Question 1

A POINT MUTATION TURNS A CODON IN THE MIDDLE OF A GENE INTO UAG. WHAT TYPE OF MUTATION IS THIS?

Accepted Answer

NONSENSE MUTATION. UAG is a stop codon, so translation ends early at that point and the protein is truncated. A change to a stop codon is a nonsense mutation.

Question 2

ONE BASE IS DELETED NEAR THE START OF A CODING SEQUENCE. WHAT IS THE MOST LIKELY EFFECT ON THE PROTEIN?

Accepted Answer

EVERY CODON AFTER THE DELETION IS READ IN A NEW FRAME. Codons are read in non-overlapping triplets from the start codon. Losing one base moves every later triplet boundary, so all downstream amino acids change and a premature stop codon usually appears soon after.

Question 3

A BASE CHANGE ALTERS THE THIRD POSITION OF A CODON, BUT THE PROTEIN SEQUENCE IS UNCHANGED. WHICH PROPERTY OF THE GENETIC CODE EXPLAINS THIS?

Accepted Answer

DEGENERACY. Most amino acids have several codons, and synonymous codons usually differ only at the third base. That redundancy (degeneracy) lets many third-position changes leave the amino acid the same.

Question 4

THE SICKLE-CELL MUTATION SWAPS GLUTAMATE FOR VALINE IN BETA-GLOBIN. THIS IS A ___ MUTATION.

Accepted Answer

MISSENSE. One base change (A to T) turns a glutamate codon into a valine codon. A change that substitutes one amino acid for another is a missense mutation.

Question 5

IN A BACTERIUM, RIBOSOMES START TRANSLATING AN mRNA BEFORE ITS TRANSCRIPTION HAS FINISHED. WHY IS THIS POSSIBLE?

Accepted Answer

PROKARYOTES HAVE NO NUCLEUS SEPARATING THE TWO PROCESSES. Without a nuclear envelope, a bacterial transcript is already in the cytoplasm as it emerges from RNA polymerase, so ribosomes can load onto its 5′ end while the 3′ end is still being made. Eukaryotes must first process and export the mRNA.

Question 6

A TOXIN BLOCKS RNA POLYMERASE II IN A HUMAN CELL. WHICH RNA STOPS BEING MADE?

Accepted Answer

PRE-mRNA. RNA polymerase II transcribes protein-coding genes into pre-mRNA. Polymerase I makes most rRNA in the nucleolus, polymerase III makes tRNA, and mitochondria use their own polymerase. Alpha-amanitin from death-cap mushrooms acts this way.

Question 7

A MUTATION IN THE ___ UPSTREAM OF A GENE CAN SILENCE IT EVEN THOUGH ITS CODING SEQUENCE IS INTACT.

Accepted Answer

PROMOTER. RNA polymerase and general transcription factors assemble at the promoter. If they cannot bind, the gene is not transcribed and no protein is made, however normal the coding sequence is.

Question 8

A EUKARYOTIC mRNA IS MADE WITHOUT A 5′ CAP. WHAT IS THE MOST LIKELY CONSEQUENCE?

Accepted Answer

IT IS DEGRADED QUICKLY AND POORLY TRANSLATED. The cap protects the 5′ end from exonucleases and is recognized by the initiation factors that bring the small ribosomal subunit to the message. Without it the mRNA is unstable and translation initiation is inefficient.

Question 9

ONE GENE PRODUCES DIFFERENT PROTEINS IN MUSCLE AND IN BRAIN TISSUE. WHICH PROCESS BEST EXPLAINS THIS?

Accepted Answer

ALTERNATIVE SPLICING. Different tissues can join different sets of exons from the same pre-mRNA, giving related proteins with different sequences. The DNA is the same in both tissues; the processing differs.

Question 10

A tRNA IS LOADED WITH THE WRONG AMINO ACID BUT HAS ITS NORMAL ANTICODON. WHAT HAPPENS DURING TRANSLATION?

Accepted Answer

THE WRONG AMINO ACID IS ADDED TO THE PROTEIN. The ribosome checks only codon-anticodon pairing, not the amino acid attached. So a mischarged tRNA delivers its wrong amino acid wherever its anticodon matches. Accuracy depends on the synthetase loading the tRNA correctly.

Question 11

DELETING TWO BASES FROM A CODING SEQUENCE CAUSES A ___ MUTATION.

Accepted Answer

FRAMESHIFT. Two is not a multiple of three, so the triplet reading frame shifts and every downstream codon changes. Deleting three bases would remove one amino acid and keep the frame.

Question 12

A DRUG INHIBITS THE PEPTIDYL TRANSFERASE ACTIVITY OF THE RIBOSOME. WHICH STEP OF TRANSLATION FAILS?

Accepted Answer

PEPTIDE BOND FORMATION. Peptidyl transferase, an activity of the large subunit rRNA, links the growing chain in the P site to the amino acid on the tRNA in the A site. Blocking it stops elongation.

Question 13

THE CODING REGION OF AN mRNA IS 300 NUCLEOTIDES LONG, COUNTING THE START AND STOP CODONS. HOW MANY AMINO ACIDS DOES THE PROTEIN CONTAIN BEFORE ANY PROCESSING?

Accepted Answer

99. The region holds 300 / 3 = 100 codons. The stop codon adds no amino acid, so the start codon plus 98 more give 99 amino acids, the first being methionine.

Question 14

ONE tRNA CAN SERVE TWO CODONS ENDING IN C AND U BECAUSE OF ___ PAIRING AT THE THIRD POSITION.

Accepted Answer

WOBBLE. The first anticodon base can pair loosely with the third codon base; a G there pairs with either C or U. This is why cells need fewer than 61 tRNAs.

Question 15

A RIBOSOME ADDS EACH NEW AMINO ACID TO THE CARBOXYL END OF THE CHAIN, SO A PROTEIN IS BUILT FROM ITS ___ TERMINUS FIRST.

Accepted Answer

AMINO. Each new peptide bond joins the incoming amino acid to the carboxyl end of the growing chain. The first amino acid placed, methionine, therefore sits at the amino (N) terminus.

Question 16

BECAUSE TRANSLATION STARTS AT AUG, EVERY NEW EUKARYOTIC POLYPEPTIDE BEGINS WITH ___.

Accepted Answer

METHIONINE. AUG codes for methionine, and the initiator tRNA carries it. Many proteins later have this first methionine trimmed off. Bacteria use formyl-methionine.

Transcription and Translation