- The code is degenerate
- The code contains 64 codons.
- The genetic code is almost universal.
- The code contains stop codons
- replication
- splicing
- transcription
- translation
- Transcription by polymerase, recognition of a consensus sequence in the promoter, and termination by a hairpin loop are conserved.
- Translation by polymerase, recognition of a consensus sequence in the promoter, and termination by a hairpin loop are conserved.
- Transcription by polymerase, recognition of a highly variable sequence in the promoter, and termination by a hairpin loop are conserved.
- Transcription by polymerase, recognition of a consensus sequence in the promoter, and elongation by a hairpin loop are conserved.
- chloroplast
- cytoplasmic membrane
- mitochondria
- nucleus
- The enzyme reverse transcriptase reverse transcribes the RNA in the genome of HIV to DNA.
- The enzyme reverse transcriptase translates the RNA of the HIV into protein and then back to DNA.
- The enzyme reverse transcriptase transcribes the DNA straight into the protein molecules.
- The enzyme reverse transcriptase transcribes DNA to RNA, then again to DNA. There is no protein synthesis.
- nucleus
- mitochondrion
- chloroplast
- plasma membrane
- There was an experimental mistake. The mRNA should have the same length as the gene.
- The mRNA should be longer than the DNA sequence because the promoter is also transcribed.
- The processed mRNA is shorter because introns were removed.
- The mRNA is shorter because the signal sequence to cross the nuclear membrane was removed.
- The globin chains produced are too long to form functional hemoglobin.
- The globin chains are too short to form functional hemoglobin.
- Fewer globin chains are synthesized because less mRNA is transcribed.
- Globin chains do not fold properly and are non-functional.
You are given three mRNA sequences:
- 5’-UCG-GCA- AAU-UUA -GUU-3’
- 5’-UCU-GCA- AAU-UUA -GUU-3’
- 5’-UCU-GCA- AAU-UAA -GUU-3’
Using the table, write the peptide encoded by each of the mRNA sequences.
- Serine-alanine-asparagine-leucine-valine
- Serine-alanine-asparagine-leucine-valine
- Serine-alanine-asparagine(-stop)
- Serine-phenylalanine-asparagine-leucine-valine
- Serine-alanine-asparagine-leucine-valine
- Serine-alanine-asparagine (-stop)
- Serine-alanine-asparagine-leucine-valine
- Serine-alanine-asparagine (-stop)
- Serine-alanine-asparagine-leucine-valine
- Serine-alanine-asparagine-leucine-valine
- Serine-arginine-asparagine-leucine-valine
- Serine-alanine-asparagine(-stop)
You are given three mRNA sequences:
- 5’-UCG-GCA- AAU-UUA -GUU-3’
- 5’-UCU-GCA- AAU-UUA -GUU-3’
- 5’-UCU-GCA- AAU-UAA -GUU-3’
Using the peptide encoded by each of the above, compare the three peptides obtained. How are peptides 2 and 3 different from 1? What would be the consequence for the cell in each case?
- There is a silent mutation in peptide 2 and peptide 3 has a stop codon due to mutation.
- There is a silent mutation in peptide 3 and peptide 2 has a stop codon due to mutation.
- There is a different amino acid in peptide 2 and peptide 3 has a stop codon due to mutation.
- There isn’t a mutation in peptide 2 and peptide 3 has a stop codon due to mutation.