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andrew-mc [135]
3 years ago
5

Some proteins are composed of two or more polypeptides. Suppose the DNA template strand sequence 3'- TACGTAGGCTAACGGAGTAAGCTAACT

-5' produces a polypeptide that joins in pairs to form a functional protein.
a) What is the amino acid sequence of the polypeptide produced from this sequence?
Biology
1 answer:
alexandr402 [8]3 years ago
4 0

Answer:

Met-his-Pro-Ile-ala-Ser-ala-asp

Explanation:

The protein molecules are synthesized from the mRNA molecule synthesized from the DNA by the transcription. The translation process then translates the mRNA to the amino acids which then join together and form the proteins.

In the given question, the template strand of the DNA is provided is

Template DNA- 3'- TAC-GTA-GGC-TAA-CGG-AGT-AAG-CTA-ACT -5

 mRNA- 5'-AUG-CAU-CCG-AUU-GCC-UCA-UUC-GAU-UGA-3'

 Protein-      Met-his-Pro-Ile-ala-Ser-ala-asp-stop codon

The translation process begins when the tRNA molecule recognizes the AUG or start codon and then reads the nucleotide in the triplets. Each triplet code for the specific amino acid.On the basis of this, amino acid chain is constructed.

Thus, Met-his-Pro-Ile-ala-Ser-ala-asp is correct.

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The following two genotypes are crossed: AaBbCc X+Xr AaBBcc X+Y, where a, b, and c represent alleles of autosomal genes and X+an
choli [55]

Answer:

1/64

Explanation:

The following genotypes were crossed: AaBbCc X+Xr and AaBBcc X+Y

If we asume that the autosomal genes are in different chromosomes, then they will assort independently during meiosis and gametogenesis. In that case, we can use the Multiplication Rule of Probability to obtain the probability of having a specific genotype in the progeny.<u> This rule states that when two or more events are independent, the probability of all them happening at the same time will be the result of the multiplication of the individual probabilities of each event.</u>

<u />

We can separate each gene in the cross to determine the genotypic ratios in the offspring, and then multiply the probabilities of the aaBbCc X+X+ genotypes to obtain the overall probability of having progeny with that genotype.

<h3><u>Aa x Aa </u></h3>

1/4 AA

2/4 Aa

1/4 aa

<h3><u>Bb x BB </u></h3>

1/2 BB

1/2 Bb

<h3><u>Cc x cc </u></h3>

1/2 Cc

1/2 cc

<h3><u> X+Xr x X+Y</u></h3>

1/4  X+X+

1/4  X+Y

1/4  X+Xr

1/4  XrY

The probability of having offspring with the aaBbCc X+X+ genotype will be: 1/4 × 1/2  × 1/2  × 1/4 = 1/64

5 0
4 years ago
PLZ HELP!!!!
sweet-ann [11.9K]

<span>Only a single amino acid could change, because the reading frame is unaffected</span>

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