Critical Thinking Questions

No, they cannot be identical because the T nucleotide in DNA is replaced with U nucleotide in RNA and AUG is the start codon.

No, they cannot be identical because the T nucleotide in RNA is replaced with U nucleotide in DNA.

They can be identical if methylation of the U nucleotide in RNA occurs and gives T nucleotide.

They can be identical if de-methylation of the U nucleotide in RNA occurs and gives T nucleotide.

2 because the minimum length of an exon is 500 base pairs. In order to fit all 200 amino acids onto the minimum exon the maximum codon length is 2.5 (500 divided by 200). However codons length must be a whole number.

3 because by the law of degeneracy there is currently 20 times fewer amino acids than are possible as most codons are redundant. There could be up to 400 amino acids with the current codon length of 3.

4 because 4 to the 4th power is 256. 4 to the 3rd power is 64; not enough combinations

5 because 4 to the 4th power is only 256. This is not enough combinations because by the law of degeneracy every amino acid must have at least one redundant codon. With 5 codons 1,024 combinations is more than enough.

The flow of information in HIV is from RNA to DNA, then back to RNA to proteins. Influenza viruses never go through DNA.

The flow of information is from protein to RNA in HIV virus, while the influenza virus converts DNA to RNA.

The flow of information is similar, but nucleic acids are synthesized as a result of translation in HIV and influenza viruses.

The flow of information is from RNA to protein. This protein is used to synthesize the DNA of the viruses in HIV and influenza.

In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted $\text{α<!-- α -->}$, $\text{α<!-- α -->}$, $\text{β<!-- β -->}$, and $\text{β<!-- β -->}$’, comprise the polymerase core enzyme. The fifth subunit, $\text{σ<!-- σ -->}$, is involved only in transcription initiation. The polymerase comprised of all five subunits is called the holoenzyme.

In prokaryotes the polymerase is composed of four polypeptide subunits, two of which are identical. These subunits, denoted $\text{α<!-- α -->}$, $\text{α<!-- α -->}$, $\text{β<!-- β -->}$, and $\text{β<!-- β -->}$’, comprise the polymerase core enzyme. There is a fifth subunit that is involved in translation initiation. The polymerase comprised of all four subunits is called the holoenzyme.

In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted $\text{α<!-- α -->}$, $\text{α<!-- α -->}$, $\text{β<!-- β -->}$, and $\text{β<!-- β -->}$’, comprise the polymerase holoenzyme. The fifth subunit, $\text{σ<!-- σ -->}$, is involved only in transcription initiation. The polymerase comprised of all five subunits is called the core enzyme.

In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted $\text{α<!-- α -->}$, $\text{α<!-- α -->}$α, $\text{β<!-- β -->}$, and $\text{β<!-- β -->}$’, comprise the polymerase core enzyme. The fifth subunit, $\text{σ<!-- σ -->}$, is involved only in termination. The polymerase comprised of all five subunits is called the holoenzyme.

Rho-dependent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in C-G nucleotides the mRNA folds into a hairpin loop that causes the polymerase to stall.

Rho-independent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-dependent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA folds into a hairpin loop that causes polymerase to stall.

Rho-dependent termination is controlled by rho protein and the polymerase begins near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA creates a hairpin loop that causes polymerase to stall.

Rho-dependent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in A-T nucleotides, the mRNA creates a hairpin loop that causes polymerase to stall.

Rho-independent termination involves the formation of a hairpin.

Rho-dependent termination involves the formation of a hairpin.

Rho-dependent termination stalls when the polymerase begins to transcribe a region rich in A-T nucleotides.

Rho-independent termination stalls when the polymerase begins to transcribe a region rich in G nucleotides.

The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.

The initiation step in prokaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.

The elongation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.

The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are not involved.

No, because they have the same $\text{α<!-- α -->}$-amanitin sensitivity in all products.

No, quantitative analysis of products is done to determine the type of polymerase.

Yes, they can be determined as they differ in $\text{α<!-- α -->}$-amanitin sensitivity.

Yes, they can be determined by the number of molecules that bind to DNA.

No, alternative splicing can lead to the synthesis of several proteins from a single gene.

Yes, alternative splicing can lead to the synthesis of several forms of mRNA from a single gene, building more complex proteins.

No, alternative splicing can lead to the synthesis of several forms of codons from a set of genes.

Yes, alternative splicing can lead to the synthesis of several forms of ribosomes from a set of genes, but only one protein per gene.

exporting the mRNA across the nuclear membrane

importing the mRNA across the nuclear membrane

the mRNA staying inside the nuclear membrane

the mRNA translating into proteins within seconds

The transcript would degrade when the mRNA moves out of the nucleus to the cytoplasm.

The mRNA molecule would stabilize and start the process of translation within the nucleus of the cell.

The mRNA molecule would move out of the nucleus and create more copies of the mRNA molecule.

The mRNA molecule would not be able to add the poly-A tail on its strand at the 5’ end.

The mRNA would be 5’-AUGGCCGGUUAUUAAGCA-3’ and the protein will be MAGY.

The mRNA would be 3’-AUGGCCGGUUAUUAAGCA-5’ and the protein will be MAGY.

The mRNA would be 5’-ATGGCCGGTTATTAAGCA-3’ and the protein will be MAGY.

The mRNA would be 5’-AUGGCCGGUUAUUAAGCA-3’ and the protein will be MACY.

rRNA has catalytic properties in the large subunit and it assembles proteins.

rRNA is a protein molecule that helps in the synthesis of other proteins.

rRNA is essential for the transcription process.

rRNA plays a major role in post-translational processes.

The anticodon will match the codon in mRNA.

The anticodon will match with the modified amino acid it carries.

The anticodon will lose the specificity for the tRNA molecule.

The enzyme amino acyl tRNA synthetase would lose control over the amino acid.