Cell & Molecular Bio Exam 1

What is the charge of the peptide A-G-S-E-K at pH 7?

0 E = glutamic acid, charge = -1 K = lysine, charge = +1 net charge = 0

For the reaction: succinate → fumarate + 2 e- + 2 H+; E = -0.03 V What is ΔG in kcal/mol? Use the formula: ΔG = -nFE, where F = Faraday constant = 23 kcal/Volt Submit only the numerical value.

1.38

Lodish et al. describes an example of a protein (P) binding to DNA (D); P + D ⇔ PD, where Kd = 10-10 M total concentration of P ([P] + [PD]) = 111 x 10-10 M (calculated concentration of 10 molecules of P per cell) total concentration of D ([D] + [PD]) = 10 x 10-10 M (just 1 molecule of DNA per cell) and calculated that the DNA molecule would have P bound to it 99% of the time. What would be the ratio of unbound to bound DNA if phosphorylation of P altered the Kd to 10-8 M?

1:1 Let's start with the equation: Kd = [P][D]/[PD] We can approximate [P] (concentration of free protein) as being about 100 x 10-10 M, because there is a 10-fold excess of P for every DNA molecule (if no protein is bound, [P] = 111 10-10 M, and if the DNA always has protein bound, [P] = 100 x 10-10 M). Then Kd = 100 x 10-10 M [D]/[PD] Then Kd of phosphoP = 10-8 M = 100 x 10-10 M = 100 x 10-10 M [D]/[PD] [D]/[PD] = 1; the DNA molecule will have P bound to it about half the time.

Tris is a common buffer used in biochemistry and molecular biology. The pKa of Tris is 8.3 Tris is available as powder in 2 forms: TrisHCl is the acid form, and Tris base as the free base. What molar ratio of TrisHCl and Tris base will form a solution of pH 8.0? Hint: Henderson-Hasselbach equation is: pH = pKa + log {[A-]/[HA]}

2:1 Henderson-Hasselbach equation: pH = pKa + log {[A-]/[HA]} log {[A-]/[HA]} = 8 - 8.3 = -0.3 [A-]/[HA] = 0.5 [HA]/[A-] = 2 or 2:1 ratio

What is the approximate pH of a 1 mM solution of a strong acid (HCl)? Enter your answer as just a number (no units).

3 pH is the negative log of the hydrogen ion concentration. A 1 mM solution of HCl, which dissociates completely, will have a H+ ion concentration of 1 mM = 1E-3. Negative log of that is 3.

DNA and RNA strands are replicated and transcribed (synthesized), respectively, in which direction? A) 5' to 3' B) 3' to 5' C) both 5' to 3' and 3' to 5' D) none of the above

A

Following is the order of packaging of eukaryotic DNA into chromosomes. What belongs in the blank? DNA double helix → 'beads-on-a-string' form of chromatin → (____) → extended scaffold associated form → condensed scaffold associated form → metaphase chromosome A) 30 nm chromatin fiber B) spermine C) spacer D) satellite DNA

A

What is true for DNA replication? A) DNA replication is semi-conservative B) Both prokaryotic and eukaryotic cells employ a uni-directional mechanism of DNA replication. C) Only the lagging strand requires a short primer for the initiation of replication. D) The replication fork has one molecule of DNA polymerase

A

Which histone modification leads to greater condensation, characteristic of heterochromatin? A) methylation B) acetylation C) phosphorylation D) ubiquitination

A

Which of the DNA repair pathways occurs 'before' DNA replication? A) Base excision repair B) Mismatch excision repair C) Nucleotide excision repar D) Non-homologous end joining repair

A

Which of the following technique can be used to evaluate the expression of many genes at one time? A)microarray B)gel electrophoresis C) restriction enzyme digestion D) PCR

A

Which post-translational modification of proteins tags them for degradation by the proteasome? A)ubiquitination B) phosphorylation C) glycosylation D) proteolytic cleavage

A

A combination of 2 or more secondary structure elements that form a characteristic 3-D structure, like EF-hands or zinc-fingers, is called ____ A) a motif B) a domain C) a supramolecular complex

A A motif is defined by Lodish as a structure that is formed by two or more secondary structures in combination to form a distinctive 3-D structure. A domain is a unit of protein structure - a module - that may contain a motif. So in general, a domain is larger than a motif. A supramolecular complex is something like a ribosome, or a viral capsid, that is assembled from multiple proteins.

