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jeka94
2 years ago
15

Transamination of an amino acid transfers an amine group to form an α‑keto acid and is catalyzed by transaminases. Some amino ac

ids can be transaminated into compounds that can be used directly in biological pathways, whereas other amino acids need to be converted through several steps. Match each amino acid to the appropriate category.
a. Yields a glycolytic or citric acid intermediate directly upon transamination

b. Requires several steps to yield a glycolytic or citric acid intermediate
Chemistry
1 answer:
iragen [17]2 years ago
7 0

Answer:

Amino acids that fall under the first category are alanine, aspartate and glutamate. While amino acids that fall under the second category are glycine, valine, proline, leucine, isoleucine, methionine, serine, threonine, cysteine, asparagine, glutamine, phenylalanine, tryptophan, tyrosine, lysine, arginine and histidine

Explanation:

Alanine, a 3-carbon amino acid reacts with α-ketoglutarate, a 5-carbon ketoacid, to produce pyruvate, a 3-carbon compound which is one of the glycolytic products in aerobic respiration, and glutamate, a 5-carbon amino acid, with the aid of alanine transaminase (ALT). The amino group from alanine is transfered from the α-carbon of alanine to the α-carbon of the α-ketoglutarate.

Aspartate, a 4-carbon amino acid also reacts with α-ketoglutarate to form oxaloacetate, a 4-carbon ketoacid which is present as an intermediate in the citric acid cycle, and glutamate. The amino group transfer occurs between the α-carbon of aspartate and the α-carbon of α-ketoglutarate.

Unlike the presence of a ketoacid in the conversion of alanine and aspartate to their corresponding amino acids, glutamate conversion to α-ketoglutarate, an intermediate in the citric acid cycle, involves no ketoacids as an amino group acceptor from glutamate. The amino group is freely released as an ammonium ion. The reaction involves the presence of a cofactor, NAD+ or NADP+, water (H2O) and glutamate dehygrogenase (GDH).

Other amino acids involve several metabolic steps to be converted into glycolytic or citric acid intermediates.

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Let us differentiate accuracy from precision. Accuracy is the nearness of the measured value to the true or exact value. On the other hand, precision is the nearness of the measured values between each other. So, for precision, select the student in which the measured values are very near to each other. That would be Student III. Now, for accuracy, let's find the average for each student.

Student I: (<span>8.72g+8.74g+8.70g)/3 = 8.72 g
Student II: (</span><span>8.56g+8.77g+8.83g)/3 = 8.72 g
Student III: (</span><span>8.50g+8.48g+8.51g)/3 = 8.50 g
Student IV: (</span><span>8.41g+8.72g+8.55g)/3 = 8.56 g

From the given results, the accurate one would be Students I and II. So, we make a compromise. Even though Student III is precise, it is not accurate. If you compare between Students I and II, the more precise data would be Student I. Therefore, the answer is Student I.</span>
6 0
3 years ago
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Which of the following pairs of reactants is not correctly listed with its product solution? A. Strong base-strong acid reactant
alexandr1967 [171]

Answer: Option B

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2 years ago
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<h3> <u>Answer</u></h3>

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Which pair shares the same empirical formula?
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Answer:- 3. CH_3 and C_2H_6

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For example, the molecular formula of benzene is C_6H_6 . The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.

In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and  the empirical formula is CH_2 . In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.

In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is CH_2 . The C to H ratio for second molecule is 1:4, so the empirical formula is CH_4 . Here also, the empirical formulas are not same and hence it is also not the right choice.

In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is CH_3 . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also CH_3 . Hence. this is the correct choice.

In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is CH_2 . As the empirical formulas are different, it is not the right choice.

So, the only and only correct pair is the third one. 3. CH_3 and C_2H_6

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3 years ago
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3.21e+11 is your answer for 321 millimeters in picometers. Hope this helps!

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