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Illusion [34]
2 years ago
9

Historically, the x chromosome had more genes assigned to it than its physical size would seem to warrant relative to the other

chromosomes. Can you suggest another explanation as to why this was the case?
Biology
1 answer:
sweet-ann [11.9K]2 years ago
5 0

Answer:

Humans contain 46 chromosome or 23 pairs of chromosomes. Among them  22 pairs are autosomes and one pair contains the sex chromosome. Humans female are XX and human males are XY.

The X chromosome determines the sex of an individual and females one X chromosome is randomly inactivated. The human X chromosome has large number of genes and contains large euchromatic regions that are highly activating gene regions. The X chromosome also contains the conserved regions that are far shorter in Y chromosome. The X chromosome is large in size and has large number of genes.

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Which option uses the monitoring method to check the growth of invasive species? A introduction of a predator species B digging
tatiyna

<span>The answer is A. Invasive species are species that thrive unregulated in an introduced environment/ habitat and affect the biodiversity of the environment. They boom unchecked mostly due to lack of a natural predator. Therefore, introducing a predator will limit their growth. </span>






4 0
1 year ago
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Drag the labels onto the equation to identify the inputs and outputs of photosynthesis.
natta225 [31]
Inputs: carbon dioxide, water, light(photon) 
<span>outputs: carbohydrate(usually glucose), oxygen, heat

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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2 years ago
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Suppose the size of a population of marmots is 300. According to genetic drift theory, what is the probability that a newly aris
arsen [322]

Answer:

E. 1/600

Explanation:

Hint:

The probability of fixation of a new neutral mutation is 1/(2N)

Given N as 300

= 1/(2×300)

=1/600

Therefore,

1/600 gives a sure fixation of one allele from the large population

5 0
1 year ago
Suppose that a scientist deletes the AAUAAA consensus sequence, the poly(A) tail, or the 5' cap from pre‑mRNAs in an mRNA. She t
Andrej [43]

Answer:

<h2>AAUAA deletion- (B ) </h2><h2> </h2><h2>Poly(A) tail deletion- (C) </h2><h2> </h2><h2>5' cap deletion-  (A) </h2>

Explanation:

(A) introns are not removed from the pre-mRNA due to 5' deletion;

(B) the pre-mRNA is not cleaved at the cleavage site due to AAUAA deletion

(C) the mRNA is not transported to the cytoplasm due to poly A tail deletion;

AAUAA deletion:  if we delete the AAUAA  sequence then there is the change of sequence of pre- mRNA that the pre-mRNA is not cleaved at the cleavage site .

Poly(A) tail deletion: Poly(A) tail is the long tail of Adenine in the 3' end of mRNA, after the deletion of Poly(A) tail, it affects the transport of mRNA from the nucleus to the cytoplasm and  the mRNA is not transported to the cytoplasm if tail is deleted.

5' cap deletion; the effect of 5' cap deletion is that introns are not removed from the pre-mRNA. The process of intron removing and exon joining is called RNA splicing.

3 0
1 year ago
1. You are studying an integral membrane protein of the plasma membrane. It has a single transmembrane domain and the N-terminal
Drupady [299]

Answer:

A 22 to 25 amino acid sequence present in the central section of the protein, which gives rise to an alpha helix in the membrane is known as the stop-transfer anchor sequence. The sequence plays an essential function in targeting the protein towards the plasma membrane. On the other hand, it also ceases targeting of the protein towards the endoplasmic reticulum, which was started by the signal peptide.  

Thus, the process of translation of the remaining of the protein occurs within the cytosol due to the tethering of the transmembrane domain. In the stop-transfer anchor sequence, the hydrophobic amino acids present are isoleucine and valine. After mutation, these amino acids get converted into arginine and lysine, thus, hydrophilic amino acids replace hydrophobic amino acids in the sequence.  

Due to this, the transmembrane domain cannot be targeted towards an integral part of the plasma membrane by the short transfer anchor sequence, and therefore, now the translocation of the protein will take place towards the endoplasmic reticulum as initiated by the signal peptide at the beginning.  

3 0
2 years ago
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