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AveGali [126]
3 years ago
9

PLEASE solve number 1 and 2 because i don’t get it , don’t mind my answers

Biology
1 answer:
Readme [11.4K]3 years ago
8 0

Answer:

1) RNA strand =  <em>UCA  GCA  AAU</em>

<em>2) </em>Amino acids:   Asparagine       Arginine             Stop

                            (ASN)               (ARG)              (STOP)

Explanation:

Some theory that you need to know to answer the question:

Each tRNA has two important sites. One of them that couples with the codon of the mRNA molecule, named <em>anticodon</em><em>.</em> The other site couples with an amino acid through the action of the aminoacyl-tRNA synthase enzyme. The whole complex, amino acid + enzyme + tRNA is named aminoacyl-tRNA. Considering that there are twenty amino acids available, there are also twenty complexes of aminoacyl-tRNA, one for each amino acid.  

tRNA allows amino acids to align according to the nucleotidic sequence in the mRNA molecule.  

Once the new amino acid joins the growing peptidic chain, the binding between the amino acid and the tRNA molecule breaks. The tRNA is now free to join another amino acid and repeat the cycle.  

Now, to fill in the boxes, you need to pay attention to two things:

• <u>Anticodon</u>: In this box, you need to place the sequence of bases that pairs with the mRNA codon.  

• <u>Amino Acid</u>: In this box, you need to place the amino acid that is codified by the<u> mRNA </u>codon.  

For instance, let us say that the codon of mRNA is CUC. The anticodon for this sequence is GAG (because guanine joins cytosine and adenine joins uracyl).  

The amino acid places on the 3´extreme of the tRNA molecule will be the one that is codified by the codon, NOT the anticodon. So, as the codon is CUC, the amino acid will be Leucine, Leu. Do not use the anticodon GAG to fill these boxes. Remember that tRNA is translating information carried by the mRNA, so we need to use this information to grow the protein.

<em />

1)

Strand of DNI  →  AGT  CGT  TTA

  • Each of these letters are bases → Adenine, A - Guanine, G - Thymine, T - Cytocine, C
  • Each of these three-lettered groups are codons.

So, we need to pair these bases, knowing that:

Adenine (A) and guanine (G) derive from purines, while Thymine (T) Uracile (U), and Cytosine (C) derive from Pyrimidines.  

In the DNA molecule, Adenine (Purine) always pairs with Timine (Pyrimidine), while Guanine (Purin) forms pairs with Cytosine.

In the RNA molecule,<em> Adenine is paired with Uracil, U. </em>So, whenever there is an Adenine in the DNI strand, it pairs with Uracil of RNA strand. If there is a Thymine in the DNI strand, it pairs Adenine of the RNA strand. And, as always, Guanine pairs with Cytosine.

  • A ⇒ U
  • T ⇒ A
  • G ⇔ C

Knowing how to pair bases, we can fill the RNA molecule boxes.

  • Strand of DNI  →  AGT                   CGT              TTA
  • <em>Strand of RNA →  UCA                  GCA              AAU</em>

<em>                                      A ⇒ U              C ⇒ G            T ⇒ A</em>

<em>                                     G ⇒ C               G ⇒ C            T ⇒ A</em>

<em>                                     T ⇒ A                T ⇒ A            A ⇒ U</em>

2)

Now we need to translate these new codons of mRNA to amino acids.

<em>To fill the boxes you should have a genetic code chart to see the codons that codify for each amino acid. </em>

Codon:                 AAU                 CGA                UAG

Amino acids:   Asparagine       Arginine             Stop

                            (ASN)               (ARG)              (STOP)

<em>The new protein is synthesized from the amino terminus to the carboxy terminus of the mRNA molecule. The added amino acids to the chain are coded by a codon formed by three bases in the mRNA. mARNs also have a start and end codon that are the signals of the synthesis initiation and finish. When the ribosome reaches a stop codon, protein synthesis is over.  </em>

<em />

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A couple in Indiana are each heterozygous for the autosomal recessive allele for albinism. They have two children. What is the p
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Probability that both children will be phenotypically identical with regards to skin color is 10/16.

Explanation:

The couple has a heterozygous genotype for the recessive allele. Representing the recessive allele with 'a' and the dominant allele with 'A', we can write the heterozygous genotype for the couple as Aa.

The gene pool of the children for this couple can be represented as:

Alleles       A        a

A              AA       Aa

a               Aa       aa

There can result 4 combinations of alleles AA,Aa,Aa and aa. The recessive allele is responsible for albinism so the child with the allele set 'aa' will have albinism while the others, AA,Aa and Aa will show no signs of albinism since the recessive allele is responsible for it.

The probability that both children will be phenotypically identical can be calculated by first considering that the children show albinism and then consider the probability that they have a normal phenotype.

Probability that a child will have albinism (aa) = 1/4

Probability that a child will not have albinism (AA, Aa and Aa) = 3/4

Probability that both children have albinism = 1/4 x 1/4 = 1/16

Probability that both children will have no albinism = 3/4 x 3/4 = 9/16

Probability that both children will be phenotypically identical = 1/16 + 9/16

                                                                                                      = 10/16

Probability that both children will be phenotypically identical with regards to skin color is 10/16.

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