Bio Unit 3 Practice Questions PDF

Unit 3 Practice Questions Lecture 1: DNA Lecture 1: DNA Lecture 1: DNA Lecture 1: DNA Sulfur but not phosphorus Lecture 1: DNA Lecture 1: DNA Lecture 1: DNA Lecture 1: DNA Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Both parent The parent strands end up strands get back together split up Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Condensation Nucleoside triphosphate to the 3’ OH group (connecting phosphate to sugar) From the hydrolysis of the outer 2 phosphates Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication Lecture 2: DNA Replication 5’ 3’ Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA DNA Polymerase RNA Polymerase Primer promoter Helicase RNA Polymerase ATCG AUCG Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 3: DNA & RNA Lecture 4: mRNA Lecture 4: mRNA Lecture 4: mRNA Lecture 4: mRNA Lecture 4: mRNA Lecture 4: mRNA Lecture 4: mRNA Lecture 4: mRNA There wouldn’t be any transcription Lecture 4: mRNA Lecture 4: mRNA Lecture 4: mRNA Lecture 4: mRNA Lecture 5: Translation Lecture 5: Translation 1. A 2. B 3. C Lecture 5: Translation Lecture 5: Translation DNA: 5’- GAT- 3’ mRNA: 3’- CUA-5’ tRNA: 5’- GAU-5’ In Class Quiz 1. Which of these processes require DNA polymerase? A. only #1 B. only #2 C. only #3 D. #1 and #2 E. #2 and #3 F. #1 and #3 G. all three In Class Quiz 1. Which of these processes require DNA polymerase? A. only #1 B. only #2 C. only #3 D. #1 and #2 E. #2 and #3 F. #1 and #3 G. all three In Class Quiz 2. Which of these processes in eukaryotic cells occur(s) in the nucleus? A. only #1 B. only #2 C. only #3 D. #1 and #2 E. #2 and #3 F. #1 and #3 G. all three In Class Quiz 2. Which of these processes in eukaryotic cells occur(s) in the nucleus? A. only #1 B. only #2 C. only #3 D. #1 and #2 E. #2 and #3 F. #1 and #3 G. all three In Class Quiz 3. If there is a significant mutation in the promoter for gene X, which will result? A. Gene X cannot be replicated but can be transcribed & translated B. Gene X can be replicated but cannot be transcribed & translated C. Gene X can be replicated and transcribed but not translated. D. Gene X cannot be replicated or transcribed but can be translated. In Class Quiz 3. If there is a significant mutation in the promoter for gene X, which will result? A. Gene X cannot be replicated but can be transcribed & translated B. Gene X can be replicated but cannot be transcribed & translated C. Gene X can be replicated and transcribed but not translated. D. Gene X cannot be replicated or transcribed but can be translated. In Class Quiz 4. If the mRNA codon for a specific amino acid is 5’ CGA 3’, what is the anticodon on the tRNA carrying that amino acid? [Note all given as 5’ – 3’ ] A. 5’ CGU 3’ B. 5’ GCU 3’ C. 5’ UCG 3’ D. 5’ UGC 3’ In Class Quiz 4. If the mRNA codon for a specific amino acid is 5’ CGA 3’, what is the anticodon on the tRNA carrying that amino acid? [Note all given as 5’ – 3’ ] A. 5’ CGU 3’ B. 5’ GCU 3’ C. 5’ UCG 3’ D. 5’ UGC 3’ In Class Quiz 5. Which of these describes why the genetic code is said to be “redundant”? A. each codon codes for more than one amino acid B. each amino acid is coded for by more than one codon C. each gene can have multiple introns D. there are multiple start codons In Class Quiz 5. Which of these describes why the genetic code is said to be “redundant”? A. each codon codes for more than one amino acid B. each amino acid is coded for by more than one codon C. each gene can have multiple introns D. there are multiple start codons In Class Quiz 6. Which of the following statements correctly describes alternative mRNA splicing? A. It is a mechanism that can increase the rate of transcription B. It causes mutations that produce abnormal proteins C. It allows the production of different protein products from a single gene D. It allows the production of similar proteins from different genes. In Class Quiz 6. Which of the following statements correctly describes alternative mRNA splicing? A. It is a mechanism that can increase the rate of transcription B. It causes mutations that produce abnormal proteins C. It allows the production of different protein products from a single gene D. It allows the production of similar proteins from different genes. In Class Quiz 7. Which of these best describes Okazaki fragments? A. Segments of single-stranded DNA made on both sides (strands) of the replication fork B. Segments of single-stranded DNA made in replication where there are multiple primers C. Segments of DNA that correspond in length to the gene for one protein D. Segments of DNA added to replace RNA primers In Class Quiz 7. Which of these best describes Okazaki fragments? A. Segments of single-stranded DNA made on both sides (strands) of the replication fork B. Segments of single-stranded DNA made in replication where there are multiple primers C. Segments of DNA that correspond in length to the gene for one protein D. Segments of DNA added to replace RNA primers In Class Quiz 8. Which of the following is NOT true about tRNA (transfer RNA)? A. It is made in the nucleus B. There are different tRNAs for different amino acids C. It aligns its anticodon to the mRNA codon by hydrogen bonding D. It attaches to its specific amino acid after aligning to the mRNA codon E. More than one of these is FALSE about tRNAs In Class Quiz 8. Which of the following is NOT true about tRNA (transfer RNA)? A. It is made in the nucleus B. There are different tRNAs for different amino acids C. It aligns its anticodon to the mRNA codon by hydrogen bonding D. It attaches to its specific amino acid after aligning to the mRNA codon E. More than one of these is FALSE about tRNAs In Class Quiz 9. Which is the main function of ribosomal RNA (rRNA)? A. Is translated to form a protein B. Holds form and structure within a ribosome C. Attaches the tRNA to its amino acid D.Identifies splice sites of introns in mRNA splicing In Class Quiz 9. Which is the main function of ribosomal RNA (rRNA)? A. Is translated to form a protein B. Holds form and structure within a ribosome C. Attaches the tRNA to its amino acid D.Identifies splice sites of introns in mRNA splicing In Class Quiz 10. Line 1 is the original DNA base sequence for a gene. Line 2 is that DNA base sequence after a mutation. line 1: A T C C G T A C T G line 2: A T C C G T T A C T G This change causes which type of mutation? A. missense mutation B. silent mutation C. nonsense mutation D. frameshift mutation In Class Quiz 10. Line 1 is the original DNA base sequence for a gene. Line 2 is that DNA base sequence after a mutation. line 1: A T C C G T A C T G line 2: A T C C G T T A C T G This change causes which type of mutation? A. missense mutation B. silent mutation C. nonsense mutation D. frameshift mutation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 6: Prokaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation 1. F 2. D (A+C) 3. F Lecture 7: Eukaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation Lecture 7: Eukaryotic Gene Regulation Lecture 8: Cell Differentiation Lecture 8: Cell Differentiation Lecture 8: Cell Differentiation Lecture 8: Cell Differentiation Genes 1, 2, 3, 4 & 5 Lecture 8: Cell Differentiation Lecture 8: Cell Differentiation Lecture 8: Cell Differentiation Lecture 8: Cell Differentiation Steroid hormones can enter the nucleus and directly turn on/off genes Lecture 8: Cell Differentiation Lecture 8: Cell Differentiation In-class Quiz A bacterial operon with four protein-coding genes has ____ promoter(s). When transcribed, the resulting mRNA contains ____ start codon(s) where translation of a protein will begin. A. one ; one B. one ; four C. four ; four D. four ; one In-class Quiz A bacterial operon with four protein-coding genes has ____ promoter(s). When transcribed, the resulting mRNA contains ____ start codon(s) where translation of a protein will begin. A. one ; one B. one ; four C. four ; four D. four ; one In-class Quiz Which of the following statements is correct regarding protein synthesis in prokaryotes and eukaryotes? A. Transcription and translation of the same mRNA can occur at the same time in both organisms. B. Translation can begin while transcription is still in progress in prokaryotes, but not in eukaryotes. C. The RNA produced during transcription is then modified in both types of organisms. D. In prokaryotes, transcription produces a protein, but in eukaryotes, it produces RNA. In-class Quiz Which of the following statements is correct regarding protein synthesis in prokaryotes and eukaryotes? A. Transcription and translation of the same mRNA can occur at the same time in both organisms. B. Translation can begin while transcription is still in progress in prokaryotes, but not in eukaryotes. C. The RNA produced during transcription is then modified in both types of organisms. D. In prokaryotes, transcription produces a protein, but in eukaryotes, it produces RNA. In-class Quiz What is an evolutionary advantage of prokaryotes having genes organized in operons? A. cells can produce fewer proteins to accomplish the same function B. cells can produce shorter messenger RNA molecules C. cells can turn on/off genes for related processes together D. cells do not need to splice their messenger RNA In-class Quiz What is an evolutionary advantage of prokaryotes having genes organized in operons? A. cells can produce fewer proteins to accomplish the same function B. cells can produce shorter messenger RNA molecules C. cells can turn on/off genes for related processes together D. cells do not need to splice their messenger RNA In-class Quiz Which is true