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WINSTONCH [101]
3 years ago
12

Blood pressure decreases below normal levels. Blood flow to the heart decreases. Heart is unable to pump as much blood. Blood pr

essure decreases even more. This is an example of _____ feedback. Multiple choice question.
Biology
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

its called systolic heart faliure or heart faliure with reduced ejection fraction

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3' GCTTAGTCCG 5' Which of the following nucleotide base sequences complements the section of DNA modeled above? A. 3' TAGGCTGAAT
Studentka2010 [4]
The answer is B. Because, The correct match is like this : Adenine-Thymine and Guanine-Cytosine. A-T and G-C
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3 years ago
Explain why serine proteases do not catalyze hydrolysis if the amino acid at the hydrolysis site is a D-amino acid. Trypsin, for
kotykmax [81]

Answer:

Explanation:

A protease is an enzyme that catalyzes the hydrolysis of the peptide bonds that tie polypeptide chains together, releasing individual amino acid subunits. The L and D nomenclature for amino acids defines the structure of the glyceraldehyde isomer through which the amino acid can be produced.

SEE BELOW FOR THE APPROPRIATE STRUCTURES.

We need to figure out why swine proteases hydrolyze L-amino acids but not D-amino acids in any way. we know that enzymatic catalysts act as polypeptides if you can recall. They must retain a very precise three-dimensional structure for a catalytic activity to occur. Substrates that do not quite match the required configuration at the active site will not be reacted to — this is a "lock and key" style.

The present exercise may be explained by the fact that the configuration and structure of D-amino acids prevent them from binding properly to the active site of the protease enzyme. Perhaps they're pointed in the wrong direction, or perhaps there happens to be missing electrical interaction that's needed to keep the substrate in position.

Nonetheless, L-amino acids, on the other hand, seem to have the right configurational aspects in the active site and are hydrolyzed.

5 0
2 years ago
ASAP Brainly, right here right now!
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4 0
2 years ago
Heat from the sun moves through space by the process of
lorasvet [3.4K]
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8 0
3 years ago
Read 2 more answers
A population is made up of individuals where 77 have the A1A1 genotype, 65 have the A1A2 genotype, and 123 have the A2A2 genotyp
lana [24]

Answer:

Frequency of allele A1- 0.41

Explanation:

In Hardy weinberg equilibrium,

P refers to the dominant allele

q refers to the recessive allele

The allele frequency will be p+q=1

The genotypic frequency is- P²+q²+2pq=1

P²= genotype of dominant trait ( A1A1)- 77

2pq= genotype of heterozygotes (2pq)- 65

q²= genotype of recessive trait (A2A2)-  123

Total number of offsprings= 77+  65+ 123

                                            = 265

Now to calculate allele frequency of A1=

\dfrac{\text{number of A1A1 offspring}}{ \text{total number of offspring}}+\dfrac{1}{2}(\dfrac{\text{number of A1A2 offspring}}{\text{total number of offspring}})

= 77/265 + 1/2( 65//265)

=  0.290+ 0.122

= 0.413

Thus, 0.41 is correct.

3 0
3 years ago
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