ap bio frq unit 6
10. Explain how a single base-pair mutant in DNA can alter the structure and, in some cases, the function of a protein.
A single base-pair mutation can alter the structure and function of a protein through either insertion or deletion because it will result in a frameshift of the codons transcripted after this point. This means that any codons after the mutation will be incorrect and coding for the wrong amino acid sequence.
1.Explain the role of each of the following in protein synthesis in eukaryotic cells. • RNA polymerase • Spliceosomes (snRNPs) • Codons • Ribosomes • tRNA
RNA polymerase-It unwinds DNA and makes mRNASpliceosomes (snRNPs)-Cuts out introns and puts together exons Codons-They code for specific amino acids Ribosomes-Help to assemble proteins tRNA-Carries amino acids to ribosome
7. (a) What types of modifications may occur to this RNA before it leaves the nucleus? (b) Once in the cytoplasm, how is the mRNA translated to a protein? (c) If the cell is a secretory cell, how is the protein from part (b) eventually targeted, packaged, and secreted to the exterior of the cell?
a)Splicing, capping, and addition of a poly-A tail b)The mRNA will attach to the small ribosomal unit and then the large ribosomal unit. Then the mRNA will be decoded with its codon's tRNA. The anticodon will bring amino acids from the cytoplasm to the mRNA. The tRNA will come in at the A site and a peptide bond will form between amino acids at the A and P sites. Now the empty tRNA will exit the ribosome and this will continue until the stop sequence is reached. c)The protein needs an srp signal which will send it to the ER, and then on to the golgi bodies, which will pack it away in vesicles, which will transfer it to the cell membrane where it can exit through exocytosis.
11. (a) Describe transcription and translation. (b) Identify similarities between transcription and translation. (c) Identify differences between transcription and translation. (d) Describe structural changes that can occur to a protein after translation to make it function properly.
a)Transcription: The synthesis of RNA on a DNA template Translation: The synthesis of a polypeptide using the genetic information encoded in an mRNA molecule b)mRNA is involved in both processes and they both aid in the synthesis of proteins c) - Transcription: products are mRNA, tRNA, rRNA, and non-coding RNA; take place in the nucleus - Translation: product is protein; takes place in the cytoplasm d)Activation into a functional protein through cleavage of certain amino acid sequences; the amino acid sequence can fold to form the secondary or tertiary structure
9. (a) Describe the three structural components of an RNA nucleotide monomer. Explain the role of RNA polymerase during transcription. (b) Identify the dependent variable in the experiments. Identify a control group missing from the second experiment. Justify the need for this control group in the second experiment. (c) Describe the effect of amanitin on the maximum elongation rate for the wild-type and modified RNA polymerases. Determine the ratio of the average maximum elongation rate for the modified RNA polymerase compared to the wild strain RNA polymerase in Figure 1. (d) State the null hypothesis for the experiment in Figure 1. Provide reasoning to justify the claim that the change in the amino acid sequence in the modified RNA polymerase affected the shape of the active site on the enzyme.
a)A description that an RNA nucleotide has the following three structural components: a five-carbon sugar (ribose), a phosphate group, and a nitrogen base (adenine, cytosine, guanine, or uracil) - RNA polymerase synthesizes a new RNA molecule based on a DNA template by matching the current DNA base with the proper RNA complement. - RNA polymerase joins/bonds the newly paired RNA nucleotide and the growing RNA strand with a covalent bond. b)The response includes all of the following criteria.The dependent variable is identified as the maximum elongation rate of the mRNA.An acceptable control group missing from the second experiment is identified. Acceptable control groups include the following.- Wild strain from the first experiment without amanitin- Experimental strain from the first experiment without amanitinThe justification that the missing control is needed because the effect of amanitin on the maximum elongation rate cannot be determined without comparison to the maximum elongation rate under the same conditions without amanitin. c) The response includes all of the following criteria.A description that amanitin decreases the maximum elongation rate for the wild strain and does not affect the rate in the experimental strainThe ratio of the average maximum elongation rate for the modified RNA polymerase compared to the wild strain RNA polymerase is determined to be 1:6 / 1/6 / 1 to 6 . d) - The modified RNA polymerase would not affect the maximum elongation rate. - Any difference between the two elongation rates is due to chance. A description that a change to the active site would explain the decrease in the elongation rate is provided as reasoning to justify the claim.