for Eukaryotes but not true for Prokaryotes? A. Many genes have introns, and many genes are in operons B. Many genes have introns, and many genes have enhancers C. Many genes are in operons but they do not have enhancers D. Many genes are in operons and most have enhancers In-class Quiz Which is true for Eukaryotes but not true for Prokaryotes? A. Many genes have introns, and many genes are in operons B. Many genes have introns, and many genes have enhancers C. Many genes are in operons but they do not have enhancers D. Many genes are in operons and most have enhancers In-class Quiz In both prokaryotes and eukaryotes RNA polymerase begins ___ and ends it at ___ A. translation at a start codon; stop codon B. transcription at a start codon; stop codon C. translation at a promoter; stop codon D. transcription at a promoter; stop codon E. transcription at a promoter; termination sequence In-class Quiz In both prokaryotes and eukaryotes RNA polymerase begins ___ and ends it at ___ A. translation at a start codon; stop codon B. transcription at a start codon; stop codon C. translation at a promoter; stop codon D. transcription at a promoter; stop codon E. transcription at a promoter; termination sequence In-class Quiz Undifferentiated human cells mature and become specialized for different functions by: A. differential gene expression B. differential mutations C. differential cell death D. differential rates of cell division In-class Quiz Undifferentiated human cells mature and become specialized for different functions by: A. differential gene expression B. differential mutations C. differential cell death D. differential rates of cell division In-class Quiz Eukaryotic DNA in which of the following chromatin states is MOST likely to be expressed? A. Methylated DNA, acetylated histone; B. Methylated DNA, non-acetylated histone C. Unmethylated DNA, acetylated histone D. Unmethylated DNA, non-acetylated histone In-class Quiz Eukaryotic DNA in which of the following chromatin states is MOST likely to be expressed? A. Methylated DNA, acetylated histone; B. Methylated DNA, non-acetylated histone C. Unmethylated DNA, acetylated histone D. Unmethylated DNA, non-acetylated histone In-class Quiz What is the advantage of a eukaryotic gene having enhancer regions? A. RNA polymerase can start transcription at the enhancer B. if activators bind there, RNA polymerase transcribes many times C. if activators bind there, no repressors can bind there In-class Quiz What is the advantage of a eukaryotic gene having enhancer regions? A. RNA polymerase can start transcription at the enhancer B. if activators bind there, RNA polymerase transcribes many times C. if activators bind there, no repressors can bind there In-class Quiz Which of these describes chromatin in a eukaryotic chromosome? A. DNA is wrapped around a cluster of histone proteins B. DNA is folded into a sphere, with a layer of histone surrounding it C. In order to be expressed, genes in chromatin are closely packed In-class Quiz Which of these describes chromatin in a eukaryotic chromosome? A. DNA is wrapped around a cluster of histone proteins B. DNA is folded into a sphere, with a layer of histone surrounding it C. In order to be expressed, genes in chromatin are closely packed In-class Quiz Which of these would contribute to a cell’s specialization? A. tissue-specific activators are present B. a signal outside the cell binds to a receptor C. a master gene transcription factor turns on other genes D. all three of these contribute to cell specialization. In-class Quiz Which of these would contribute to a cell’s specialization? A. tissue-specific activators are present B. a signal outside the cell binds to a receptor C. a master gene transcription factor turns on other genes D. all three of these contribute to cell specialization. Lecture 9: Cell Cycle Mitosis Lecture 9: Cell Cycle Mitosis Lecture 9: Cell Cycle Mitosis Lecture 9: Cell Cycle Mitosis 2. B 3. C - Has two double helices Lecture 9: Cell Cycle Mitosis Lecture 9: Cell Cycle Mitosis Lecture 9: Cell Cycle Mitosis Lecture 9: Cell Cycle Mitosis Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Mitosis promoting factor Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Phosphorylation activates a kinase Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Lecture 10: Cell Cycle and Signaling Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: Meiosis Lecture 11: In class quiz Which of the following is NOT true about the two members of a homologous pair of chromosomes? A. each member is a double helix B. they are identical DNA molecules C. they will be replicated during the S phase D. they contain the same gene loci in the same order Lecture 11: In class quiz Which of the following is NOT true about the two members of a homologous pair of chromosomes? A. each member is a double helix B. they are identical DNA molecules C. they will be replicated during the S phase D. they contain the same gene loci in the same order Lecture 11: In class quiz The term “sister chromatids” (or “twin chromatids”) refers to which of these? A. the two complementary strands in one DNA molecule B. the two identical molecules from DNA replication C. the two members of a homologous pair of chromosomes D. any two chromosomes that originally came from the same parent Lecture 11: In class quiz The term “sister chromatids” (or “twin chromatids”) refers to which of these? A. the two complementary strands in one DNA molecule B. the two identical molecules from DNA replication C. the two members of a homologous pair of chromosomes D. any two chromosomes that originally came from the same parent Lecture 11: In class quiz If a cell with 23 homologous pairs of chromosomes undergoes one round of cell division by mitosis, the result will be: A. Two cells, each with 23 chromosomes; B. Two cells, each with 46 chromosomes C. Four cells, each with 23 chromosomes D. Four cells, each with 46 chromosomes Lecture 11: In class quiz If a cell with 23 homologous pairs of chromosomes undergoes one round of cell division by mitosis, the result will be: A. Two cells, each with 23 chromosomes; B. Two cells, each with 46 chromosomes C. Four cells, each with 23 chromosomes D. Four cells, each with 46 chromosomes Lecture 11: In class quiz Staining chromosomes in a cell early in mitosis reveals structures that resemble elongate X’s. What is true about one of those X structures? A. it shows two double-helix DNA molecules B. it shows one homologous pair of chromosomes, one from mom and one from dad C. Many of its genes are currently being transcribed D. Two of these are true E. All three of these are true Lecture 11: In class quiz Staining chromosomes in a cell early in mitosis reveals structures that resemble elongate X’s. What is true about one of those X structures? A. it shows two double-helix DNA molecules B. it shows one homologous pair of chromosomes, one from mom and one from dad C. Many of its genes are currently being transcribed D. Two of these are true E. All three of these are true Lecture 11: In class quiz Which of these is NOT something happens early in Prophase of Mitosis? A. chromosomes condense B. DNA replicates C. spindle forms D. nuclear membrane dissociates Lecture 11: In class quiz Which of these is NOT something happens early in Prophase of Mitosis? A. chromosomes condense B. DNA replicates C. spindle forms D. nuclear membrane dissociates Lecture 11: In class quiz In the eukaryotic cell cycle, DNA replication occurs A. only during G1 phase B. only during G2 phase C. only at the beginning of M phase D. only during S phase E. DNA replication can occur at two of these times. Lecture 11: In class quiz In the eukaryotic cell cycle, DNA replication occurs A. only during G1 phase B. only during G2 phase C. only at the beginning of M phase D. only during S phase E. DNA replication can occur at two of these times. Lecture 11: In class quiz In G2 phase of the cell cycle, one eukaryotic chromosome consists of: A. one homologous pair B. one twin chromatid C. one double helix model of DNA D. two twin chromatids, attached Lecture 11: In class quiz In G2 phase of the cell cycle, one eukaryotic chromosome consists of: A. one homologous pair B. one twin chromatid C. one double helix model of DNA D. two twin chromatids, attached Lecture 11: In class quiz Which of these would be an important check that should be completed before the cell begins S phase? A. Attach DNA to microtubules B. Repair any broken DNA C. Nuclear membrane dissociates D. Condense the chromosomes Lecture 11: In class quiz Which of these would be an important check that should be completed before the cell begins S phase? A. Attach DNA to microtubules B. Repair any broken DNA C. Nuclear membrane dissociates D. Condense the chromosomes Lecture 11: In class quiz The reason we are studying cancer at this week in the semester is: A. cancer cells often undergo apoptosis B. many cancer-causing oncogenes speed up the cell cycle C. cancer cells typically exit the cell cycle and enter G0 phase D. cancer cells pass slowly through the cell cycle E. two of these are true of cancer cells Lecture 11: In class quiz The reason we are studying cancer at this week in the semester is: A. cancer cells often undergo apoptosis B. many cancer-causing oncogenes speed up the cell cycle C. cancer cells typically exit the cell cycle and enter G0 phase D. cancer cells pass slowly through the cell cycle E. two of these are true of cancer cells Lecture 11: In class quiz The reason we are reviewing signal transduction at this week in the semester is: A. signal transduction pathways can