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deff fn [24]
3 years ago
15

Animals cannot synthesize tyrosine from shikimate-3P because they lack EPSP synthase. However, tyrosine is listed as a non-essen

tial amino acid in most tables. What is the explanation?
Biology
1 answer:
Marina86 [1]3 years ago
8 0

Tyrosine is listed as a non-essential amino acid because it can be synthesised from the essential amino acid phenylalanine.

<u>Explanation:</u>

Amino acids are classified as essential and non essential according to whether they can be synthesized in the body itself or should be supplied by the diet. Essential amino acids cannot be synthesized in the body of the organism and thus can be obtained only from the diet. In the case of human beings, valine, phenylalanine, methionine etc are examples of essential amino acids.

Non essential amino acids can be synthesized within the body of the organism. Thus they need not be supplied by the diet. Alanine, aspartic acid etc are some of the non essential amino acids in human beings. It is given here that animals cannot synthesize tyrosine from shikimate-3p due to the lack of EPSP synthase.

But the amino acid tyrosine can be synthesized from the amino acid phenylalanine which is an essential amino acid. Thus tyrosine will be a non-essential amino acid.

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<span>The propositions are:
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d. stores vitamin D
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The right answer is: B. </span>does all of these

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4 0
3 years ago
I WILL GIVE BRAINLIEST!!!
Nesterboy [21]

Answer:

25%

Explanation:

When looking at a pedigree remember that:

  • squares are males
  • circles are females
  • the solid colored figure represents an individual affected by a disease
  • the empty figure represents a healthy individual

Let us assign the symbol X⁺ to represent the dominant allele linked to the X-chromosome and expressing healthiness, and X⁻ to represent the recessive allele expressing the dissease.

According to this pedigree

  • I1 is a man affected by the disease, YX⁻
  • I2 is a healthy woman X⁺X⁻
  • we can see that among the progeny (generation II) there are two individuals affected (a boy and a girl) and one healthy girl. This means that the mother I2 is heterozygous for the trait.

So, having their genotypes we can know what are the probabilities of getting a son with DMD

Parentals)    YX⁻     x     X⁺X⁻  

Gametes)   Y     X⁻      X⁺     X⁻

Punnett square)

                        X⁺             X⁻

            X⁻      X⁺X⁻         X⁻X⁻    

            Y        X⁺Y           X⁻Y

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  • The probabilities of getting a healthy daughter X⁺X⁻ are 25%
  • The probabilities of getting a healthy son X⁺Y are 25%
  • The probabilities of getting a daughter with DMD X⁻X⁻ are 25%
  • The probabilities of getting a son with DMD X⁻Y are 25%
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In short, Your Answer would be 50x

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Answer:

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