What characteristic of the genetic code points to a common ancestry for all organisms?

1. The code is degenerate.
2. The code contains 64 codons.
3. The genetic code is almost universal.
4. The code contains stop codons.

1. replication
2. splicing
3. transcription
4. translation

When comparing transcription of heritable information in prokaryotes and eukaryotes, which events are the same?

1. The transcription by polymerase, the recognition of a consensus sequence in the promoter, and the termination by a hairpin loop are conserved.
2. The translation by polymerase, the recognition of a consensus sequence in the promoter, and the termination by a hairpin loop are conserved.
3. The transcription by polymerase, the recognition of a highly variable sequence in the promoter, and the termination by a hairpin loop are conserved.
4. The transcription by polymerase, the recognition of a consensus sequence in the promoter, and the elongation by a hairpin loop are conserved.

1. chloroplast
2. cytoplasmic membrane
3. mitochondria
4. nucleus

How does the enzyme reverse transcriptase violate the central dogma of molecular biology in HIV?

1. The enzyme reverse transcriptase reverse-transcribes the RNA in the genome of HIV to DNA.
2. The enzyme reverse transcriptase translates the RNA of the HIV into protein and then back to DNA.
3. The enzyme reverse transcriptase transcribes the DNA straight into the protein molecules.
4. The enzyme reverse transcriptase transcribes DNA to RNA, then again to DNA. There is no protein synthesis.

Radioactive deoxythymidine triphosphate is supplied to the protist Euglena. After an interval of time, the cells are homogenized, and different fractions are analyzed for radioactivity content in large nucleic acid molecules. Which fraction will not be labeled?

1. nucleus
2. mitochondrion
3. chloroplast
4. plasma membrane

You sequence a gene of interest and isolate the matching mRNA. You find that the mRNA is considerably shorter than the DNA sequence. Why is that?

1. There was an experimental mistake. The mRNA should be the same length as the gene.
2. The mRNA should be longer than the DNA sequence because the promoter is also transcribed.
3. The processed mRNA is shorter because introns were removed.
4. The mRNA is shorter because the signal sequence to cross the nuclear membrane was removed.

A mutation in the promoter region of the gene for beta-globin can cause beta-thalassemia, a hereditary condition that causes anemia. Why would mutations in the promoter region lead to low levels of hemoglobin?

1. The globin chains produced are too long to form functional hemoglobin.
2. The globin chains are too short to form functional hemoglobin.
3. Fewer globin chains are synthesized because less mRNA is transcribed.
4. Globin chains do not fold properly and are nonfunctional.

You are given three mRNA sequences:

1. 5′-UCG-GCA-AAU-UUA-GUU-3′
2. 5′-UCU-GCA-AAU-UUA-GUU-3′
3. 5′-UCU-GCA-AAU-UAA-GUU-3′

Using the given table, write the peptide encoded by each of the mRNA sequences.

1. (i) Serine-alanine-asparagine-leucine-valine; (ii) serine-alanine-asparagine-leucine-valine; (iii) serine-alanine-asparagine(-stop)
2. (i) Serine-phenylalanine-asparagine-leucine-valine; (ii) serine-alanine-asparagine-leucine-valine; (iii) serine-alanine-asparagine(-stop)
3. (i) Serine-alanine-asparagine-leucine-valine; (ii) serine-alanine-asparagine(-stop); (iii) serine-alanine-asparagine-leucine-valine
4. (i) Serine-alanine-asparagine-leucine-valine; (ii) serine-arginine-asparagine-leucine-valine; (iii) serine-alanine-asparagine(-stop)

Refer to the table.

You are given three mRNA sequences:

1. 5′-UCG-GCA-AAU-UUA-GUU-3′
2. 5′-UCU-GCA-AAU-UUA-GUU-3′
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 the consequence be for the cell in each case?

1. There is a silent mutation in peptide 2, and peptide 3 has a stop codon due to a mutation.
2. There is a silent mutation in peptide 3, and peptide 2 has a stop codon due to a mutation.
3. There is a different amino acid in peptide 2, and peptide 3 has a stop codon due to a mutation.
4. There isn’t a mutation in peptide 2, and peptide 3 has a stop codon due to a mutation.