show amplification B. signal transduction pathways can turn on genes C. signal transduction kinases control passing through the cell cycle D. cyclic AMP can activate transcription factors Lecture 11: In class quiz The reason we are reviewing signal transduction at this week in the semester is: A. signal transduction pathways can show amplification B. signal transduction pathways can turn on genes C. signal transduction kinases control passing through the cell cycle D. cyclic AMP can activate transcription factors Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 12: Meiosis and Mendel Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance The principle of segregation says that the 2 alleles for a trait end up in 2 different gametes during meiosis Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 13: Mendelian Inheritance Lecture 14: Inheritance & Review Lecture 14: Inheritance & Review Lecture 14: Inheritance & Review Lecture 14: Inheritance & Review Lecture 14: Inheritance & Review Start with all-heavy DNA. After one replication with only new light nucleotides as building material, according to the semi-conservative model, the results are 2 DNA molecules: A. one DNA all heavy and the other all light B. two DNA both strands all heavy C. two DNA both strands all light D. two DNA with one strand heavy and one strand light Lecture 14: Inheritance & Review Start with all-heavy DNA. After one replication with only new light nucleotides as building material, according to the semi-conservative model, the results are 2 DNA molecules: A. one DNA all heavy and the other all light B. two DNA both strands all heavy C. two DNA both strands all light D. two DNA with one strand heavy and one strand light Lecture 14: Inheritance & Review If a mutation in the lac operon promoter, proteins from which genes will be made? A. Gene A B. Gene B C. Gene C D. all of these E. none of these 6. If a mutation in the start codon for Gene C, which proteins will be made? A. Gene A only B. Genes A and B C. all three proteins D. none of these proteins Lecture 14: Inheritance & Review If a mutation in the lac operon promoter, proteins from which genes will be made? A. Gene A B. Gene B C. Gene C D. all of these E. none of these 6. If a mutation in the start codon for Gene C, which proteins will be made? A. Gene A only B. Genes A and B C. all three proteins D. none of these proteins Lecture 14: Inheritance & Review In this liver cell, mRNA for albumin is being made but mRNA for crystallin is not. What is the explanation for this? A. Liver cells have the albumin gene but not the crystallin gene B. Liver cells have enhancers for the albumin gene but not enhancers for the crystallin gene A. Liver cells contain activator proteins that bind to albumin enhancers, but not activator proteins for crystallin enhancers. Lecture 14: Inheritance & Review In this liver cell, mRNA for albumin is being made but mRNA for crystallin is not. What is the explanation for this? A. Liver cells have the albumin gene but not the crystallin gene B. Liver cells have enhancers for the albumin gene but not enhancers for the crystallin gene A. Liver cells contain activator proteins that bind to albumin enhancers, but not activator proteins for crystallin enhancers. Lecture 14: Inheritance & Review A certain mouse species has a genome of 20 homologous pairs. For that species, n = ? A. 10 B. 20. C. 30. D. 40. E. 80 For that species, each somatic (body) cell nucleus contains how many chromosomes? A. 10 B. 20 C. 30. D. 40 E. 80 A sperm of that species would have how many chromosomes? A. 10 B. 20 C. 30. D. 40 E. 80 Lecture 14: Inheritance & Review A certain mouse species has a genome of 20 homologous pairs. For that species, n = ? A. 10 B. 20. C. 30. D. 40. E. 80 For that species, each somatic (body) cell nucleus contains how many chromosomes? A. 10 B. 20 C. 30. D. 40 E. 80 A sperm of that species would have how many chromosomes? A. 10 B. 20 C. 30. D. 40 E. 80 Lecture 14: Inheritance & Review This image is of identical DNA molecules still attached to each other following replication. Which term or terms is/are NOT accurate? A. Sister (twin) chromatids A. Homologous pair A. One replicated chromosome A. Two of these are not accurate Lecture 14: Inheritance & Review This image is of identical DNA molecules still attached to each other following replication. Which term or terms is/are NOT accurate? A. Sister (twin) chromatids A. Homologous pair A. One replicated chromosome A. Two of these are not accurate Lecture 14: Inheritance & Review If a cell with 30 homologous pair of chromosomes undergoes one round of cell division by mitosis, the result will be__?_; if it undergoes meiosis the result will be _?__. L. Two cells, each with 30 chromosomes M. Two cells, each with 60 chromosomes N. Four cells, each with 30 chromosomes O. Four cells each with 60 chromosomes. A. L ; N B. M ; N C. L ; O D. M ; O E. O ; N F. N ; O Lecture 14: Inheritance & Review If a cell with 30 homologous pair of chromosomes undergoes one round of cell division by mitosis, the result will be__?_; if it undergoes meiosis the result will be _?__. L. Two cells, each with 30 chromosomes M. Two cells, each with 60 chromosomes N. Four cells, each with 30 chromosomes O. Four cells each with 60 chromosomes. A. L ; N B. M ; N C. L ; O D. M ; O E. O ; N F. N ; O Lecture 14: Inheritance & Review If a cell with 24 pairs of chromosomes undergoes meiosis, how many cells result, A. Two B. Four How many DNA molecules are in each resulting cell? A. 12 B. 24 C. 48 D. None of these Lecture 14: Inheritance & Review If a cell with 24 pairs of chromosomes undergoes meiosis, how many cells result, A. Two B. Four How many DNA molecules are in each resulting cell? A. 12 B. 24 C. 48 D. None of these Unit 3AEs AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 D AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 7 AE 8 AE 8 AE 8 AE 8 AE 8 AE 8 less less less AE 8 AE 8 AE 8 AE 8 AE 8 AE 8 AE 9 Any mRNA may be translated many times to make many copies of the protein before being degraded. mRNAs in bacterial cells usually last 5-10 minutes, while mRNAs in eukaryotic cells last 10-15 hours. Which explains the difference? A. Bacterial mRNA can have simultaneous transcription and translation. B. Eukaryotic mRNAs have their introns spliced out. C. Eukaryotic mRNAs have a cap and poly-A tail added D.Eukaryotic mRNAs must be transferred out of the nucleus before translation. AE 9 Any mRNA may be translated many times to make many copies of the protein before being degraded. mRNAs in bacterial cells usually last 5-10 minutes, while mRNAs in eukaryotic cells last 10-15 hours. Which explains the difference? A. Bacterial mRNA can have simultaneous transcription and translation. B. Eukaryotic mRNAs have their introns spliced out. C. Eukaryotic mRNAs have a cap and poly-A tail added D.Eukaryotic mRNAs must be transferred out of the nucleus before translation. AE 9 What is the advantage of alternate mRNA splicing in the development of specialized cells? A. different gene products can be spliced together to form different proteins B. slightly different protein forms can be made from the same gene C. certain mRNA’s can be translated many times and others quickly destroyed AE 9 What is the advantage of alternate mRNA splicing in the development of specialized cells? A. different gene products can be spliced together to form different proteins B. slightly different protein forms can be made from the same gene C. certain mRNA’s can be translated many times and others quickly destroyed AE 9 How is acetylated histone different from non-acetylated (typical) histone? A. it is negatively charged B. it is positively charged C. it is electrically neutral AE 9 How is acetylated histone different from non-acetylated (typical) histone? A. it is negatively charged B. it is positively charged C. it is electrically neutral AE 9 DNA in which of the following chromatin states is LEAST likely to be expressed? A. Methylated DNA, unpacked from histone B. Methylated DNA, packed close to histone C. Unmethylated DNA, unpacked from histone D. Unmethylated DNA, packed close to histone AE 9 DNA in which of the following chromatin states is LEAST likely to be expressed? A. Methylated DNA, unpacked from histone B. Methylated DNA, packed close to histone C. Unmethylated DNA, unpacked from histone D. Unmethylated DNA, packed close to histone AE 9 Refer to the previous figure (15.10) and rank the following cells according to the total amount of polypeptide they will make from that gene (from most to least). [Will “least” be zero?] 1. Cells with enhancer DNA & promoter DNA & activator proteins & general transcription factors 2. Cells with enhancer DNA & promoter DNA but no control proteins. 3. Cells with enhancer DNA & promoter DNA & general transcription factors 4. Cells with enhancer DNA & promoter DNA & activator proteins & general transcription factors & mediator proteins A. 1, 4, 2, 3 B. 4, 1, 3, 2 C. 3, 4, 2, 1 D. 4, 3, 1, 2 AE 9 Refer to the previous figure (15.10) and rank the following cells according to the total amount of polypeptide they will make from that gene (from most to least). [Will “least” be zero?] 1. Cells with enhancer DNA & promoter DNA & activator proteins & general transcription factors 2. Cells with enhancer DNA & promoter DNA but no control proteins. 