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Bogdan [553]
4 years ago
14

The barrel of a gun is also known as what? Rifle Caliber Bore Gauge

Biology
2 answers:
Trava [24]4 years ago
6 0
I think the correct answer from the choices listed above is the third option. The barrel of a gun is also known as bore. The barrel is the metal tube which the bullet is fired and the bore is the inside of the barrel. It is the one closest to the barrel so it must be the answer.
Yuki888 [10]4 years ago
3 0

Answer:

The barrel is also known as the Bore.

Explanation:

Rifle- is a large type of gun, not a part of one.

Caliber- is the type of bullet fired from a gun, or the size of the bullet.

Gauge- used to show magnitude, amount, or contents of something, typically with a visual display.

None of these except bore would make sense.

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Pleas help will mark brainly need done by 12am
KiRa [710]

Answer:

the answer is juvenile

Explanation:

Turtles' life cycles progress from being an egg, hatchling or baby turtle, into being a juvenile.

6 0
3 years ago
How high can creatinine and bun levels go before death?
wel
Creatinine is a by-product of muscular metabolism. In the natural and normal scheme of things, this substance or waste product can be eliminated from the body. A high-serum creatinine level may cause kidney damage. In relation to the above question as to how high can creatinine levels go before death, it must be noted that kidneys have strong compensatory ability and by that as long as its still 50 percent functional, creatinine level won't be that high. Which leads us to a conclusion that, the lesser the kidney function level is, the higher the creatinine level.

On the other hand for the blood urea nitrogen (BUN) is the most stable blood constituent following death as it reaches antemortem (before death) levels and even after moderate decomposition.

Therefore, as long as the kidneys are functional, regardless of other factors such as the patient's condition/ diagnosis, these blood constituents are nearly normal.
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
The second generation phenotypes resulting from the cross of purebred monohybrid pollination will display a ratio of _____. 2:2
Art [367]
The answer is 3:1.

If we imagine that plant has two alleles for the trait, we can dominant allele represent with P represents and recessive allele with p. To get purebred monohybrid in the first generation, parents must be a dominant homozygote (PP) and a recessive hetero<span>zygote (pp):

Parental generation: PP   x   pp
The first generation: Pp  Pp  Pp  Pp

Pp represents a heterozygote.
If we cross these heterozygotes:

The first generation: Pp   x   Pp
The second generation: PP  Pp  Pp  pp

If dominant allele determines the phenotype, there will be 3 plants (one PP and two Pp) with one phenotype and only 1 plant </span><span>(pp)</span> with another phenotype and vice versa.
5 0
3 years ago
Read 2 more answers
What mining type is aluminum
kykrilka [37]

Extraction of aluminium metal takes place in three main stages¿mining of bauxite ore, refining the ore to recover alumina and smelting alumina to produce aluminium. Bauxite is mined by surface methods (open-cut mining) in which the topsoil and overburden are removed by bulldozers and scrapers.
8 0
3 years ago
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