RNA & Protein Synthesis (13.1-13.2)
A researcher identifies a new RNA molecule that is 1,200 nucleotides long, contains no introns, and has a 5' cap and poly-A tail. Is this most likely from a prokaryotic or eukaryotic cell? Explain your reasoning.
Eukaryotic. The presence of a 5' cap and poly-A tail indicates eukaryotic mRNA processing. Prokaryotic mRNA does not undergo these modifications.
RNA polymerase binds to a promoter region in a prokaryotic cell, but no transcription occurs. Which regulatory protein is most likely bound and where is it bound? What would need to happen to allow transcription to proceed?
The repressor protein is bound to the operator region. To allow transcription, the inducer molecule (e.g., allolactose for lac operon) must bind to the repressor, causing a conformational change that releases it from the operator, allowing RNA polymerase to proceed.
A point mutation changes the codon UUU to UUC. Using the genetic code, determine what type of mutation this is and explain whether this mutation would likely affect protein function.
Silent mutation (both UUU and UUC code for phenylalanine). No effect on protein function since the amino acid sequence remains unchanged.
A man with blood type AB (genotype I^A I^B) and a woman with blood type O (genotype ii) have a child. Calculate the probability that their child will have blood type A. Show your Punnett square.
50% probability of blood type A.
A PCR reaction starts with a single DNA molecule. After 10 cycles of PCR, calculate how many copies of the target DNA sequence exist. Show your calculation.
2^10 = 1,024 copies of the target sequence. Formula: N = 2^n where n = number of cycles (starting from 1 molecule).
A DNA template strand has the sequence 3'-TAC GCA CCT AGG-5'. What is the sequence of the mRNA transcript produced? What amino acid sequence would result (use codon chart knowledge)?
mRNA: 5'-AUG CGU GGA UCC-3'. Amino acids: Methionine-Arginine-Glycine-Serine (Met-Arg-Gly-Ser). (AUG=Met, CGU=Arg, GGA=Gly, UCC=Ser)
In the lac operon, both glucose and lactose are absent. Draw the state of the operon and explain whether transcription occurs, referencing the roles of the repressor, operator, and RNA polymerase.
No transcription occurs. Both glucose and lactose absent = operon off. Repressor protein is active (no lactose to inactivate it) and binds to operator, blocking RNA polymerase from transcribing the structural genes. This conserves energy because lactose is not available to metabolize.
A DNA sequence reads: 5'-ATG GGC CTA TTC-3'. If a single nucleotide is inserted after the first nucleotide, determine: (a) the new mRNA sequence, (b) the new amino acid sequence, and (c) explain what type of mutation this is.
(a) Original: 5'-ATG GGC CTA TTC-3' → mRNA: 5'-AUG GGC CUA UUC-3'. Insertion after first nucleotide: New DNA: 5'-AAT GGG CCT ATT C-3' → New mRNA: 5'-AAU GGG CCU AUU C-3'. (b) New amino acids: Asparagine-Glycine-Proline-Isoleucine. (c) Frameshift mutation (insertion of 1 nucleotide shifts reading frame, changing all downstream codons).
A woman who is a carrier for red-green color blindness (X^A X^a) has children with a color-blind man (X^a Y). Calculate the probability that: (a) a son will be color blind, and (b) a daughter will be color blind. Show your reasoning.
(a) Sons: 50% affected (X^a Y). (b) Daughters: 50% carriers (X^A X^a), 0% affected (need X^a X^a, but father passes X^a to all daughters, mother passes X^A or X^a with 50% probability).
A restriction enzyme recognizes the sequence GAATTC and cuts between G and A on both strands. If a linear DNA fragment with the sequence ...GAATTC... is cut with this enzyme, how many fragments will result and what type of ends (blunt or sticky) will be produced? Explain what would happen if DNA ligase was then added.
2 fragments (linear DNA cut once). Sticky ends will be produced (single-stranded overhangs). If DNA ligase is added, it will catalyze the formation of sugar-phosphate bonds, joining the sticky ends back together (re-ligation), potentially recreating the original fragment if the correct ends join.
If a ribosome reads an mRNA molecule in the 3'→5' direction instead of 5'→3', how would this affect polypeptide synthesis? Explain in terms of codon reading and tRNA binding.
The ribosome would read codons incorrectly in the 3'→5' direction (antiparallel to the tRNA anticodon pairing). Since tRNA anticodons are complementary to mRNA codons read 5'→3', the wrong amino acids would be incorporated. The peptide bonds could still form between adjacent amino acids, but the primary structure would be incorrect and the protein would likely be nonfunctional.
An E. coli cell is placed in an environment rich in tryptophan. Using your knowledge of the trp operon, explain in detail whether this cell will synthesize tryptophan and justify why this regulatory mechanism is evolutionarily advantageous.
With tryptophan abundant, the trp operon is turned off (repressible). Tryptophan acts as a corepressor, binds to the repressor protein, activating it, causing it to bind the operator and block transcription. This is evolutionarily advantageous because it conserves energy: the cell doesn't waste resources synthesizing tryptophan when it can obtain it from the environment.
