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stepan [7]
3 years ago
6

A greenhouse is filled with air that contains more carbon dioxide than normal air has. how might photosynthesis and plant growth

be affected?
Biology
1 answer:
____ [38]3 years ago
3 0

The greenhouse filled with air containing more carbon dioxide than the normal air will promote the growth of the plant by increasing the rate of the photosynthesis. The water loss by the process of transpiration is reduced considerably, and the water-use efficiency is increased. This will result in a elevated growth of the greenhouse plants.

Hence, the answer is 'the photosynthesis rate will be increased which will increase the plant growth'.

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Taya2010 [7]
The organ is small intestine.
6 0
3 years ago
Question 12:
Ugo [173]
Hey there!

Number 3, "has more people than ever before" is correct. As there are more people, there is a greater need more resources.

Hope this helped, have a good night!
8 0
3 years ago
A white rooster mates with a black hen. The offspring is gray in color. What is this phenomenon called? A. codominance B. coinhe
Alona [7]

I am pretty sure the answer is B

5 0
4 years ago
Explain how we know that DNA breaks and rejoins during recombination.
alisha [4.7K]

Answer:

It occurs through homologous recombination

Explanation:

GENERAL RECOMBINATION OR HOMOLOGIST

           Previously we defined its general characteristics. We will now describe a molecular model of this recombination, based on the classic Meselson and Radding, modified with the latest advances. Do not forget that we are facing a model, that is, a hypothetical proposal to explain a set of experimental data. Not all points of this model are fully clarified or demonstrated:

           Suppose we have an exogenote and an endogenote, both consisting of double helices. In recombination models, the exogenote is usually referred to as donor DNA, and the endogenote as recipient DNA.

1) Start of recombination: Homologous recombination begins with an endonucleotide incision in one of the donor double helix chains. Responsible for this process is the nuclease RecBCD (= nuclease V), which acts as follows: it is randomly attached to the donor's DNA, and moves along the double helix until it finds a characteristic sequence called c

Once the sequence is recognized, the RecBCD nuclease cuts to 4-6 bases to the right (3 'side) of the upper chain (as we have written above). Then, this same protein, acting now as a helicase, unrolls the cut chain, causing a zone of single-stranded DNA (c.s. DNA) to move with its 3 ’free end

2) The gap left by the displaced portion of the donor cut chain is filled by reparative DNA synthesis.

3) The displaced single chain zone of the donor DNA is coated by subunits of the RecA protein (at the rate of one RecA monomer per 5-10 bases). Thus, that simple chain adopts an extended helical configuration.

4) Assimilation or synapse: This is the key moment of action of RecA. Somehow, the DNA-bound RecA c.s. The donor facilitates the encounter of the latter with the complementary double helix part of the recipient, so that in principle a triple helix is formed. Then, with the hydrolysis of ATP, RecA facilitates that the donor chain moves to the homologous chain of the receptor, and therefore matches the complementary one of that receptor. In this process, the chain portion of the donor's homologous receptor is displaced, causing the so-called "D-structure".

It is important to highlight that this process promoted by RecA depends on the donor and the recipient having great sequence homology (from 100 to 95%), and that these homology segments are more than 100 bases in length.

Note that this synapse involves the formation of a portion of heteroduplex in the double receptor helix: there is an area where each chain comes from a DNA c.d. different parental (donor and recipient).

5) It is assumed that the newly displaced chain of the recipient DNA (D-structure) is digested by nucleases.

6) Covalent union of the ends originating in the two homologous chains. This results in a simple cross-linking whereby the two double helices are "tied." The resulting global structure is called the Holliday structure or joint.

7) Migration of the branches: a complex formed by the RuvA and RuvB proteins is attached to the crossing point of the Holliday structure, which with ATP hydrolysis achieve the displacement of the Hollyday crossing point: in this way the portion of heteroduplex in both double helices.

8) Isomerization: to easily visualize it, imagine that we rotate the two segments of one of the DNA c.d. 180o with respect to the cross-linking point, to generate a flat structure that is isomeric from the previous one ("X structure").

9) Resolution of this structure: this step is catalyzed by the RuvC protein, which cuts and splices two of the chains cross-linked at the Hollyday junction. The result of the resolution may vary depending on whether the chains that were not previously involved in the cross-linking are cut and spliced, or that they are again involved in this second cutting and sealing operation:

a) If the cuts and splices affect the DNA chains that were not previously involved in the cross-linking, the result will be two reciprocal recombinant molecules, where each of the 4 chains are recombinant (there has been an exchange of markers between donor and recipient)

b) If the cuts and splices affect the same chains that had already participated in the first cross-linking, the result will consist of two double helices that present only two portions of heteroduplex DNA.

8 0
4 years ago
Leigh's disease is a mitochondrially inherited disease with symptoms that include seizures, fatigue, impaired reflexes, breathin
Ahat [919]

About the question:

You will find the pedigree in the attached files

Answer:

  • individual 4 → Affected woman → Black circle
  • individual 5 → Affected woman or man → Black circle or square
  • individual 7 → Healthy woman → Empty circle
  • individual 10 → Affected boy → Black Square

Explanation:      

Mitochondrial inheritance is the transmission of a disease or a trait from the maternal line.

<em>Most of the DNI is in the nucleus, but there is also DNI in mitochondria. Sperm cells hardly transfer mitochondrial genes to the progeny</em>, so mitochondrial DNI is mostly inherited from the maternal side. If there exists any mutation in this mitochondrial DNI, the whole progeny of the mutated woman will be affected, as they will get the mother genotype carrying the mutation. On the contrary, if an affected man is carrying a mutation in mitochondrial DNI, non of their descendants will get the disease.

Before answering the question, let us remember the pedigree symbols.

  • Squares represent Males/Men
  • Circles represent Females/Women
  • Empty symbols represent healthy/non-affected individuals
  • Solid black symbols represent sick/affected individuals

In the exposed pedigree, we can see that the mother is affected by the disease (individual number 2), so all her children are also affected (individuals 4, 5, and 6) because the <em>disease is mitochondrially inherited</em>.

Individual 3 is a healthy man, so individual 4 must be an affected woman (Black circle). As she is the one affected, then all her children will also be affected. This couple <em>had one boy and two girls</em>. Individuals 8 and 9 are girls (circle), so individual 10 must be the affected boy (black square).  

On the other hand, individual 6 is an affected man (black square), son of individuals 1 and 2. This man couples with a woman, and they have all healthy children. So this woman (individual 7) must be healthy. Even though the man is affected, all their children are not because their mother (7) is not. Remember that sperm cells do not transmit the mitochondrial genes to the progeny.      

And finally, individual number 5 might be either a man or a woman. In any case, this person is also affected by Leigh's disease.  

5 0
3 years ago
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