What do you know about ∆G for the energetically favored reaction in the previous question? A) it is negative B) it is positive C) it = zero D) none of the above

A By definition, any energetically favored reaction must have a negative deltaG

This figure shows 3 different types of complexity in transcription of a protein-coding gene. Match the letter to the correct description of the outcome.

A - mRNAs encode proteins that differ in an internal domain or motif of the protein B - mRNAs encode proteins that have different C-terminal domains C - mRNAs encode proteins that have different N-terminal domains In A, alternative splicing produces mRNAs that have different internal exons. Therefore, their proteins will differ in the amino acid sequence of an internal portion of the protein, although their N-termini and C-termini will be the same. In B, the two mRNAs have different 3' exons, so their protein products will have different carboxy terminal regions. In C, the two mRNAs have different 5' exons, so their proteins will have different N-terminal regions.

Select all true statements about chemical equilibrium A) Keq = kf/kr (rate constant for forward reaction/rate constant for reverse reaction) B) For the reaction A⇔ B, Keq = [B]/[A] at equilibrium C) Reactions in cells tend to be at chemical equilibrium

A and B These statements correctly state or arise from the definition of the equilibrium constant. Cells maintain steady-state (homeostasis), which is different from chemical equilibrium. See Fig. 2-23 - the product of one reaction is used as the reactant for another reaction, so reactions never actually reach equilibrium. This is really important from an energetics point of view, because at equilibrium ΔG = 0. No free energy is released at equilibrium, so no energy will be available for cells to do any work.

What would be the consequence(s) to a cell of a loss of telomeric sequences from the ends of its chromosomes? A) chromosomes will shorten at the ends with each round of DNA replication B) chromosomes will randomly join end-to-end during interphase, but then break during anaphase of mitosis C) telomerase will repair the chromosomes by adding new telomere sequence

A, B

what are the possible outcomes after CRISPR-Cas9 mediated cleavage of DNA? Select all that apply A) The cleavage site is joined by non-homologous end joining, usually resulting in a mutation at the joint B) The cleavage site is repaired by recombination repair with a homologous sequence

A, B

What are the minimum structural or functional elements required for a eukaryotic chromosome? Select all that apply. A) telomeres B) centromere C) origin of DNA replication D) LTRs (long terminal repeats)

A, B, C

The eukaryotic translational initiation factor eIF4 binds to which of the following? select all that apply A) ribosomal small subunit in a pre-initiation complex B) ribosomal large subunit C) tRNAiMet D) 5' cap on the mRNA E) polyA-binding protein (PABP) bound to the 3' polyA tail on the mRNA

A, D, E

Supplementary Fig. 1c from Ma et al. 2017 shows the DNA constructs used for their experiments to edit the genome of human embryos. These are co-injected into fertilized human eggs (zygotes). The human embryos containing these plasmids will express the guide RNA and the Cas9 protein. The Cas9 protein will then cut the DNA of the embryo genome at a site or sites where the sgRNA binds. The single-stranded oligonucleotides (ssODN) provides a template for homology-directed recombination repair of the dsDNA breaks generated by the Cas9 protein. This homology-directed recombination repair will copy the sequence of the ssODN and replace the original DNA sequence near the site of the cut by Cas9 protein. The two plasmids with sgRNA and spCas9 (Streptococcus pyogenes Cas9) are maintained, grown and isolated from E. coli. cultures. Identify 3 DNA sequence features that these plasmids must have, that are not shown in this diagram?

An origin of DNA replication for E. coli A selectable marker (antibiotic resistance gene) A multiple cloning site for insertion of the sgRNA or spCas9 DNA sequences

If the product of a mutated gene still interacts with the same elements as the wild-type product, but blocks some aspect of its function, it acts as a ( ). A) recessive B) dominant negative C) haploinsufficient D) all of the above

B

In the CRISPR-Cas9 system, what is the function of the guide RNA? A) it carries out the catalytic cleavage of the target DNA molecule B) it binds to the target DNA sequence by base-pairing C) it holds CRISPR and Cas9 together as a complex D) it initiates homologous recombination with the target DNA sequence E) it serves as the template for the break repair of the DNA after cleavage by Cas9