A mutation in the HB gene changes codon 6 from GAG (glutamic acid) to GTG (valine). Calculate the change in net charge of the amino acid at this position (glutamic acid has a charge of -1, valine is neutral). Explain how this single amino acid change leads to the polymerization of hemoglobin molecules and the resulting sickle shape.
Net charge change: from -1 to 0 (loss of one negative charge). Polymerization: Valine is hydrophobic/nonpolar while glutamic acid is charged/polar. The mutation causes hemoglobin to become less soluble. When oxygen levels drop, the hydrophobic valine residue on one hemoglobin molecule binds to a complementary site on another, causing long rigid chains to form. These chains distort RBCs into sickle shape, blocking capillaries and reducing oxygen transport efficiency.
A calico cat (female) has black and orange patches. Explain the cellular and molecular mechanism that produces this pattern. If this calico cat is crossed with a black male, what is the probability of producing a male calico kitten? Justify your answer.
Mechanism: X chromosome inactivation in females. Each cell randomly inactivates one X chromosome (forms Barr body). If heterozygous for fur color gene (X^B = black, X^O = orange), some cells express black allele (X^O inactivated), others express orange allele (X^B inactivated), producing patches. Male calico probability: ~0% (male has XY, needs two X chromosomes to be calico; XXY males [Klinefelter] can be calico but rare).
A researcher wants to insert a human insulin gene into a bacterial plasmid. The plasmid contains an ampicillin resistance gene. After transformation, bacteria are grown on media containing ampicillin. (a) Which bacteria will survive? (b) How would the researcher determine which surviving bacteria actually contain the insulin gene? (c) Explain the role of both restriction enzymes and DNA ligase in this process.
(a) Bacteria containing the plasmid (with ampicillin resistance gene) will survive. (b) Screen for presence of the insulin gene using antibiotic selection with a second marker, colony hybridization with a labeled probe, or PCR screening. (c) Restriction enzyme cuts both plasmid and insulin gene at specific sequences, creating complementary sticky ends. DNA ligase joins the fragments by forming phosphodiester bonds, creating the recombinant plasmid.
A eukaryotic gene contains 4 exons and 3 introns. Alternative splicing can produce multiple mRNA variants. How many distinct polypeptide products could theoretically be produced from this single gene if all exons can be variably included? Show your reasoning.
With 4 exons, each can either be included or excluded (2^4 = 16 possible combinations). However, at least the first and last exons are typically required. If all 4 exons can be variably included and order is maintained, 2^4 = 16 theoretical products. If requiring at least one exon and maintaining order, 2^4 - 1 = 15 products. (Either 15 or 16 with justification accepted.)
A researcher treats eukaryotic cells with a drug that inhibits the Dicer enzyme. What specific gene regulation pathway would be disrupted and what would be the immediate molecular consequence? Name three potential downstream effects on the cell.
RNA interference (RNAi) pathway would be disrupted. Without Dicer, small RNA molecules (siRNAs and miRNAs) wouldn't be processed from longer double-stranded RNA precursors. Consequences: (1) reduced ability to degrade complementary mRNA targets, (2) loss of post-transcriptional gene silencing, (3) potential overexpression of genes normally regulated by RNAi, (4) loss of defense against viruses, (5) potential cancer development from unregulated oncogenes.
A chromosomal mutation results in the following sequence change: ABCDEFG → ABEDCFG. Identify the specific type of chromosomal mutation. Calculate how many different gamete genotypes would be produced if crossing over occurs within the inverted region in a heterozygous individual, explaining your reasoning.
Inversion. If crossing over occurs within the inverted region in a heterozygous individual (one chromosome normal, one inverted), it produces: 2 viable gametes (normal and inverted) and 2 inviable gametes (with duplications and deletions due to recombination within the inverted loop). Without crossing over: 2 gamete genotypes. With crossing over in inversion heterozygotes: 2 viable, 2 inviable = effectively 2 viable gamete types, but if considering recombinant chromatids produced, 4 total products from meiosis.
A pedigree shows an affected father with an affected daughter and an unaffected son. The affected daughter has an affected son. This trait appears in every generation and affects both sexes equally. Identify the most likely inheritance pattern and explain why the other patterns (X-linked dominant, X-linked recessive, autosomal recessive) can be ruled out using specific evidence from the pedigree described.
Autosomal dominant. Evidence: affects every generation (no skipping), affects both sexes equally, affected father has affected daughter and unaffected son (rules out X-linked dominant, which would make all daughters affected). Autosomal recessive ruled out because affected individuals appear in every generation. X-linked recessive ruled out because affected father would pass X to all daughters (making them carriers, not necessarily affected) and males would be more commonly affected.
Design a 3-step protocol to clone a sheep using somatic cell nuclear transfer. For each step, identify the cell types used and the key manipulation performed. Then explain why Dolly the sheep was genetically identical to the donor sheep rather than the surrogate mother.
