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Amanda [17]
3 years ago
9

Three-carbon molecules of PGA are converted to G3P small sugar molecules by _____, which come from the light reaction.

Biology
1 answer:
anyanavicka [17]3 years ago
5 0
<h3><u>Answer;</u></h3>

<em>ATP and NADPH</em>

<h3><u>Explanation;</u></h3>

Three-carbon molecules of PGA are converted to G3P small sugar molecules by <em><u>ATP and NADPH</u></em>, which come from the light reaction.

  • <em><u>Photosynthesis</u></em>, a process in which green plants use energy from the sun to make their own food, involves two types of reactions, light reactions in light stage and light independent reactions in dark stage.
  • During <em><u>the light reactions ATP and NADH are generated from the electron transport chains</u></em>, which are then taken to the second stage (light independent stage) of photosynthesis. During the Light independent stage the energy stored in ATP and NADH is <em><u>used to convert 3-carbon PGA to G3P</u></em>.

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Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the
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Complete question:

Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited.

A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time.

Assuming that both parents are heterozygous for the gene that causes the disease, what is the probability that the second child will also have the disease? Express your answer as a fraction using the slash symbol and no spaces (for example, 1/2).

Answer:

The probability that the second child will also have the disease is 1/16.

Explanation:

<u>Available data:</u>

  • Two genes independently inherited: one for blood type, the other for disease
  • Man with type A blood  x  Woman with type B blood
  • Both parents are heterozygous for the gene that causes the disease; Dd

If the man has A blood, and the woman has B blood, and they already have an affected child, this means that they must be heterozygous for blood type too.

Cross:  

Parentals)           AiDd         x      BiDd

Gametes)  AD  Ad  iD  id       BD  Bd  iD  id

Punnett square)      AD       Ad       iD       id

                     BD   ABDD  ABDd  BiDD BiDd

                     Bd   ABDd   ABdd   BiDd Bidd

                     iD    AiDD    AiDd    iiDD   iiDd

                     id    AiDd     Aidd     iiDd    iidd

F1) <u>Genotype</u>:

1/16 ABDD

2/16 ABDd

1/16 ABdd

1/16 AiDD

1/16 BiDD

2/16 AiDd

2/16 BiDd

1/16 Aidd

1/16 Bidd

1/16 iiDD

2/16 iiDd

1/16 iidd

<u>Phenotype:</u>

3/16 A/B normal

4/16 A normal

4/16 B normal

3/16 0 normal

1/16 0 affected by the disease.

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