3. Cells with enhancer DNA & promoter DNA & general transcription factors 4. Cells with enhancer DNA & promoter DNA & activator proteins & general transcription factors & mediator proteins A. 1, 4, 2, 3 B. 4, 1, 3, 2 C. 3, 4, 2, 1 D. 4, 3, 1, 2 AE 9 Cells in your big toe and cells in your eye contain genes for making toenails. Which of these could explain why toenails do not grow on your eyes? A. Eye cells have mutations in toenail genes. B. Toenail gene DNA is methylated in eye cells C. Toenail gene histone is acetylated in eye cells D. Activators in eye cell bind to toenail enhancer E. Two of these are explanations F. Three of these are explanations G. All four of these (A-D) are explanations AE 9 Cells in your big toe and cells in your eye contain genes for making toenails. Which of these could explain why toenails do not grow on your eyes? A. Eye cells have mutations in toenail genes. B. Toenail gene DNA is methylated in eye cells C. Toenail gene histone is acetylated in eye cells D. Activators in eye cell bind to toenail enhancer E. Two of these are explanations F. Three of these are explanations G. All four of these (A-D) are explanations AE 10 AE 10 AE 10 AE 10 AE 10 AE 10 AE 10 AE 10 In the middle of meiosis 1, the homologous pairs haven’t split up yet. This means that the cell has a full set of duplicated DNA or double the amount of DNA content/molecules than a cell in G1 AE 10 AE 10 Unit 3 HWs HW 7 1. Purines are double-ring nitrogenous bases, while pyrimidines are single-ring nitrogenous bases. Chargaff’s ratios of bases found in the DNA of different species suggested that A. a base pair consisted of two purines B. a base pair consisted of two pyrimidines C. a base pair consisted of one purine and one pyrimidine D. DNA replication is conservative E. DNA replication is semi-conservative HW 7 1. Purines are double-ring nitrogenous bases, while pyrimidines are single-ring nitrogenous bases. Chargaff’s ratios of bases found in the DNA of different species suggested that A. a base pair consisted of two purines B. a base pair consisted of two pyrimidines C. a base pair consisted of one purine and one pyrimidine D. DNA replication is conservative E. DNA replication is semi-conservative HW 7 2. If the DNA helix is viewed as a structure similar to a ladder, the uprights of the ladder (not the steps) would consist of A. complementary base pairs linked by hydrogen bonds B. sugars and phosphates linked covalently in a chain C. sugars and bases linked covalently in a chain D. phosphates and bases linked covalently in a chain HW 7 2. If the DNA helix is viewed as a structure similar to a ladder, the uprights of the ladder (not the steps) would consist of A. complementary base pairs linked by hydrogen bonds B. sugars and phosphates linked covalently in a chain C. sugars and bases linked covalently in a chain D. phosphates and bases linked covalently in a chain HW 7 3. Thymine (T) makes up 22% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be guanine (G)? A. 11% B. 22% C. 28% D. 44% E. 78% HW 7 3. Thymine (T) makes up 22% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be guanine (G)? A. 11% B. 22% C. 28% D. 44% E. 78% HW 7 4. Bacteriologists tried to determine whether the molecule of heredity was protein or DNA by experimenting with the extracted molecule that could transform the genetics of bacterial cells. They found that when the molecule of heredity A. was treated with protein-digesting enzymes it could still transform the cells B. was treated with DNA-digesting enzymes it could still transform the cells C. was treated with RNA-digesting enzymes it no longer could transform the cells HW 7 4. Bacteriologists tried to determine whether the molecule of heredity was protein or DNA by experimenting with the extracted molecule that could transform the genetics of bacterial cells. They found that when the molecule of heredity A. was treated with protein-digesting enzymes it could still transform the cells B. was treated with DNA-digesting enzymes it could still transform the cells C. was treated with RNA-digesting enzymes it no longer could transform the cells HW 7 5. Hershey & Chase tried to demonstrate whether the molecule of heredity was protein or DNA by experimenting with bacterial viruses made with radioactive labels. They found which of these? A. Since radioactive phosphorus ended up inside the bacterial cells, DNA must be the genetic material B. Since radioactive sulfur ended up inside the bacterial cells, DNA must be the genetic material C. Since radioactive carbon ended up inside the bacterial cells, DNA must be the genetic material HW 7 5. Hershey & Chase tried to demonstrate whether the molecule of heredity was protein or DNA by experimenting with bacterial viruses made with radioactive labels. They found which of these? A. Since radioactive phosphorus ended up inside the bacterial cells, DNA must be the genetic material B. Since radioactive sulfur ended up inside the bacterial cells, DNA must be the genetic material C. Since radioactive carbon ended up inside the bacterial cells, DNA must be the genetic material HW 7 6. To test whether DNA replication was semi-conservative or conservative, Investigators used bacteria grown in either: A. radioactive S or radioactive P B. labeled RNA or labeled DNA C. heavy or light isotopes of N HW 7 6. To test whether DNA replication was semi-conservative or conservative, Investigators used bacteria grown in either: A. radioactive S or radioactive P B. labeled RNA or labeled DNA C. heavy or light isotopes of N HW 7 7. DNA polymerase adds new nucleotides to: A. the 3’ sugar of the growing chain B. the 5’ phosphate of the growing chain C. the 3’ sugar of the complementary (original) chain D. the 5’ phosphate of the complementary (original) chain HW 7 7. DNA polymerase adds new nucleotides to: A. the 3’ sugar of the growing chain B. the 5’ phosphate of the growing chain C. the 3’ sugar of the complementary (original) chain D. the 5’ phosphate of the complementary (original) chain HW 7 8. In DNA replication, DNA polymerase requires which of these in order to begin? A. promoter B. primer C ribosome D start codon E more than one of these HW 7 8. In DNA replication, DNA polymerase requires which of these in order to begin? A. promoter B. primer C ribosome D start codon E more than one of these HW 7 9. In transcription of DNA by RNA polymerase, which of the following is required? A. promoter B. primer C. ribosome D. start codon E. more than one of these HW 7 9. In transcription of DNA by RNA polymerase, which of the following is required? A. promoter B. primer C. ribosome D. start codon E. more than one of these HW 7 10. Which of the following is the same for both RNA polymerase and DNA polymerase? A. requires a primer to start B. proofreads and corrects itself C. uses the same four nucleotide bases D. adds to the 3’ end of the growing chain HW 7 10. Which of the following is the same for both RNA polymerase and DNA polymerase? A. requires a primer to start B. proofreads and corrects itself C. uses the same four nucleotide bases D. adds to the 3’ end of the growing chain HW 7 11. In a certain cell there is a mutation in the promoter for a gene. Because of that particular mutation, the gene A. cannot be replicated B. cannot be transcribed C. cannot be translated D. cannot be replicated or transcribed HW 7 11. In a certain cell there is a mutation in the promoter for a gene. Because of that particular mutation, the gene A. cannot be replicated B. cannot be transcribed C. cannot be translated D. cannot be replicated or transcribed HW 7 12. A particular triplet of bases in the template strand of DNA is 5′-AGT-3′. What would be the corresponding codon for the mRNA that is transcribed from that DNA? A. 3'-UGA-5' B. 3′-UCA-5′ C. 5′-TCA-3′ D. 3′-ACU-5′ HW 7 12. A particular triplet of bases in the template strand of DNA is 5′-AGT-3′. What would be the corresponding codon for the mRNA that is transcribed from that DNA? A. 3'-UGA-5' B. 3′-UCA-5′ C. 5′-TCA-3′ D. 3′-ACU-5′ HW 7 13. This messenger RNA molecule is being translated at a ribosome in a direction from the 5’ end (left side). If you begin at the start codon, how many amino acids will be in the polypeptide that has been made by the time it gets to the last base shown (A at the right)? Begin with the start codon—what’s that? There are no stop codons in this sequence. You do not need the codon chart 5’ G C C A G G A U G C G A G G U A U G C C C G U C G C A 3' A. 3 B. 4 C. 6 D. 7 E. 9 HW 7 13. This messenger RNA molecule is being translated at a ribosome in a direction from the 5’ end (left side). If you begin at the start codon, how many amino acids will be in the polypeptide that has been made by the time it gets to the last base shown (A at the right)? Begin with the start codon—what’s that? There are no stop codons in this sequence. You do not need the codon chart 5’ G C C A G G A U G C G A G G U A U G C C C G U C G C A 3' A. 3 B. 4 C. 6 D. 7 E. 9 HW 7 14. Why does a mutation that inserts 1 base at a point in the DNA for a protein has a greater effect on the encoded protein than a mutation that has 1 base substitution at that point? A. base insertion increases the chance of causing an early stop codon B. base insertion increases the chance of causing a mutation in an intron C. base insertion causes a frameshift mutation affecting all following codons D. base insertion likely to cause numerous silent mutations HW 7 14. Why does a mutation that inserts 1 base at a point in the DNA for a protein has a greater effect on the encoded protein than a mutation that has 1 base substitution at that point? A. base insertion increases the chance of causing an early stop codon B. base insertion increases the chance of causing a mutation in an intron C. base insertion causes a frameshift mutation affecting all following codons D. base insertion likely to cause numerous silent mutations HW 8 1. In order for the correct amino acid to be added in protein synthesis (translation) at the ribosome, all of the following except one are required. Which one does NOT apply? A. mRNA codon must be complementary to tRNA anticodon B. mRNA codon must be antiparallel to tRNA anticodon C. tRNA must previously be attached to the correct amino acid D. tRNA molecule is entirely unfolded and single-stranded HW 8 1. In order for the correct amino acid to be added in protein synthesis (translation) at the ribosome, all of the following except one are required. Which one does NOT apply? A. mRNA codon must be complementary to tRNA anticodon B. mRNA codon must be antiparallel to tRNA anticodon C. tRNA must previously be attached to the correct amino acid D. tRNA molecule is entirely unfolded and single-stranded HW 8 2. Rank the following one-base point mutations 1-4 with respect to their likelihood of affecting the structure of the corresponding polypeptide (from most likely to change the protein to least likely to change it). 