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Leviafan [203]
3 years ago
9

If there are 12 sister chromatids in a cell, how many centromeres are there?

Biology
1 answer:
Ahat [919]3 years ago
3 0
Each pair of sister chromatids is bound by one centromere, so there should be 6 centromeres.
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In these imprinted cells, the SNRPN transcript overlaps with another gene, called UBE3A, which is transcribed in the opposite di
Radda [10]

Answer:

a. The maternal copy of UBE3A is expressed and the paternal copy is silenced.

Explanation:

When UBE3A chromosome is in neurons, paternal allele is silent and maternal allele is expressed in process of genomic imprinting. In some cases UBE3A is expressed from both paternal and maternal chromosomes. The paternal chromosome is blue whereas maternal chromosome is magenta.

5 0
2 years ago
True or false: alleles for the same trait are at different locations on homologous chromosomes​
sergejj [24]

Answer:

<h3>TRUE</h3>

Explanation:

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8 0
3 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
3 years ago
Rna splicing is catalyzed by a complex of rna and protein referred to as a
Vilka [71]
You mean what is the name of this complex? snRNP - small nuclear ribonucleoprotein
4 0
3 years ago
In noncyclic photophosphorylation, O2 is released from
jonny [76]

The correct answer is: B) H2O

Non-cyclic photophosphorylation is a light reaction and it occurs in the thylakoid membrane and it involves two different chlorophyll photosystem.

Non-cyclic photophosphorylation is a reaction where a water molecule is broken down into 2H+ + 1/2 O2 + 2e− . This process is called photolysis. The  2H+ and 1/2O2 from the water are left out for further use, while two electrons are kept in photosystem II.

3 0
3 years ago
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