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Zarrin [17]
3 years ago
12

Imagine that you've discovered a regulatory sequence thousands of base pairs away from the promoter and that when this sequence

is lost due to mutation, levels of gene expression decrease sharply. This sequence is most likely ______.
A. A silencerB. An enhancerC. A promoter-proximal elementD. A promoter
Biology
1 answer:
Tresset [83]3 years ago
4 0

Answer:

An Enhancer

Explanation:

Enhancers are the regulatory sequences and are generally located 80-160 base pairs upstream of the transcription start site. Enhancers have bidirectional elements that are involved in the initiation of transcription. Enhancers serve as the binding sites for activators to increase the concentration of activators in the vicinity of the promoter and thereby increase the rate of transcription of the gene.

According to the given information, the loss of a regulatory sequence led to a reduced level of transcription of the gene. This means that the regulatory sequence was involved in the upregulation of transcription. The presence of the sequence far away from the promoter makes it an enhancer.

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MariettaO [177]

Answer:

temperature and pressure is the answer

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2 years ago
Which of the following is NOT a benefit of GMOs They can resist disease. They require less fertilizer. They grow faster. They cr
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They create super weeds would be the correct answer
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Both food webs and energy pyramids show relationships between organisms in an ecosystem. A pyramid is preferred when showing the
dolphi86 [110]

Answer:

individual feeding relationships between organisms.

Explanation:

When we try to explain the abundance accumulated by each link in the food chain, it can be observed that both the magnitude of individuals and biomass decrease rapidly from plants to predators. By sorting the information in this way, we get a graph in the form of a "pyramid".

3 0
3 years ago
A proton has a speed of 3.0 × 106 m/s in a direction perpendicular to a uniform magnetic field, and the proton moves in a circle
jeyben [28]

Answer:

The magnitude of the magnetic field is 0.156 T.

Explanation:

Given that,

Speed of a proton, v=3\times 10^6\ m/s

Radius of the circular path, r = 0.2 m

When the proton enters in the circular path, the centripetal force is balanced by the magnetic force such that :

\dfrac{mv^2}{r}=qvB\sin\theta

q is the charge on proton

Here, \theta=90^{\circ}

\dfrac{mv^2}{r}=qvB\\\\B=\dfrac{mv}{qr}\\\\B=\dfrac{1.67\times 10^{-27}\times 3\times 10^6}{1.6\times 10^{-19}\times 0.2}\\\\B=0.156\ T

So, the magnitude of the magnetic field is 0.156 T. Hence, this is the required solution.

3 0
3 years ago
If a DNA sequence is TTA, what would be the anticodon the tRNA would bring to the ribosome?
agasfer [191]

Answer:

So the central dogma of molecular biology describes the journey from DNA to protein product:

DNA --transcription--> mRNA --translation--> Protein

Assuming the DNA sequence provided is the template strand (rather than the complimentary coding strand), we start by transcribing the sequence into mRNA starting on the 3' end of the DNA towards the 5' end (which would build the mRNA 5' to 3'). This process involves the enzyme "RNA polymerase," which can only add nucleotides to the 3' end of the mRNA, just like how DNA polymerase can only synthesize DNA in the 5' to 3' direction. The RNA polymerase will bind to the template DNA strand and synthesize the complimentary mRNA, substituting uracil for thymine (since RNA does not contain thymine like DNA).

In terms of transcribing the sequence given to you, we'll have to work backwards + flip it around to get the 5' to 3' mRNA since the DNA is given 5' to 3' rather than 3' to 5'. Due to the length and the fact that we'll have to use triplets in translation anyways, it can help to break the sequence into triplet codons now.

5’-AAG | TTA | ATG | AGA | AAT | CGA | CAT | GGG | GCG | CCG | AAA | GTA | TAA | CCG | TCT | TAG | AAT | AGC-3’

We can then cross out each codon as we transcribe it and flip the sequence to be 5'-3' mRNA:

mRNA: 5'- GCU | AUU | CUA | AGA | CGG | UUA | UAC | UUU | CGG | CGC | CCC | AUG | UCG | AUU | UCU | CAU | UAA | CUU -3'

Normally, mRNA sequences start with "AUG" which is the start codon (and codes for Methionine), but I'll assume this is just for practice translating + transcribing in general. There's also a stop codon before the end but I'll assume the same again.

Translation involves three main steps - initiation, elongation, and termination. Initiation involves the translation ribosome assembling around the mRNA starting at the 5' end start codon, and tRNA carrying an amino acid binding to the complimentary section of the mRNA. As each tRNA attaches and the ribosome moves along the mRNA, the amino acids on each tRNA are bonded into a longer and longer peptide chain and the now amino acid-less tRNA are ejected (elongation). Termination occurs when a stop codon is reached, the ribosome will end elongation and help fold the protein into its final structure.

To translate the mRNA sequence here we'll need an amino acid/mRNA codon chart. I don't believe I can attach an image here, but looking up those exact words should yield the right results in images.

5'- GCU | AUU | CUA | AGA | CGG | UUA | UAC | UUU | CGG | CGC | CCC | AUG | UCG | AUU | UCU | CAU | UAA | CUU -3'

Ala - Ile - Leu - Arg - Arg - Leu - Tyr - Phe - Arg - Arg - Pro - Met - Ser - Ile - Ser - His - STOP - Leu

Amino acids are often abbreviated into three letters (Ala = alanine, Met = methionine, etc), and sometimes are abbreviated as single letters, though I've only seen that for sequencing databases.

In terms of locations for each of these processes, transcription occurs in the nucleus for eukaryotes and translation in the ribosomes/cytoplasm.

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3 years ago
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