1. Base insertion mutation in the middle of an intron 2. Base substitution mutation at the third position of a codon in an exon 3. Base substitution mutation at the second position of a codon in an exon 4. Base deletion mutation within the first exon of the gene A. 1, 2, 3, 4 B. 4, 3, 2, 1 C. 3, 4, 2, 1 D. 3, 1, 4, 2 HW 8 2. Rank the following one-base point mutations 1-4 with respect to their likelihood of affecting the structure of the corresponding polypeptide (from most likely to change the protein to least likely to change it). 1. Base insertion mutation in the middle of an intron 2. Base substitution mutation at the third position of a codon in an exon 3. Base substitution mutation at the second position of a codon in an exon 4. Base deletion mutation within the first exon of the gene A. 1, 2, 3, 4 B. 4, 3, 2, 1 C. 3, 4, 2, 1 D. 3, 1, 4, 2 HW 8 3. Which one of these types of RNA is NOT found in Bacteria? [think about what these RNA’s do, and about bacterial genes, and you’ll have the answer.] A. mRNA B. tRNA C. rRNA D. snRNA HW 8 3. Which one of these types of RNA is NOT found in Bacteria? [think about what these RNA’s do, and about bacterial genes, and you’ll have the answer.] A. mRNA B. tRNA C. rRNA D. snRNA HW 8 4. A bacterial operon with 1 regulatory region and 3 protein-coding genes will have how many promoters? A. zero B. 1 C. 2 D. 3 E. 4 HW 8 4. A bacterial operon with 1 regulatory region and 3 protein-coding genes will have how many promoters? A. zero B. 1 C. 2 D. 3 E. 4 HW 8 5. A bacterial operon with 1 regulatory region and 3 protein-coding genes will have how many DNA sequences for the start codon (AUG)? A. zero B. 1 C. 2 D. 3 E. 4 HW 8 5. A bacterial operon with 1 regulatory region and 3 protein-coding genes will have how many DNA sequences for the start codon (AUG)? A. zero B. 1 C. 2 D. 3 E. 4 HW 8 6. A certain bacterial operon contains genes that code for enzymes controlling a pathway for synthesis of amino acid X from precursor molecule B. Choose the optimal – most efficient - pattern for regulation of that operon. A. operon genes should be generally off and only active when X is abundant B. operon genes should be generally on and only inactivated when X is abundant C. operon genes should be generally off only activated when B is absent HW 8 6. A certain bacterial operon contains genes that code for enzymes controlling a pathway for synthesis of amino acid X from precursor molecule B. Choose the optimal – most efficient - pattern for regulation of that operon. A. operon genes should be generally off and only active when X is abundant B. operon genes should be generally on and only inactivated when X is abundant C. operon genes should be generally off only activated when B is absent HW 8 7. What is an evolutionary advantage of prokaryotes having genes organized in operons? A. cells can produce fewer proteins to accomplish the same function B. cells can produce shorter messenger RNA molecules C. cells can turn on/off genes for related functions together D. cells do not need to splice their messenger RNA HW 8 7. What is an evolutionary advantage of prokaryotes having genes organized in operons? A. cells can produce fewer proteins to accomplish the same function B. cells can produce shorter messenger RNA molecules C. cells can turn on/off genes for related functions together D. cells do not need to splice their messenger RNA HW 8 8. Which of the following sets of genes in prokaryotes are likely to be found in inducible operons? (inducible—usually off, can be turned on) A. genes for proteins that use glucose for making ATP B. genes for proteins that use sugars only occasionally available to the cell C. genes for enzyme pathways for making essential amino acids HW 8 8. Which of the following sets of genes in prokaryotes are likely to be found in inducible operons? (inducible—usually off, can be turned on) A. genes for proteins that use glucose for making ATP B. genes for proteins that use sugars only occasionally available to the cell C. genes for enzyme pathways for making essential amino acids HW 8 9. If a particular operon encodes enzymes for making an essential amino acid and is regulated like the trp operon, then which would happen? [question from the text, end of chapter 15 question 1, answers in the appendix] A. The amino acid inactivates the repressor B. the repressor is active in the absence of the amino acid C. the amino acid acts as a co-repressor, activates the repressor D. the amino acid turns on transcription of the operon. HW 8 9. If a particular operon encodes enzymes for making an essential amino acid and is regulated like the trp operon, then which would happen? [question from the text, end of chapter 15 question 1, answers in the appendix] A. The amino acid inactivates the repressor B. the repressor is active in the absence of the amino acid C. the amino acid acts as a co-repressor, activates the repressor D. the amino acid turns on transcription of the operon. HW 8 10. What would occur if the repressor protein of an inducible operon were mutated so it could not bind the operator? [question from the text, end of chapter 15, question 4, answers in the appendix] A. irreversible binding of the repressor to the promoter B. reduced transcription of the operon’s genes C. buildup of a substrate for the pathway controlled by the operon D. continuous transcription of the operon’s genes HW 8 10. What would occur if the repressor protein of an inducible operon were mutated so it could not bind the operator? [question from the text, end of chapter 15, question 4, answers in the appendix] A. irreversible binding of the repressor to the promoter B. reduced transcription of the operon’s genes C. buildup of a substrate for the pathway controlled by the operon D. continuous transcription of the operon’s genes HW 8 11. Which of the following statements best describes the significance of the TATA box in the promoters of eukaryotes? A. It is the recognition site for the binding of a specific transcription factor. B. It sets the reading frame of the mRNA during translation. C. It is the recognition site for ribosomal binding during translation D. It is the recognition site for ribosomal binding during transcription HW 8 11. Which of the following statements best describes the significance of the TATA box in the promoters of eukaryotes? A. It is the recognition site for the binding of a specific transcription factor. B. It sets the reading frame of the mRNA during translation. C. It is the recognition site for ribosomal binding during translation D. It is the recognition site for ribosomal binding during transcription HW 8 12. Which of these is an adaptive system in Eukaryotes that allows a specific gene in a cell to be transcribed at a very high rate and thus produce more gene product? A. Enhancer B. Promoter C. Primer D. Spliceosome HW 8 12. Which of these is an adaptive system in Eukaryotes that allows a specific gene in a cell to be transcribed at a very high rate and thus produce more gene product? A. Enhancer B. Promoter C. Primer D. Spliceosome HW 8 13. Which is true for Eukaryotes but not for Prokaryotes? A. Many genes have introns, and many genes are in operons B. Many genes have introns, and many genes have enhancers C. Many genes are in operons but they do not have enhancers D. Many genes are in operons and most have enhancers HW 8 13. Which is true for Eukaryotes but not for Prokaryotes? A. Many genes have introns, and many genes are in operons B. Many genes have introns, and many genes have enhancers C. Many genes are in operons but they do not have enhancers D. Many genes are in operons and most have enhancers HW 8 14. In the chromatin of eukaryotic cells, DNA is packed with histone proteins. Which chromatin modifications results in most expression of the genes in that region? A. histone acetylation B. DNA methylation C. highly condensed chromatin HW 8 14. In the chromatin of eukaryotic cells, DNA is packed with histone proteins. Which chromatin modifications results in most expression of the genes in that region? A. histone acetylation B. DNA methylation C. highly condensed chromatin HW 9 1. Which of the following is true about the two members of a homologous pair of chromosomes? A. they are identical DNA molecules B. they form the two sides of one double helix C. they separate from one another during meiosis II D. they contain the same gene loci in the same order E. two of these are true. HW 9 1. Which of the following is true about the two members of a homologous pair