B

Plants have protein kinases (enzymes that phosphorylate other proteins) that are activated by calcium ions. The portion of the protein that binds calcium resembles calmodulin, a small calcium-binding protein that is ubiquitous in eukaryotes, with 4 EF-hands. The calmoduline-like part of this enzyme is a _____; each EF-hand is a ______ A) motif; domain B) domain; motif C) subunit; domain D) domain; cofactor

B

Supplementary Fig. 1b from Ma et al. 2017 shows one of their CRISPR-Cas9 sgRNAs and the single-stranded oligodeoxynucleotide-2 (ssODN-2). What would be the outcome of a non-homologous end-joining event after Cas9 cuts at the site indicated by the lightning bolt symbol? A) recreation of the original mutant allele B) a new mutant allele that contains an additional deletion around the cut site C) formation of a wild-type allele that contains the BstB1 site D) insertion of the 100 bp ODN-2 sequence at the cut site

B

Suppose that this graph shows Michaelis-Menten kinetics for two enzymes. The dotted horizontal line intersecting the Y-axis at 1.0 represents the rate that both enzymes approach at higher substrate concentrations. The enzyme represented by the red curve has _____ A) a higher Km than the enzyme represented by the blue curve B) a lower Km than the enzyme represented by the blue curve C) a higher Vmax than the enzyme represented by the blue curve D) a lower Vmax than the enzyme represented by the blue curve

B

Telomerase contains an RNA template and has ( ). A) DNA degrading activity B) reverse transcriptase activity C) RNA ligating activity D) RNA priming activity

B

The base in the wobble position of a codon A) is the 5 ́ (first) base. B) is the 3 ́ (third) base. C) is the second base. D) often contains adenine.

B

The sequence of the 5' end of an mRNA is: 5'-CCAUAUGGGAAUCAGA....3' What are the first 3 amino acids that will be translated from this mRNA? A) Pro-Tyr-Gly (PYG in single letter code) B) Met-Gly-Ile (MGI in single letter code) C) can't be determined because the 3' end of the mRNA sequence is not given D) Gly-Ile-Pro (GIP in single letter code)

B

What do chaperons such as Hsp70 and Hsp90, and chaperonins have in common? A) they all form barrel-shaped chambers to fold denatured proteins B) they all bind and hydrolyze ATP in the process of folding proteins C) they are all associated with ribosomes D) all of the above

B

Which of these DNA elements can jump (relocate) and proliferate in the genome by a "copy and paste" mechanism? A)DNA transposons B) retrotransposons C) microsatellite sequences D) Alu sequence

B

what would you predict to be the structural/functional consequences of a genetic mutation that results in a K -> E substitution at the site indicated by the box above? A) little or no effect B) likely disruption of protein A binding to protein B C) binding of proteins A and B to each other will probably be stronger D) cannot even guess

B

Flavins and pyridine nucleotides are major mediators of redox reactions in cells. The standard redox potential for them are: Reaction (OXIDIZED « REDUCED) ∆Eo (volts) FAD + 2H+ + 2e- ⇔ FADH2 (in flavoproteins) 0 NAD+ + H+ + 2e- ⇔ NADH -0.315 Given these data, which of the following reactions is energetically favored? A) the reduction of NAD+ by FADH2 (NAD+ + FADH2 ⇔ FAD + NADH) B) the reduction of FAD by NADH (FAD + NADH ⇔ NAD+ + FADH2) C) neither of these electron transfers can occur

B A positive deltaE for a reaction is energetically favored: the deltaG is negative. The reduction of NAD+ by FADH2 will have a deltaE of -0.315 V; deltaG is positive and the reaction is not favored The reduction of FAD+ by NADH will have a deltaE of +0.315 V because the half-reaction shown in the table is reversed (now NADH <=> NAD+ + H+ + 2e-). With a positive deltaE, this reaction has a negative deltaG and will occur spontaneously.