Protocol: Step 1: Obtain an enucleated egg cell (remove nucleus from donor egg cell). Step 2: Obtain a somatic cell (mammary cell) from sheep to be cloned; starve to arrest in G0 phase. Step 3: Fuse the somatic cell with the enucleated egg cell (electrofusion or chemical fusion), stimulate cell division, implant embryo into surrogate mother. Dolly was genetically identical to the donor sheep because the nuclear DNA came entirely from the donor's somatic cell, not the surrogate's egg (enucleated) or the surrogate mother's cells.
A mutation in the gene encoding rRNA's catalytic site prevents peptide bond formation. If a cell has this mutation, at what specific step of translation would the process be blocked? Would tRNA molecules still be able to bind to the ribosome? Justify both answers.
Elongation would be blocked specifically at peptide bond formation. tRNA molecules could still bind to the A and P sites (charging and anticodon-codon pairing occur at different sites), but the ribozyme activity of rRNA (peptidyl transferase) would be nonfunctional, so the growing polypeptide chain couldn't be transferred to the incoming amino acid.
Compare and contrast the regulation of the lac operon and trp operon in terms of: (a) whether they are inducible or repressible, (b) the role of the effector molecule, (c) the default state of the operon, and (d) the biological rationale for each system existing as it does.
(a) Lac operon is inducible; trp operon is repressible. (b) Lac operon: inducer (allolactose) inactivates repressor; trp operon: corepressor (tryptophan) activates repressor. (c) Lac operon default = OFF; trp operon default = ON. (d) Lac operon: only useful when lactose present, so off by default conserves energy; trp operon: tryptophan is essential, so on by default ensures synthesis, shut off when available to conserve energy.
A gene is 1,500 base pairs long and contains 3 exons separated by 2 introns. The first exon is 200 bp, the second exon is 350 bp, and the third exon is 400 bp. A deletion mutation removes 15 nucleotides from the second exon. Calculate the length of the mRNA transcript produced from the mutated gene (in nucleotides, excluding modifications). Then determine how many amino acids the resulting polypeptide would contain (assuming no stop codon changes), and explain whether this could be a frameshift mutation and why.
Exon 2 originally 350 bp, after deletion: 350 - 15 = 335 bp. Total mRNA = 200 + 335 + 400 = 935 nucleotides (no introns transcribed). Amino acid count = 935/3 = 311.67, so 311 amino acids (assuming no stop codon changes). This is NOT a frameshift mutation if 15 nucleotides (multiple of 3) were deleted. It's an in-frame deletion removing 5 amino acids, not shifting the reading frame.
A population has the following allele frequencies for a recessive autosomal disorder: q = 0.01 for the disease allele. (a) Calculate the frequency of carriers in this population assuming Hardy-Weinberg equilibrium. (b) If two carriers have children, calculate the probability that their first child will be affected, and the probability that their first two children will both be affected. (c) Explain why this disorder might persist in the population despite being deleterious, providing at least two distinct mechanisms. A population has the following allele frequencies for a recessive autosomal disorder: q = 0.01 for the disease allele. (a) Calculate the frequency of carriers in this population assuming Hardy-Weinberg equilibrium. (b) If two carriers have children, calculate the probability that their first child will be affected, and the probability that their first two children will both be affected. (c) Explain why this disorder might persist in the population despite being deleterious, providing at least two distinct mechanisms.
(a) q = 0.01, so p = 0.99. Carrier frequency = 2pq = 2(0.99)(0.01) = 0.0198 = 1.98%. (b) Two carriers (Aa × Aa): affected child probability = 1/4 = 25%. Two affected children in a row = (1/4)(1/4) = 1/16 = 6.25%. (c) Persistence mechanisms: (1) heterozygote advantage (e.g., sickle cell trait confers malaria resistance), (2) mutation-selection balance (new mutations continuously introduce disease alleles), (3) founder effects in isolated populations, (4) late-onset diseases (affected individuals reproduce before symptoms appear).
A patient has cystic fibrosis caused by a deletion of 3 bases in the CFTR gene. (a) Calculate how many amino acids are missing from the CFTR protein as a result of this deletion. (b) Explain the cascade of effects from this molecular change to the symptoms, including: protein folding, chloride transport, water movement, and mucus production. (c) Explain the heterozygote advantage for the CF allele, identifying the specific pathogen and mechanism involved.
(a) Deletion of 3 bases removes exactly 1 amino acid (phenylalanine). (b) Effects: CFTR protein missing phenylalanine → improper folding → protein degraded by cell → no chloride channels on membrane → chloride cannot cross membrane → chloride gradient disrupted → water movement impaired → thick, heavy mucus accumulates in lungs, pancreas, liver, reproductive tract → respiratory and digestive dysfunction. (c) Heterozygote advantage: carriers (one CF allele, one normal) have resistance to typhoid fever. S. Typhi bacteria enter the body through M cells in digestive system by binding to normal CFTR protein. Nonfunctional CFTR from mutated allele blocks bacterial entry. Carriers have ~50% normal CFTR, enough to avoid CF symptoms but resistant to typhoid.