of chromosomes? A. they are identical DNA molecules B. they form the two sides of one double helix C. they separate from one another during meiosis II D. they contain the same gene loci in the same order E. two of these are true. HW 9 2. Which two of these (A-D) describes the makeup of the human nuclear genome? A. haploid number of 23 B. haploid number of 46 C. 46 homologous pairs D. 23 homologous pairs E. A and C are true F. A and D are true G. B and C are true H. B and D are true HW 9 2. Which two of these (A-D) describes the makeup of the human nuclear genome? A. haploid number of 23 B. haploid number of 46 C. 46 homologous pairs D. 23 homologous pairs E. A and C are true F. A and D are true G. B and C are true H. B and D are true HW 9 3. If an organism has a diploid chromosome number of 50, its sperm will have how many chromosomes? A. 25 B. 50 C. 100 D. 500 HW 9 3. If an organism has a diploid chromosome number of 50, its sperm will have how many chromosomes? A. 25 B. 50 C. 100 D. 500 HW 9 4. If a cell with 30 homologous pairs of chromosomes undergoes one round of cell division by mitosis, the result will be: A. Two cells, each with 30 chromosomes B. Two cells, each with 60 chromosomes C. Four cells, each with 30 chromosomes D. Four cells, each with 60 chromosomes HW 9 4. If a cell with 30 homologous pairs of chromosomes undergoes one round of cell division by mitosis, the result will be: A. Two cells, each with 30 chromosomes B. Two cells, each with 60 chromosomes C. Four cells, each with 30 chromosomes D. Four cells, each with 60 chromosomes HW 9 5. During which stages of the cell cycle does a chromosome consist of two identical chromatids? A. entire G1 stage B. entire S stage C. entire G2 stage D. entire M stage HW 9 5. During which stages of the cell cycle does a chromosome consist of two identical chromatids? A. entire G1 stage B. entire S stage C. entire G2 stage D. entire M stage HW 9 6. If the DNA content of a diploid cell in the G1 phase of the cell cycle is represented by x, then the DNA content of the same cell in the middle of meiosis I would be: A. 0.25x B. 0.5x C. x D. 2x HW 9 6. If the DNA content of a diploid cell in the G1 phase of the cell cycle is represented by x, then the DNA content of the same cell in the middle of meiosis I would be: A. 0.25x B. 0.5x C. x D. 2x HW 9 7. If we continued to follow the cell lineage from the previous question, the DNA content of a cell in the middle of meiosis II would be: A. 0.25x B. 0.5x C. x D. 2x HW 9 7. If we continued to follow the cell lineage from the previous question, the DNA content of a cell in the middle of meiosis II would be: A. 0.25x B. 0.5x C. x D. 2x HW 9 8. Which of these is NOT something happens early in Prophase of Mitosis? A. chromosomes condense B. DNA replicates C. spindle forms D. nuclear membrane dissociates HW 9 8. Which of these is NOT something happens early in Prophase of Mitosis? A. chromosomes condense B. DNA replicates C. spindle forms D. nuclear membrane dissociates HW 9 9. Which of these best describes the kinetochore? A. a region of DNA attached to spindle fibers B. a set of proteins attached to the centromere C. a certain type of microtubule D. a centrosome that organizes microtubules HW 9 9. Which of these best describes the kinetochore? A. a region of DNA attached to spindle fibers B. a set of proteins attached to the centromere C. a certain type of microtubule D. a centrosome that organizes microtubules HW 9 10. When a cell in S phase was fused with a cell in G1, the G1 nucleus immediately entered S phase, began DNA synthesis. This is most likely because A. S phase automatically follows G1 phase B. the chromosomes were duplicated in S phase but not in G1 phage C. control molecules in the S phase cell stimulated DNA synthesis in the G1 phase cell. D. control molecules in the G1 phase cell stimulated more DNA synthesis in the S phase cell. HW 9 10. When a cell in S phase was fused with a cell in G1, the G1 nucleus immediately entered S phase, began DNA synthesis. This is most likely because A. S phase automatically follows G1 phase B. the chromosomes were duplicated in S phase but not in G1 phage C. control molecules in the S phase cell stimulated DNA synthesis in the G1 phase cell. D. control molecules in the G1 phase cell stimulated more DNA synthesis in the S phase cell. HW 9 11. Which of these would be an important check that should be completed before the cell begins S phase? A. Attach DNA to microtubules B. Repair any broken DNA C. Nuclear membrane dissociates D. Condense the chromosomes HW 9 11. Which of these would be an important check that should be completed before the cell begins S phase? A. Attach DNA to microtubules B. Repair any broken DNA C. Nuclear membrane dissociates D. Condense the chromosomes HW 9 12. Which of these would be an important check that should be completed before the cell begins Anaphase? A. Attach DNA to microtubules B. Repair any broken DNA C. Nuclear membrane dissociates D. Condense the chromosomes HW 9 12. Which of these would be an important check that should be completed before the cell begins Anaphase? A. Attach DNA to microtubules B. Repair any broken DNA C. Nuclear membrane dissociates D. Condense the chromosomes HW 9 13. In a diploid cell with 5 chromosome pairs (2n = 10), how many DNA molecules will be found in a nucleus at the beginning of mitosis? A. 5 B. 10 C. 20 D. 40 HW 9 13. In a diploid cell with 5 chromosome pairs (2n = 10), how many DNA molecules will be found in a nucleus at the beginning of mitosis? A. 5 B. 10 C. 20 D. 40 HW 9 14. Which of the following does NOT occur during Mitosis? A. chromosomes condense B. DNA replicates C. sister chromatids separate D. spindle formation HW 9 14. Which of the following does NOT occur during Mitosis? A. chromosomes condense B. DNA replicates C. sister chromatids separate D. spindle formation Unit 3 LOs Lecture 1: DNA 3-1. Using principles of hypothesis testing and experimental design, outline and interpret data that demonstrated that DNA is the molecule of heredity. How did experiments on transformed pneumonia-causing bacteria and radioactively-labeled bacterial viruses (Hershey & Chase) inform scientists that the genetic material was indeed DNA and not protein? Hershey and Chase labeled the protein casing of the bacterial viruses with radioactive sulfur and the DNA inside the casing with radioactive phosphorus. They then watched to see if the DNA or the protein was injected into the host cell to determine which was the genetic material. Lecture 1: DNA 3-2. Describe the double helix model of DNA, reviewing the evidence that supports this model. What were some specific types of data used by Watson & Crick to determine the structure of DNA? Watson and Crick used X-ray diffraction pattern to determine DNA structure. The image resembled helical proteins and suggested a uniform diameter. They also used Chargaff’s ratios to determine which bases paired. Lecture 1: DNA 3-3. Describe the structure of DNA nucleotide components (phosphate, sugar, A T G C bases), their arrangement in DNA including complementary base pairing, and the types of bonds (hydrogen or covalent) between various components. What are 3’ and 5’ ends of DNA? What is meant by the “antiparallel” structure of DNA? The phosphate is covalently bonded to the 5’ carbon on the sugar and the nitrogenous base is covalently bonded to the 1’ carbon on the sugar. In a DNA strand, the phosphate of one nucleotide covalently bonds to the 3’ carbon on the sugar. The bases are paired up A to T with 2 hydrogen bonds and G to C with 3 hydrogen bonds. The 5’ end has a phosphate while the 3’ end has a sugar. The antiparallel structure of DNA means that one strand runs 5’ → 3’ and the other strand runs 3’ → 5’ Lecture 2: DNA Replication 3-4. Tell what is meant by “semi-conservative replication”. How did Meselson & Stahl use bacteria and heavy N to determine that DNA replication is semi-conservative, not conservative? Semi-conservative replication means that the parent strands are separated and the daughter strands end up one parent strand and one daughter strand. Meselson and Stahl used heavy N to make the parent strands really dense. They then let it replicate and centrifuged the DNA to separate the different densities. Because they saw all the DNA at the middle, they were able to conclude that DNA replication is semi-conservative. If DNA replication was conservative, they would have seen a line at the bottom and top rather than just in the middle. Lecture 2: DNA Replication 3-5. Describe the steps of DNA replication beginning at the “origin of replication” site, including the role of these major enzymes (helicase, primase, DNA polymerase, ligase). Include the role of primers for starting, the use of tri-phosphate nucleosides as building blocks (& energy), and the addition of new nucleotides to the 3’ end of the growing strand. Helicase unwinds the helix, which creates the replication bubble. From there, primase creates the RNA primer which acts as a starting block for the DNA polymerase to start adding nucleotides to the 3’ end of the growing strand. The polymerase adds nucleotides to the template strand and ligase connects the okazaki fragments on the lagging strand. DNA replication is an energy coupled process with the endergonic process being adding the nucleotides and the exergonic being the hydrolysis of the 2 outer phosphates on the nucleotide triphosphates. Lecture 2: DNA Replication 3-6. Why does DNA’s anti-parallel arrangement require a difference between replication processes at the leading & lagging strands? What are Okazaki