Chapter 38

Question 1

Ribonucleic Acid Types

Answer 1

rRNA (ribosomal): made up by RNA polymerase I
- ribosomes

tRNA (transfer): made by RNA polymerase III
- cloverleaf shape w/ anticodon for mRNA and accepter stem for amino acid
- will facilitate process of translation

mRNA (messenger): made by RNA polymerase II
- genetic sequence of protein from DNA

Question 2

RNA Processing

Answer 2

RNA polymerase I synthesizes a large precursor RNA (45S) that is subsequently processed to yield 18s, 28s, and 5.8s rRNAs ~ components of the ribosome

Prior to cleavage, the precursor is alerted by modifications of some bases and ribose’s that are catalyzed by RNA-protein complex called small nuclear ribonucleoproteins (snoRNPs)

Question 3

Transfer RNA Processing

Answer 3

1. RNA polymerase III: catalyzes the synthesis of precursors to tRNA
2. RNase P: removes nucleotides from the 5’ end
3. RNase Z: removes nucleotides from the 3’ end
4. tRNA nucleotidyltransferase: adds nucleotides to the 3’ end
5. bases and riboses are modified
6. Intron is removed by an endonuclease, resulting products are joined by a ligase

Question 4

Messenger RNA Processing

Answer 4

 Messenger RNA precursors (pre-mRNA) are synthesized by RNA polymerase II and then subsequently processed in several ways:

1. The 5’ end is modified by the addition of the 5’ cap in which a GTP is added to the precursor in an unusual 5’-5’ linkage

 The cap may also be methylated

2. The 3’ end is cleaved by a specific endonuclease, and a stretch of polyadenylate is added by poly(A) polymerase to form the poly(A) tail about 250 nucleotides long
3. Noncoding stretches of RNA called introns are removed and the products ligated to form mature mRNA

Question 5

Messenger RNA Processing – Splicing

Answer 5

 Most genes in eukaryotes consist of exons (coding regions) and introns (noncoding regions)

 Introns vary in length from 50-10,000 nucleotides

 The exons are joined by splicing

 Intron-exon junctions have several common features:

1. The 5’ end of the junction has the consensus sequence 5’ AGGUAAGU 3’ w/ the first GU from the 5’ end demarcating the beginning of the intron
2. The 3’ end of the intron is marked by a stretch of pyrimidines (polypyrimidine tract) followed by any base, a C, and then the intron ends w/ AG
3. The branch site is located 20-50 nucleotides from the 3’ end of the intron