You are interested in an enzyme that uses ATP for a critical biological process. You wonder if the enzyme loses a substantial amount of its activity (and hence the biological proces is affected) when cells become partially hypoxic (i.e., have reduced access to oxygen) which decreases the cellular ATP concentration. You know that the enzyme has a Km for ATP of 0.1 mM, and you look up in the literaturę that the normal ATP concentration of the cells is 5 mM and decreases to 1 mM when the cells are partially hypoxic. Based on this information, you would predict that: A) the activity of the enzyme will decrease by >50% due to this 5-fold decrease in cellular ATP concentration B) the activity of the enzyme will decrease by much less than 50% due to this 5-fold decrease in cellular ATP concentration C) the activity of the enzyme will INCREASE due to this 5-fold decrease in cellular ATP

B Since the enzyme has a Km for ATP of 0.1 mM, it will be operating at near Vmax at 1 mM (10x Km). So a decrease in ATP concentration from 5 mM to 1 mM will have only a small effect on enzyme activity

The glucose molecule shown in this figure is in which anomeric form? A) alpha B) beta

B The hydroxyl on the #1 carbon (the only carbon with bonds to 2 oxygen atoms) is in the "up" position, so this is beta-D-glucose. In the ring form, the hydroxyl can be either in the down (axial) positon or up (equatorial) position. Axial is alpha. Cellular enzymes distinguish between the two forms and their polymers have different properties. Starch is made of glucose linked in their alpha forms, whereas cellulose is made of glucose linked in their beta forms. Starch is easily digestible by humans, cellulose is not.

A genetic mutation that results in a V ⇒ E substitution at the site indicated by the arrow labeled "II" in the diagram on the right would be predicted to: A) have little or no effect B) likely cause disruption of the tertiary structure of the mutant protein C) likely disrupt quaternary structure or the interaction between the two proteins II = hydrophobic and van der waals interactions region

B and C

This part of Figure 6-37a from Lodish et al shows the Gene X replacement construct with a neomycin gene (shaded green) replacing a part of gene X (orange and white). The construct also contains the thymidine kinase (tk) gene from Herpes Simplex virus. If cultured mammalian cells are transfected with this construct and grown in medium containing neomycin (G418), what cells will survive and grow? Select all that apply. A) cells that did not take up the construct (non-transfected cells) B) cells that took up the construct and the construct inserted into an off-target site (not into gene X) C) cells where the Gene X replacement construct underwent homologous recombination with a chromosomal copy of Gene X D) no cells will survice in medium containing G418

B, C

Which mobile genetic elements encode a reverse transcriptase? Choose all that apply. A) DNA transposable elements B) LTR retrotransposons C) LINEs D) SINEs

B, C

In this image, which colored nucleotides form a pseudoknot? A) the blue-colored nucleotides B) the light green-colored nucleotides C) the dark gold-colored nucleotides

C

Suppose an enzyme has a Km for substrate of 10 micromolar. The concentration of the substrate changes from 50 micromolar to 100 micromolar. What happens to the rate of the enzyme-catalyzed reaction? Try using the equation and express in terms of the Vmax (the maximal velocity of the enzyme): v = Vmax[S]/([S] + Km) Then choose the best answer below. A) the rate of the enzyme-catalyzed reaction will increase 2x when the substrate concentration increases from 5x Km to 10x Km B) the rate of the enzyme-catalyzed reaction will increase more than 2x when the substrate concentration increases from 5x Km to 10x Km C) The rate of the enzyme-catalyzed reaction will increase by less than 2x when the substrate concentration increases from 5x Km to 10x Km

C

What information does mass spectrometry provide most directly? A)the shape of the peptide fragment B) the amino acid sequence of a peptide fragment C) the mass to charge ratio of an ion D) the length of a molecule

C

This part of Figure 6-37a from Lodish et al shows the Gene X replacement construct with a neomycin gene (shaded green) replacing a part of gene X (orange and white). The construct also contains the thymidine kinase (tk) gene from Herpes Simplex virus. If this gene replacement construct undergoes homologous recombination with the chromosomal copy of Gene X (a crossover in each of the white regions), the cell will____ A) have the tk gene and be resistant to ganciclovir B) have the tk gene and be sensitive to ganciclovir C) lose the tk gene and be resistant to ganciclovir D) lose the tk gene and be sensitive to ganciclovir

C The TK gene makes cells sensitive to ganciclovir. A homologous recombination even will insert the neomycin resistance gene into the chromosomal copy of gene X, and exclude the TK gene, which will be degraded and lost. Thus cells where homologous recombination has inserted the neomycin resistance gene into Gene X, disrupting Gene X, will be resistant to both neomycin and ganciclovir.