4. (a) The genetic material in one eukaryotic cell is copied and distributed to two identical daughter cells (b) A gene in a eukaryotic cell is transcribed and translated to produce a protein (c) The genetic material from one bacterial cell enters another via transformation, transduction, or conjugation
a)DNA copied, helicase unzips/unwinds, RNA primer, semiconservative relationship among bases; P: condensed chromosomes M: alignment to the middle A: chromatids pulled to opposite poles T: cell cleavage furrow b)- Transcription:DNA to RNA, promoter recognition, complementary bases, 5' to 3' - Translation:base sequence to amino acid sequence, complementary (codon to anticodon), peptide formation (aa joined by peptide bonds = polypeptide), stop codon + release of polypeptide c)RNA
3. (a) Provide ONE piece of evidence that would indicate new genetic variation has occurred in the engineered flies. (b) Describe ONE mechanism that could lead to genetic variation in the engineered strain of flies. (c) Describe how genetic variation in a population contributes to the process of evolution in the population.
a)Different mRNA sequence and DNA sequence b)Sexual reproduction produces offspring with new combinations of alleles/traits c)Without genetic variation, there is no phenotypic variation on which natural selection can act
6. (a) Assuming that species I is the ancestral species of the group, explain the most likely genetic change that produced the polypeptide in species II and the most likely genetic change that produced the polypeptide in species III. (b) Predict the effects of the mutation on the structure and function of the resulting protein in species IV. Justify your prediction.
a)In specie 2, the most likely genetic change was point mutation, in which a single nucleotide is altered, in which changes the amino acid being coded for.In specie 3, the most likely gentic change would be a point mutation that coded for a STOP codon, which halted production of polypeptide prematurely. b)The protein produced in Specie 4 will have a much different structure, as well as the function. The frameshift mutation that resulted in the polypeptide completely altered the polypeptide chain, which in turn will change the interaction of the amino acids, so the structure is changed. The altered amino acid sequence will cause the protein's function to shift.
2. (a) Explain how ligand A and ligand B can cause identical cellular responses in a cell. (b) Predict the most likely effect of a two-nucleotide deletion in the middle of the intron located between exons 4 and 5 on the structure of protein A. Justify your prediction.
a)Ligand A and ligand B can cause identical cellular responses in a cell because they use the same secondary messenger and activate the same signal transduction pathway. b)There is no change because the intron does not contain protein coding information.
5. (a) Discuss the organization of the genetic material in prokaryotes and eukaryotes. (b) Contrast the following activities in prokaryotes and eukaryotes: • Replication of DNA • Transcription or translation • Gene regulation • Cell division
a)PROKARYOTES -no introns -not in nucleus -circular -no histones -usually one chromosome -plasmids common -supercoiled DNA EUKARYOTES -introns -in nucleus -linear -histones -usually more than one chromosome -plasmids rare -chromatin DNA b)PROKARYOTES -Dna: found freely in the cytoplasm -Transcription: no RNA processing, occurs in the cytoplasm -Ribosome: small ribosome EUKARYOTES -Dna: membrane bound in the nucleus -Transcription: RNA processing, occurs in the nucleus -Ribosome: larger ribosome
8. (a) Describe the role of THREE of the following in the regulation of protein synthesis: • RNA splicing • repressor proteins • methylation • siRNA (b) Information flow can be altered by mutation. Describe THREE different types of mutations and their effect on protein synthesis. (c) Identify TWO environmental factors that increase the mutation rate in an organism, and discuss their effect on the genome of the organism. (d) Epigenetics is the study of heritable changes in the phenotype caused by mechanisms other than changes in the DNA sequence. Describe ONE example of epigenetic inheritance.
a)RNA Splicing- take out introns and splices the exons together. Repressor proteins prevent transcription and translation and regulate gene expression (turning them off). siRNA degrade the mRNA and inhibit translation. b) - Silent mutation: when a one nucleotide is changed and there is no change in the amino acid/protein sequence - Substitution (missense): when one nucleotide change causes a new codon, therefore a different amino acid/protein sequence - Substitution (nonsense): when one nucleotide change causes a stop codon, therefore the protein is truncated or not formed at all c) - Carcinogens (ex. cigarette smoke)Effect: DNA is altered/damaged which affects the products of the affected genes - Viruses: disrupt gene sequence d) Histone acetylation: The acetylation of histone tails in the nucelosomes loosens the structure of a chromsome, therefore increasing the rate of expression of the more exposed genes.