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makkiz [27]
3 years ago
13

What do most experiments test? 1:Scientific theories 2:Falsifiability of hypotheses 3:Accuracy of information 4:Predictions base

d on hypotheses
Biology
1 answer:
saul85 [17]3 years ago
6 0
2 is the correct answer
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A molecule that can be used as a molecular clock has a neutral mutation rate of one mutation per 5 million years. How many years
zubka84 [21]

Answer:

20 million years

Explanation:

If we have a neutral mutation rate of one mutation per 5 million years, then the total of eight mutation between the two different species would be 20 million years. This is because both species will have 4 mutations in those 20 million years, so combined, both by 4, will have 8 mutations between them. So few mutations on so much time will result in two species that are very similar to each other even after 20 million years of evolution, even making them hardly distinguishable, especially if it comes to defining fossil records from them both. A nice example of this are the members of the felidae (cat) family, which are all very closely related, and are almost identical, thus making it extremely hard to distinguish two species of the same or similar size by their fossils.

7 0
3 years ago
4. Which of the following statements about magnetic materials TRUE?
Pani-rosa [81]

Answer:

All of them seems true tho

4 0
3 years ago
Read 2 more answers
Brainliest Answer Contest Whoever answers with the best explanation gets Brainliest Answer and 30 Points
posledela
Isolation can be due to behavioral, geographical, or temporal barriers. <span>Speciation can take place in two general ways. A single species may change over time into a new form that is different enough to be considered a new species. This process is known as anagenesis. More commonly, a species may become split into two groups that no longer share the same gene pool. This process is known as cladogenesis. There are several ways in which anagenesis and cladogenesis may take place. In all cases, reproductive isolation occurs. Hope this helps! </span>
8 0
3 years ago
The mitochondrial matrix has what charge compared with the intermembrane space?
Solnce55 [7]
The matrix (inside mito) has a negative charge compared to the intermembrane space (outside of mito but still inside) because positively charged protons are constantly pumped from the matrix to the intermembrane space. A cell oxidizes an organic molecule producing 3 NADH molecules and 3 FADH2 molecules..
7 0
2 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
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