This biology question covers important biological concepts and processes. The step-by-step explanation below helps you understand the underlying mechanisms and reasoning.
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You're on a roll — Question 2.1 2.1.1 The stage of protein synthesis shown is Translation. 2.1.2 Molecule X is amino acids. Molecule Y is mRNA (messenger RNA). 2.1.3 The term for the group of three nitrogenous bases indicated by V (on tRNA) is an anticodon. 2.1.4 If molecule Y (mRNA) has the bases UAU, then the complementary bases on the DNA template strand would be ATA. (Adenine pairs with Uracil in RNA, Thymine pairs with Adenine in DNA). 2.1.5 The mRNA codon shown in the diagram for amino acid W is UAU. From the table, the tRNA anticodon AUA codes for tyrosine. 2.1.6 The process that occurs in the nucleus to produce molecule Y (mRNA) is Transcription. During transcription, the DNA double helix unwinds and unzips. One strand of the DNA acts as a template. Free RNA nucleotides in the nucleus pair with their complementary bases on the DNA template strand (A with U, T with A, C with G, G with C). These RNA nucleotides then join together to form a single strand of mRNA, which then detaches from the DNA and leaves the nucleus. Question 2.2 2.2.1 *a) Molecule X is DNA. *b) Organelle Y is a ribosome. 2.2.2 In Process W, the mRNA codon shown is AGA. The tRNA anticodon that pairs with AGA would be UCU. Therefore, nitrogenous base 1 is Uracil (U) and nitrogenous base 3 is Uracil (U). 2.2.3 During transcription, DNA serves as a template for the synthesis of messenger RNA (mRNA). It carries the genetic code that determines the sequence of amino acids in a protein. 2.2.4 Process W is Translation, which occurs at organelle Y (the ribosome). During translation, the mRNA molecule attaches to a ribosome. Transfer RNA (tRNA) molecules, each carrying a specific amino acid, move to the ribosome. The anticodon on the tRNA pairs with the complementary codon on the mRNA. As the ribosome moves along the mRNA, amino acids are linked together by peptide bonds, forming a polypeptide chain (protein). 2.2.5 A DNA mutation involves a change in the sequence of nitrogenous bases in the DNA. This change can lead to a different codon being transcribed onto the mRNA. A different mRNA codon may result in a different amino acid being incorporated into the polypeptide chain during translation. This alteration in the amino acid sequence can change the protein's primary structure, which in turn affects its folding (secondary, tertiary, and quaternary structures), ultimately leading to a change in the protein's overall three-dimensional shape and function. 2.2.6 Step 1: Apply Chargaff's rules for DNA base pairing. In DNA, the amount of Adenine (A) is equal to the amount of Thymine (T), and the amount of Guanine (G) is equal to the amount of Cytosine (C). Given that Thymine (T) = 28%. Therefore, Adenine (A) = 28%. Step 2: Calculate the total percentage of A and T. Percentage of (A + T) = 28% + 28% = 56%. Step 3: Calculate the remaining percentage for G and C. Percentage of (G + C) = 100% - 56% = 44%. Step 4: Calculate the percentage of Cytosine (C). Since G = C, the percentage of Cytosine is half of the (G + C) total. Percentage of C = (44\%)/(2) = 22\%. The percentage of nucleotides which contain cytosine is 22\%. Send me the next one 📸