Which of the following statements are generally true about lipids in cell membranes? Select all true statements. A) cell membranes have only phospholipids B) fatty acid chains in phospholipids frequently have both cis- and trans- double bonds between carbons C) phospholipid can have different polar "head" groups covalently linked via phosphodiester bonds to diacyl glycerol (glycerol plus 2 fatty acid chains) D) phospholipids can have a polar headgroup (e.g., phosphocholine) covalently linked via phosphodiester bonds to ceramide (sphingosine plus a fatty acyl).

C and D Although phospholipids are a main component, cell membranes also have other lipids such as sterols (e.g. cholesterol). Cell membranes also have significant amounts of proteins, and glycosyl groups. Naturally occurring fatty acids with double bonds almost always have the cis-configuration; trans-fatty acids in foods occur as a result of chemical processing (hydrogenation). Most common phospholipids consist of glycerol with two fatty acid chains, a phosphate and a polar "head group" linked to the phosphate - see Fig 2-20 and Table 2-5.

Which of the following non-coding RNAs never exit the nucleus? Select all that apply. A) rRNAs B) tRNAs C) snoRNAs D) snRNAs E) miRNAs

C, D rRNAs and tRNAs both function in translation, which occurs in the cytoplasm, so these two RNAs must exit the nucleus. miRNAs also exit the nucleus where they bind with mRNAs to regulate translation and mRNA degradation. snoRNAs function in the nucleolus to process ribosomal RNAs, so they stay in the nucleus snRNAs function to splice introns out of pre-mRNAs, so they also stay in the nucleus.

The anti-codon of a tRNA is 5'-IGC-3'. What mRNA codon(s) will pair with this tRNA? Select all that apply A) 5'-AGC-3' B) 5'-ACG-3' C) 5'-GCA-3' D) 5'-GCU-3' E) 5'-UCG-3' F) 5'-GCC-3' G) 5-CCG-3' H) 5-UGC-3'

C, D, F The tricky part of this question is that the anti-codon is complementary to, and has the opposite directionality from, the mRNA codon. We are given that the third position in the anticodon is C. This C must pair with a G in the first position of the mRNA codon. Only 3 of the answer choices have a G in the first position of the mRNA codon. The G in the second position of the anticodon will pair with a C in the second position of the mRNA codon. All three of the answer choices that start with G also have a C in the second position. The first position in the anticodon (the wobble position) is I (inosine). The figure shows that at the wobble position, inosine can pair with C, A or U in the third position of the mRNA codon.

Proteasomes ___ A) are supramolecular complexes that synthesize proteins B) are supramolecular complexes that refold proteins C) are supramolecular complexes that degrade proteins bound to chaperones D) are supramolecular complexes that degrade ubiquitinated proteins

D

Which of the following is NOT a necessary feature of plasmid cloning vectors? A) replication origin B) antibiotic resistance gene C) polylinker D) inverted repeats

D

Which of the following is NOT the component of DNA replication fork? A)okazaki fragment B)primase C)helicase D)reverse transcriptase

D

Which of these bonds or interactions requires the least amount of energy to disrupt or break? A) a covalent C-C bond B) a covalent C=O double bond C) a hydrogen bond between a water molecule and an -OH group D) van der Waals interaction between two carbon atoms

D

Molecular chaperones, such as Hsp70, ____. A) bind ATP B) stabilize unfolded or partly folded proteins C) help newly synthesized proteins fold into their proper conformation D) help unfolded proteins to fold back into their proper conformation E) all of the above

E

The sequence of a portion of the template strand of DNA is: 5'-...GTATAGCGAATGCAGT....3' What is the sequence of the RNA that is transcribed from this DNA sequence? A) 5'-...GTATAGCGAATGCAGT....3' B) 5'-...GUAUAGCGAAUGCAGU....3' C) 5'-...CATATCGCTTACGTCA....3' D) 3'-...CATATCGCTTACGTCA....5' E) 3'-...CAUAUCGCUUACGUCA....5' F) 3'-...GUAUAGCGAAUGCAGU....5'

E

Match the proteins to their function in DNA replication

Helicase - unwinds DNA ahead of replication fork topoisomerase - relieves torsional stress PCNA - sliding clamp that stabilizes DNA pol binding to template strand RPA - binds single-stranded DNA primase - synthesizes short RNA segments complementary to template DNA strand DNA ligase - seals phosphodiester bonds between Okazaki fragments clamp loader (Rfc) - loads PCNA onto end of primer:template

Suppose that Gene X is expressed in a variety of different tissues and organs during mammalian embryonic development. Mutants with no Gene X die early in embryonic development, before differentiation of eyes. Describe how the Cre-lox recombinase system could be used to study the function of Gene X specifically in eye development in laboratory mice, without disrupting the function of Gene X in other tissues or organs. The transgenic mouse strain that contains the Cre recombinase gene should express the Cre recombinase with what kind of promoter? Describe the floxed allele in the mouse strain that has the recombinant Gene X targeted for disruption. What happens when these two mouse strains are mated (Cre recombinase containing mouse X mouse with floxed Gene X allele)?

