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ZanzabumX [31]
2 years ago
9

Cells spend time making their organelles and completing tasks such as producing protein. When during the cell cycle would a cell

most likely perform these tasks? A.G1 phase B.S phase C.cytokinesis D.mitosis
Biology
2 answers:
notsponge [240]2 years ago
5 0
D. Mitosis, During<span> the mitotic phase, the duplicated chromosomes are segregated ... reserves to </span>complete<span> the </span>task<span> of replicating each chromosome in the nucleus</span>
ki77a [65]2 years ago
4 0

It’s not D. mitosis it’s A. G1 phase

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Which of the following statements about DNA (deoxyribonucleic acid) is NOT true?
makkiz [27]

D. Seems the most likely.

8 0
2 years ago
HELP <br><br> Why is sea temperature important?
Snezhnost [94]

Explanation:

Sea surface temperature provides fundamental information on the global climate system. ... SST is an essential parameter in weather prediction and atmospheric model simulations, and is also important for the study of marine ecosystems. SST data are especially useful for identifying the onset of El Niño and La Niña cycles

8 0
2 years ago
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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
3 years ago
What do the human body cell do with the carbon dioxide
Ber [7]

Answer: When the oxygen-rich blood gets to the cells, the cells receive the oxygen and release the carbon dioxide. The blood with less oxygen and a lot of carbon dioxide returns to the heart. Then the heart returns this blood to the lungs where carbon dioxide is released and oxygen is received.

Explanation:

3 0
2 years ago
Read 2 more answers
In humans, to fold the tongue is a dominant trait (L), and the straight tongue is the recessive trait. What are the probabilitie
kaheart [24]

The given question is incomplete as the genotype of the parents is not given, so the answer is providing in the followings case:

1. dominant parent and recessive parent

2. heterozygous parents

Answer:

1. dominant parent and recessive parent:

dominant parents can be represented by LL and recessive parent is represented by ll, so the gametes would be L, L and l, l.

so,

 L    L

l  Ll   Ll

l  Ll   Ll

so there are all offspring in heterozygous condition as we known one or two dominant allele masks the recessive allele for the trait so 100% offspring can fold their tongue.

2. heterozygous parents

In this case, parents have Ll genotype and gametes would be L and l for each parent so,

    L    l

L   LL  Ll

l    Ll    ll

In this case, one is pure dominant and two heterozygous whereas only one is recessive so, the phenotype of offspring that cant fol the tongue would be:

3/4 = 75%

6 0
3 years ago
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