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laila [671]
2 years ago
13

How many unique gametes could be produced aabbccddee?.

Biology
1 answer:
Kazeer [188]2 years ago
8 0

The number of different possible gametes produced by the diploid genotype (AaBbCcDdEe) is 2 x 2 x 2 x 2 x 2 = 32 (2 for each pair of heterozygous genes).

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What definition goes with what cell?
snow_tiger [21]

━━━━━━━☆☆━━━━━━━

▹ Answer

<em>See below.</em>

▹ Step-by-Step Explanation

Genes - parts of the DNA which control the traits of an organism

DNA - molecule that carries the genetic instructions for all living organisms. Found in the chromosome.

Nucleus - stores the cell's hereditary material or DNA

Chromosomes - Thread-like structures that are found in the nucleus of a cell that contains DNA.

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

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8 0
3 years ago
A marine fish has about 1% salt in his body. The ocean water is 3.5%. Will water move into or out of the fish?
timurjin [86]
The water will move out of the fish by osmosis  due to existence of concentration gradient .
6 0
3 years ago
Read 2 more answers
imagine a cell-surface receptor protein is being newly synthesized. How can it get into the membrane that it will ultimately be
motikmotik

Answer:

the steps for a new receptor synthesis:

Explanation:

Do not forget the steps vary from an eukarityc cell to prokarityc one.  

IN AN EUKARYOTIC CELL:

1- Transcription: in the nucleus, mRNA is synthesized from the DNA sequence that codifies the protein. Exons are spliced if needed by spliceosomes.  

2- From nucleus into cytoplasm: mRNA goes through nuclear membrane into the cytoplasm were ribosomes, substrates for protein synthesis and other organelles are.

3- Ribosomes recognize a signal in mRNA and start reading the genetic code, three consecutive nucleotides at a time and adjust the correspondent tRNAs (tRNAs carry the aminoacids)  

4- Peptide bonds are formed between consecutives aminoacids, and a polypeptide is build up according to the mRNA sequence.

5- (The mRNA has a signal that is recognized by the rough endoplasmic reticulum) The peptide goes into this reticulum (RER) and then into Golgi apparatus. (more chemical modifications are made if needed). By this moment the protein has an inserted portion in a lipid membrane.

6- The protein is transported in a vessel that finally fusions with cytoplasmic membrane and the receptor is in destiny. Hydrophobic aminoacid sequences are inserted in membrane.

IN A PROKARYTIC CELL:

1- Transcription (no splicing is needed, there are not exons)

2- There are not nucleus and cytoplasm divided (no transport)

3- Ribosomes recognize a signal in mRNA and start reading the genetic code, three consecutive nucleotides at a time and adjust the correspondent tRNAs (tRNAs carry the aminoacids)

4- Peptide bonds are formed between consecutives aminoacids, and a polypeptide is build up according to the mRNA sequence.

5- The mRNA has a signal that is guides the growing protein to cellular membrane. Hydrophobic aminoacid sequences are inserted in membrane.

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 sole of a gecko's foot is covered with millions and millions of small, dry "hairs" that make direct contact with surfaces, a
tamaranim1 [39]

Answer: Van der Waals forces

Explanation:

Van der Waals forces are weak intermolecular forces that depend on the distance between two particles. They are caused by correlations in the change in polarization between two nearby particles. To put it in other words, when a particle changes its polarization (becomes more positive on one end and more negative on the other), so does the adjacent particle, and the next one, and so on. This causes these particles to stick together weakly.

The tiny "hairs" increase the surface area of the gecko's feet in contact with the wall, which makes the bond stronger and allows it to support all of its weight.

Because experiments have shown that geckos stick well to both hydrophobic and hydrophilic surfaces, we can assume there aren't any hydrogen bonds present.

Ionic bonds can't be present either because geckos wouldn't stick to electrically neutral surfaces, as these bonds require charged molecules.

6 0
3 years ago
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