fragments? Why is DNA ligase needed only in replication on the lagging strand? DNA polymerase can only add nucleotides to the 3’ end of the new strand. This means that the leading strand can replicate continuously but to replicate the DNA towards the 5’ end of the RNA primer, new RNA Primers have to be added and the lagging strand is replicated in fragments called Okazaki fragments. DNA Ligase is only needed on the lagging strand because the leading strand is made without okazaki fragments and doesn’t need the sugar phosphate backbone to be joined together. Lecture 3: DNA & RNA 3-7. Name 2 ways in which RNA differs chemically from DNA. RNA has ribose instead of deoxyribose and uracil instead of thymine. ○ On the ribose, there is one more OH group at the 2’ carbon, whereas deoxyribose has a H at the 2’ carbon. Lecture 3: DNA & RNA 3-8. RNA polymerase always begins transcription of a gene at a DNA sequence called what? It will stop transcription at a place called what? Comare/contrast each pair of terms: (a) action of DNA polymerase vs RNA polymerase action, (b) primer vs promoter; (c) replication vs transcription RNA Polymerase begins transcription at a promoter and ends it at a terminator DNA Polymerase requires a primer and helicase to unwind DNA for it to begin adding complementary bases to the 3’ end of the growing strand. RNA Polymerase unwinds the DNA and doesn’t need a primer to add complimentary bases to the 3’ end of the growing strand. A primer is used in DNA replication while a promoter is used in DNA transcription into mRNA Replication is used to make more DNA while transcription is used to express genes by making mRNA to translate into proteins. Lecture 3: DNA & RNA 3-9. Explain what is meant by a “triplet code”. Examine the mRNA codon chart and explain what is meant by “redundant code”. From a mRNA sequence, how do you use the codon chart to translate mRNA into a protein? Triplet code refers to the fact that amino acids are coded by groups of 3 bases called codons Redundant code refers to the fact that multiple codons can code for the same amino acid Using an mRNA sequence, find your starting codon (AUG) and begin translating the mRNA from there until you hit a stop codon Lecture 5: mRNA 3-10. Tell how a point mutation like a base substitution or insertion/deletion can affect the resulting protein. What’s an example of a frameshift mutation? Distinguish silent mutation, missense mutation, and nonsense mutations. A point mutation like a base substitution can change the amino acid the codon codes for and change the resulting protein. An insertion or deletion can change the all the codons that come afterwards by shifting the reading frame and affecting all the amino acids. A frameshift mutation results from a 1-2 base insertion/deletion which changes the reading frame for the codons and amino acids. Silent mutation- a change in DNA that doesn’t change the amino acid Missense mutation- a change in DNA that changes the amino acid Nonsense mutation- change in DNA that results in a stop codon being introduced Lecture 5: mRNA 3-11. In what three ways do eukaryotic cells modify the messenger RNA (pre-mRNA) after transcription? Tell the main function of the 3’ poly-A tail and the 5’ cap Eukaryotic cells add the poly-A tail to the 3’ end and a 5’ cap to the pre-mRNA and remove introns from the pre-mRNA using spliceosomes. The 3’ poly-A tail helps mRNA resist digestion by hydrolytic enzymes as it leaves the nucleus. The 5’ cap helps protect the end of the mRNA and helps the 5’ end bind to the ribosome. Lecture 5: mRNA 3-12. What are introns and exons? What complex of proteins & snRNA’s splice introns out of the mRNA? Does this splicing occur in the nucleus or in the cytosol? Is mRNA splicing a form of mutation? How can alternate mRNA splicing allow different cells to make slightly different proteins from 1 gene? Introns are the sections of the non-coding mRNA that are removed; exons are the coding parts of the mRNA that are expressed. The spliceosome is made up of proteins and snRNAs that splice introns out of mRNA Splicing occurs in the nucleus and isn’t a form of mutation. Alternate splicing can allow cells to make slightly different proteins from one gene by changing which sections of the mRNA are translated and expressed. Lecture 6: Translation 3-13. Distinguish the functions of each of the following types of RNA: rRNA, tRNA, mRNA, snRNA. Remember that each is made in the nucleus by transcription from DNA. rRNA is ribosomal RNA and is used for making ribosomal subunits tRNA is transfer RNA and brings amino acids to the ribosome to make proteins during translation. mRNA is messenger RNA and brings the message from DNA to the ribosome for protein synthesis during translation snRNA is small nuclear RNA and binds to splice sites on introns so the spliceosomes know where to cut Lecture 6: Translation 3-14. What is the general function of a tRNA? How does a tRNA become bonded to its specific amino acid? Does this occur at the mRNA or before the tRNA arrives at the mRNA? How do the mRNA codon and tRNA anticodon interact with each other? tRNA brings the corresponding amino acids to the ribosome when its translating the mRNA. tRNA become bonded with its specific amino acid by an enzyme in an ATP-requiring reaction before reaching the mRNA. The mRNA codon and tRNA anticodon bind to each other via complementary base pairing. They are also antiparallel to each other. Lecture 6: Translation 3-15. What is translation? How does translation begin? Tell what happens in translation elongation to synthesize the protein. What happens during translation when the mRNA stop codon appears at the A Site on the ribosome? How could the same mRNA be translated multiple times? Translation is the synthesis of proteins from the mRNA. It begins when the 5’ cap of mRNA binds to a small ribosomal subunit; from there, the first tRNA binds its anticodon to the start codon on the mRNA and then the large ribosomal subunit binds and translation continues. When a stop codon appears at the A site, the completed polypeptide is released and the ribosome breaks apart into its subunits. The same mRNA can be translated multiple times if it has multiple ribosomes on it. One ribosome starts translating it and as soon as there is space for another, it binds and begins translating as well. Lecture 7: Prokaryotic Gene Regulation 3-16. Bacteria can have simultaneous transcription and translation of the same mRNA molecule. Why can’t this occur in eukaryotes? In bacteria, there is no nucleus or nuclear envelope meaning that both transcription and translation happen in the cytoplasm. Because of this, transcription and translation can occur simultaneously. In eukaryotes, transcription is confined to inside the nucleus and translation is in the cytoplasm. Lecture 7: Prokaryotic Gene Regulation 3-17. A bacterial operon consists of what? Are human genes organized in operons? Bacterial operons with 1 regulatory region and 3 protein-coding genes will have how many promoters? Explain how bacteria can make one mRNA molecule that contains the code for several proteins. An operon consists of several protein-coding genes and 1 shared promoter. Human genes don’t have operons because eukaryotes don’t have operons. 1 regulatory region and 3 protein coding genes will have 1 promoter. Operons code for 1 long mRNA strand but within the operon, there are multiple proteins coded for. Because there are start and stop codons in between each of the protein coding regions, ribosomes can bind to each of the protein coding regions and make multiple proteins from the 1 mRNA. Lecture 7: Prokaryotic Gene Regulation 3-18. Genes in a certain bacterial operon are generally turned off, except when a certain food sources is present when they turn on. Is this a represible or an inducible operon? Name a specific example of this type. This is an inducible operon because it can be turned on but is generally off. The Lac operon is an example of this. Lecture 7: Prokaryotic Gene Regulation 3-19. The trp operon genes code for enzymes in a pathway that make the amino acid tryptophan from a precursor molecule. What are two different ways (one enzymatic and one genetic) a bacterial cell can turn off making tryptophan? Why is the trp operon a good example of end-product feedback inhibition for both enzyme and gene activity? Tryptophan can act as an inhibitor for the 1st enzyme in the pathway that makes tryptophan or as an activator for the repressor protein of the trp operon and prevent production of the enzymes in the tryptophan making pathway. The trp operon is a good example of end-product feedback inhibition because the final product is the inhibitor in both scenarios. It can prevent enzymes from making more or prevent the production of the enzymes that make it. Lecture 8: Eukaryotic Gene Regulation 3-20. What is the structure of a eukaryotic chromosome- how is DNA arranged with histone proteins? Chromosomes are made up of chromatin. The base unit of chromatin is a nucleosome, which consists of DNA wrapped around 8 histone proteins. Around a histone, DNA is wrapped around twice and held tightly to the histone. Lecture 8: Eukaryotic Gene Regulation 3-21. In each pair of items, which DNA is more likely to be transcribed and why? a) DNA in a tightly-packed region of chromatin or DNA in loose, unwould chromatin. i) Loose because it is more accessible for the transcription factors and RNA polymerase b) DNA with nucleotide bases that are methylated or DNA that is un-methylated i) Unmethylated because methyl acts as an inhibitor to transcription c) DNA