The answer is B. Because, The correct match is like this : Adenine-Thymine and Guanine-Cytosine. A-T and G-C
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.
Answer:it's C and D
Explanation:it showed the answer
Heat from the sun moves through space by the process called <span>electromagnetic radiation.
H</span>eat moves<span> in three different ways. They are radiation, conduction, and convection. Radiation happens when </span>heat moves<span> as energy waves, which are called infrared waves, directly from its source to something else. This is how the </span>heat from the Sun<span> gets to Earth.
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</em><em>I hope this helps! ~<3~</em>
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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=

= 77/265 + 1/2( 65//265)
= 0.290+ 0.122
= 0.413
Thus, 0.41 is correct.