteriophage that is composed of the T2 phage protein coat and T3 phage DNA. If a bacterium is infected by this phage, the new phages produced would be expected to have:
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Survivorship patterns within a species are pre-set by life history characteristics and can never change in response to variable environmental conditions.a. Trueb. False
Suppose that you began a one-year study of tuberculosis (TB) in St. Petersburg on January 1st, 2018. You enrolled 200 residents in your study and checked on their TB status on a monthly basis. At the start of your study on January 1st, you screened all 200 residents. Upon screening, you found that 10 of the healthy residents were immigrants who were vaccinated for TB and so were not at risk. Another 10 residents already had existing cases of TB on January 1st. On February 1st, two residents developed TB. On April 1st, three more residents developed TB. On July 1st, three of the residents who had existing TB on January 1st died from their disease. The study ended on December 31, 2018. Assume that once a person gets TB, they have it for the duration of the study, and assume that all remaining residents stayed healthy and were not lost to follow-up.What was the prevalence of TB in the screened community on January 1st?

Give 3 examples
of other organ systems in the human body.

Answers

Answer:

nervous system, skeletal system, muscular system, cardiovascular system, urinary system and respiratory system are just a few

Explanation:

Answer:

All systems with short explanation

Circulatory system / Cardiovascular system:

Circulates blood around the body via the heart, arteries and veins, delivering oxygen and nutrients to organs and cells and carrying their waste products away.

Keeps the body's temperature in a safe range.

Digestive system and Excretory system:

System to absorb nutrients and remove waste via the gastrointestinal tract, including the mouth, esophagus, stomach and intestines.

Eliminates waste from the body.

Endocrine system:

Influences the function of the body using hormones.

Integumentary system / Exocrine system:

Skin, hair, nails, sweat and other exocrine glands

Immune system and lymphatic system:

Defends the body against pathogens that may harm the body.

The system comprising a network of lymphatic vessels that carry a clear fluid called lymph.

Muscular system:

Enables the body to move using muscles.

Nervous system:

Collects and processes information from the senses via nerves and the brain and tells the muscles to contract to cause physical actions.

Renal system and Urinary system

The system where the kidneys filter blood to produce urine, and get rid of waste.

Reproductive system:

The reproductive organs required for the production of offspring.

Respiratory system:

Brings air into and out of the lungs to absorb oxygen and remove carbon dioxide.

Skeletal System:

Bones maintain the structure of the body and its organs.

After 3 half lives how much of a radioactive substance will remain

Answers

354/4561 and that is I think the answer

Cells responsible for histamine release

Answers

Answer:

Mast Cells

Explanation:

Mast cells release histamine in the blood stream, when they detect a substance that triggers an allergic reaction, also called an allergen.

Which of the following is the semi-fluid transition zone of the mantle between the lithosphere and the mesosphere?Asthenosphere
Atmosphere
Biosphere
Geosphere

Answers

The correct answer is A. Asthenosphere.

Explanation

The earth is composed of three large layers: the nucleus, the mantle, and the crust. However, there are more layers within the main layers; in the case of the the asthenosphere, this is a layer of semi-molten materials that is located under the lithosphere (50 to 150km of depth) and has a thickness of around 650km. Additionally, this is located above the mesosphere, another layer of the crust that is part of the mantle, and extends to approximately 2700km deep. Additionally, this layer serves as a transition between the lithosphere and the mesosphere which are two major layers. So, the correct answer is A. Asthenosphere.
I’m guessing asthenosphere

Landforms is a physical feature on earth's surface true or false

Answers

The correct answer is True

Explanation:

In geography, the term "landform" refers to a natural shape or form that can be found on the surface of Earth such as mountains, hills or canyons. All of these are formed through different processes such as movement in tectonic plates. Additionally, these are considered as a physical feature of Earth as landforms define the shape and features of the Earth's surface and also influenced other physical phenomena such as weather or wind patterns. According to this, it is true landforms is a physical feature on Earth's surface.

Answer:

True*

Explanation:

Fine spines (s), smooth fruit (tu), and uniform fruit color (u) are three recessive traits in cucumbers whose genes are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross. The progeny from this testcross are:S U Tu 2
s u Tu 70
S u Tu 21
s u tu 4
S U tu 82
s U tu 21
s U Tu 13
S u tu 17__
Total 230
a. Determine the order of these genes on the chromosome.
b. Calculate the map distances between the genes.
c. Determine the coefficient of coincidence and the interference among these genes.
d. Draw the chromosomes of the parents used in the testcross.

Answers

Answer and Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

  • S U Tu 2
  • s u Tu 70
  • S u Tu 21
  • s u tu 4
  • S U tu 82
  • s U tu 21
  • s U Tu 13
  • S u tu 17

The total number, N, of individuals is 230.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the thrid not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

  • s u TU (70 individuals)
  • S U tu (82 individuals)

Double recombinant)

  • S U Tu (2 individuals)
  • s u tu (4 individuals)

Comparing them we will realize that between

s u TU (parental)

s u tu (double recombinant)

and

S U tu (Parental)

S U TU (double recombinant)

They only change in the position of the alleles TU/tu. This suggests that the position of the gene TU is in the middle of the other two genes, S and U, because in a double recombinant only the central gene changes position in the chromatid.

So, the order of the genes is:

---- S ---- TU -----U ----

In a scheme it would be like:

Chromosome 1:

---s---TU---u--- (Parental chromatid)

---s---tu---u--- (Double Recombinant chromatid)

Chromosome 2

---S---tu---U--- (Parental chromatid)

---S---TU---U--- (Double Recombinant chromatid)

Now we will call Region I to the area between S and TU and Region II to the area between TU and U.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between S and TU genes, and P2 to the recombination frequency between TU and U.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.  So:

  • P1 = (R + DR) / N

        P1 = (21+17+4+2)/230

        P1 = 44/230

        P1 = 0.191

  • P2= = (R + DR) / N

        P2 = (21+13+4+2)/230

        P1 = 40/230

        P1 = 0.174

Now, to calculate the recombination frequency between the two extreme genes, S and U, we can just perform addition or a sum:

P1 + P2= Pt

0.191 + 0.174 = Pt

0.365=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.191 x 100 = 19.1 MU

GD2= P2 x 100 = 0.174 x 100 = 17.4 MU

GD3=Pt x 100 = 0.365 x 100 = 36.5 MU

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:

  • observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals
  • expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

CC= ((2 + 4)/230)/0.174x0.191

CC=(6/230)/0.0332

CC=0.7857

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.7857

I = 0.2143            

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