One mouse strain should express the Cre-recombinase with a promoter that is eye-specific (such as a lens protein promoter). The floxed strain should have loxP sites in introns flanking an exon of Gene X. When these two strains are mated, the progeny will express Cre-recombinase in their eyes, which will mediate recombination between the loxP sites, removing the exon in between them. Gene X will thereby be inactivated. The consequences for eye development in this mouse will yield insights into the role of Gene X in eye development.

Above is an image of an agarose gel stained with ethidium bromide. Lane 2 contains a sample of DNA purified from chromatin that has been partially digested with micrococcal nuclease. Lane 1 has a 100 bp DNA ladder. Explain the banding pattern of the nuclease-digested chromatin sample in lane 2.

The nuclease is cutting the DNA between nucleosomes. DNA wrapped around the nucleosomes is protected from nuclease digestion. Since about 150 bp of DNA is wrapped around a single nucleosome, the smallest DNA fragment is about 150 bp. Other fragments represent 2 nucleosomes (150 bp + 200 bp), 3 nucleosomes, etc.

What are 2 odd or unnatural features of this molecule as shown? Molecule = fatty acid

The saturated fatty acid has an odd number of carbons. Most natural fatty acids have an even number of carbons because they are synthesized 2 carbons at a time, from acetyl-CoA. The monounsaturated fatty acid chain has a trans configuration, whereas naturally occuring fatty acids have cis double bonds.

Supplementary Fig. 1b from Ma et al. 2017 shows one of their CRISPR-Cas9 sgRNAs and the single-stranded oligodeoxynucleotide-2 (ssODN-2). The mutant allele is missing the 4 bases, GAGT as shown in larger letters above the dashes in the figure. Identify any oddities or possible errors in this figure.

The sgRNA portion that matches the target DNA sequence is shown with T's instead of U's.

Rank the predicted G+C content of the genomic DNAs of the bacteria listed below, from lowest to highest A) a bacterium like E. coli that lives in a mammal's intestines B) a thermophilic bacterium that lives in hot springs C) a cryophilic bacterium that lives under the arctic ice

a cryophilic bacterium that lives under the arctic ice < a bacterium like E. coli that lives in a mammal's intestines < a thermophilic bacterium that lives in hot springs Organisms that live in cold temperatures will have DNA with lower Tm and be A:T-rich, so they can replicate and transcribe their DNA at the cold temperatures. Conversely, organisms that live in high temperatures will have DNA with high Tm and be G:C-rich so their DNA stays double-stranded at the high temperatures.

What are the differences in hydrogen bonds that stabilize alpha-helices, beta-sheets, and tertiary/quarternary structures? Match the description of the hydrogen bonds to their corresponding structural element

alpha-helix - hydrogen bonds between peptide backbone atoms of amino acids spaced 3 residues apart along the polypeptide chain beta-sheet - hydrogen bonds between peptide backbone atoms of amino acids that are far apart along the polypeptide chain tertiary and quaternary structure - hydrogen bonds between side chain atoms of amino acids that are far apart along the polypeptide chain

Match the DNA repair system to the appropriate description

nucleotide excision repair - repairs thymine dimers caused by UV irradiation mismatch repair - repairs base mismatches and small insertions or deletions after DNA replication base excision repair - repairs base mismatches caused by cytidine deamination before DNA replication non-homologous end joining - repairs double-strand DNA breaks, very error-prone recombination repair - repairs double-strand DNA breaks using a homologous DNA molecule as a template

Match each RNA to its function

rRNA - catalyzes peptide bond formation during protein synthesis tRNA - matches codons in the mRNA to their amino acids snoRNA - required for processing rRNAs in the nucleolus miRNA - post-transcriptional regulation of gene expression snRNAs - catalyzes pre-mRNA splicing to remove introns