Question

Explain the role of NADH and FADH2 in cellular respiration

Explain the orientation of the large and small ribosomal subunits, as well as the initiator tRNA during initiation of translation

Describe how the release factor separates the amino acid chain and ribosomal subunits to terminate translation

Describe why mitosis would not be effective for the production of new gametes (

Answer 1

Answer-

According to the given question-

Question 1-

Role of NADH and FADH2 in cellular respiration-

We know that During oxidative phosphorylation NADH and FADH2 Molecule get oxidized in the electron transport chain with 4 protein complexes found in electron transport chain. 2 NADH molecule was produced during Glycolysis, 2 NADH from oxidation of 2 molecule of pyruvate to 2 molecule of Acetyl Co A and 6 molecule of NADH and 2 Molecule of FADH2 are produced during citric acid cycle. now these molecule of NADH and FADH2 enters in Electron transport chain and where they get oxidized into NAD+ and FADH respectively. Both NADH and FADH2 donate electrons to electron transport chain which are used for reduction of oxygen into water.

Question 2-

orientation of the large and small ribosomal subunits, as well as the initiator tRNA during initiation of translation-

We know that during translation the information present in mRNA in the form of codon are converted into amino acid sequence.  adaptor molecules transfer RNA or tRNA recognizes  the codon on mRNA and also at the same time recognizes as well as binds with amino acid .Enzyme called  Aminoacyl-tRNA synthetases covalently attaches amino acid to the tRNA molecule by using ATP.   ribosome generally have four binding sites in which one was bind by mRNA and remaining three other site called E-site, P-site and A-site, for binding of tRNA. tRNA carry amino acids to mRNA , ribosome’s , initiation factors, elongation factors, as well as release factors are needed for translation. Initiation of translation starts when small subunit of ribosomes binds with the charged initiator tRNA carrying initiator amino acid Methionine. this complex binds at the 5’ end of mRNA and start finding the start codon with the help of initiation factor followed by binding of large subunit of ribosomal and this causes releases of initiation factors. Large subunit of ribosome’s have A-site where new tRNA carry molecule with charged amino-acid. peptide bond formation takes place at P-site and also for  growing polypeptide chain while E-site is where tRNA exit .

Question 3-

Role of release factor in translation termination-

The elongation of Polypeptide chain occur till the ribosome find termination codon such as either from UAA, UAG or UGA in  mRNA template strand. We know that tRNAs does not have anti-codons which are complementary to these termination codons. So when ribosome recognizes the termination codon, then this codon is also recognized by release factors or RF which is a protein and bind to the A site of ribosome. Release Factor 1 responsible for identifying UAA and UAG termination codons and Release Factor 2  UGA codon. while Release Factor 3 helps Release Factor 1 and Release Factor 2. and this causes termination of translation, as well as release of polypeptide chain from tRNA with the help of enzyme  peptidyl transferase , followed by dissociation large and small  subunits of ribosome from mRNA with the help of Initiation Factor 3.

Question 4-

We know that a cell can divide by two ways in humans as well as in most animals such as  mitosis and meiosis. If a cell divides by following mitosis then it produces two copies of identical clone which means that both the cell have same chromosome number. but when a cell following meiosis division then  it produces four cells and they termed as gametes and these gametes are not copy or clones to their parent cell because they have half number of chromosomes than their parent cell. We know that gametes are haploid so for there formation meiosis is necessary.