that is ripped closely to histone or DNA that is not i) DNA that isn’t wrapped closely to the histone because it’s more accessible for transcription factors and RNA polymerase d) DNA that is associated with acetylated histones or DNA with non-acetylated histones i) DNA with acetylated histones because the acetyl groups open up the chromatin which is easier to transcribe Lecture 8: Eukaryotic Gene Regulation 3-22. In eukaryotic cells, RNA polymerase cannot begin transcription at the promoter until what proteins are there bound to the DNA in the region of the promoter? Transcription factors Lecture 8: Eukaryotic Gene Regulation 3-23. Eukaryotic genes that will be transcribed a great deal have enhancer DNA regions. What is the role of the enhancer? Explain how transcription factors and activator proteins interact with enhancer and promoter regions of the DNA to initiate the process of transcription (major control of eukaryotic gene expression). The enhancer makes the gene get transcribed really quickly and really often. Transcription factors interact with the promoter region and highlight where the promoter is for RNA polymerase to bind to. Activator proteins bind to the enhancer region of DNA and lots of transcription happens. Lecture 8: Eukaryotic Gene Regulation 3-24. If (almost) all cells in a person’s body have exactly the same genes, then how do various cells differentiate (i.e. become specialized) into different cell types in different body tissues? Briefly describe some specific mechanisms that determine eukaryotic cell specialization, including tissue-specific gene expression. Different cells have specific activator and transcription factor proteins that help determine which genes get expressed. In embryos, cells can develop different receptors that bind and respond to specific signals that begin turning on and off different genes. Cells also have alternate gene splicing and cut out some of the exons to prevent expression. There are also master regulatory genes that turn on multiple genes that are part of the same function using a master transcription factor. Different genes can be acetylated or methylated to promote transcription of specific genes. In tissue-specific gene expression, only the specific activator and transcription factors for specific genes are present, so only the tissue specific genes are expressed in the specific tissue. Lecture 9: Cell Differentiation 3-25. Tell how responses to extracellular signals can possibly lead to different responses in a cell turning on/off specific genes. Signals that bind to the outside of the cell can trigger signal transduction pathways that turn on and off different genes by activating or deactivating certain transcription factors within the nucleus. Lipid soluble directly turn on and off genes in the nucleus. Lecture 9: Cell Differentiation 3-26. A nucleic acid probe is a short length of single stranded nucleic acid color-labeled with a fluorescent tag. How can probes be used to measure gene expression (mRNA Production) and locate where, within a cell or in an embryo, a certain mRNA is located? Probes are complementary to the mRNA that is being sought and because it has a fluorescent tag, it glows when it binds to the mRNA that is being studied. Probes can measure gene expression by showing where certain mRNA is located, as well as showing how much mRNA is located there. Lecture 10: Cell Cycle and Mitosis 3-27. Name an example of how a small, non-coding RNA can affect and even block the expression of specific genes. Small noncoding RNA includes micro RNA and small interfering RNA. Micro RNA binds to the small ribosomal subunit and stops translation to prevent expression. Small interfering RNA packs chromatin to prevent transcription and therefore translation. Lecture 10: Cell Cycle and Mitosis 3-28. Distinguish/relate these terms: DNA Molecule, double helix, gene, chromosome, sister chromatids, pair of homologous chromosomes, human genome (46 chromosomes in 23 homologous pairs). DNA molecule- one chromosome worth of DNA Double helix- shape of DNA, twisted ladder Chromosome- unit of packed DNA, one DNA molecule sister chromatids- 1 chromosome made up of 2 identical DNA molecules, connected by centromere Pair of homologous chromosomes- chromosomes with similar shape and banding that have same locus for genes (1 from each parent); contain information for the same genes but aren’t identical Human genome- the 46 chromosomes/23 pairs of chromosomes that make up the entirety of a human’s DNA Lecture 10: Cell Cycle and Mitosis 3-29. Describe the eukaryotic cell cycle, an tell what happens in each phase, G1, S, G2, M. During what parts of a cell cycle (G1, S, G2, M) would a “chromosome” consist of 2 identical chromatids? G1- cell grows, S- DNA Replication, G2- Grows more and preps for cell division, M-cell division. A chromosome would have 2 identical chromatids during G2, after DNA replication. Lecture 10: Cell Cycle and Mitosis 3-30. What happens in mitosis? State changes in chromosome form and movements. RElate chromosome movements to cell changes, spindle fibers (microtubules), and cytokinesis. Prophase- chromatin condenses into chromosomes (appease as sister chromatids) and the mitotic spindle forms (made of centrosomes that pull to opposite sides of the cell with microtubules extending from them) Prometaphase- Microtubules extend from the centrosomes and enter nucleus to attach to kinetochores that form at each centromere and nuclear envelope dissolves Metaphase- centrosomes (spindles) are at opposite ends of the cell and chromosomes are lined up with their centromeres at the metaphase plate (down the middle of the cell) Anaphase- sister chromatids pulled apart to opposite sides of the cell and the cell begins to elongate; both sides of the cell have equal and complete collections of chromosomes by the end of anaphase Telophase- 2 daughter nuclei and nuclear envelopes form, chromosomes become less condensed, spindle fibers dissolve Cytokinesis- division of cytoplasm, starts with cleavage furrow which pinches the cell into 2 Lecture 11: Cell Cycle and Signaling 3-31. A checkpoint in the cell cycle is a control point where “stop” and “go-ahead” signals can regulate the cycle. Name some “signals” that could affect whether the cell progresses through the cycle or not.\ Cyclins, Mitosis promoting factor (super kinase complex made of cyclin and cdk), growth hormone, nutrient levels, presence/lack of specific activated kinases (cdk-cyclin) Lecture 11: Cell Cycle and Signaling 3-32. The cell cycle is regulated by specific control proteins called cyclins, which must be phosphorylated in order to become active. Recall how cells respond in signaling pathways: (a) distinguish these: kinase, phosphatase, cAMP, (b) give an example of signal amplification in cells. Kinase- activates proteins by phosphorylating them; Phosphatase- deactivates proteins by dephosphorylating them; cAMP- 2nd messenger made of modified ATP Amplification is a multi-fold increase in the number of activated molecules at each step of the transduction pathway. The signal is amplified by each signal molecule can going and activating multiple molecules in the next step of the pathway. For example, one kinase might be able to activate 10 more kinases for a 10 fold increase. Lecture 11: Cell Cycle and Signaling 3-33. What is apoptosis? Why is it called ‘programmed cell death’? Give an example of its role in normal development.. Apoptosis is programmed cell death caused by a sequence of gene expression changes that degrades the DNA and cause the nucleus to condense. The cell is eventually devoured by surrounding cells. In human embryos, there is webbing between their fingers that eventually disappears via apoptosis. As frogs change from tadpoles, their tails disappear via apoptosis. Lecture 12: Meiosis 3-34. Cancer cells are characterized by uncontrolled cell growth. Cancer results from an accumulation of mutations (many steps) within a cell line. One mutation permanently switches ON a G-protein (so it is always activated by GTP). Another mutation causes tyrosine kinase signal receptors to auto-phosphorylate even when there is no signal bound to the receptor. Tell how each of these could contribute to cancer. If the G-protein is permanently on, the growth hormone can keep stimulating the cell cycle and division, leading to cancer or uncontrolled cell growth. If the tyrosine kinase signal receptors auto-phosphorylate without a signal, the cell can divide continuously without a signal. Lecture 12: Meiosis 3-35. p53 is a tumor-suppressor gene. How do mutations in the p53 contribute to cancer? P53 is a gene that activates other genes to stop the cell cycle and repair DNA before proceeding. If the DNA is irreparable, the p53 gene will trigger apoptosis to prevent tumors from mutated DNA. If the p53 gene is mutated, the cell’s ability to regulate the cell cycle and prevent the multi-hit mutations that cause cancer is greatly diminished. Lecture 12: Meiosis 3-36. Distinguish diploid (2n) from haploid (n) chromosome number, and tell how they fit together in a sexual reproduction life cycle that includes both meiosis and fertilization. Diploid (2n)- total number of chromosomes in a cell, have both members of a homologous pair Haploid (n)- have one member of each homologous, same number as the number of homologous pairs in the cell Haploids are created so that during fertilization, the number of chromosomes in the newly fertilized egg doesn’t keep doubling from generation to generation. The newly fertilized egg is a diploid cell that was created by the combination of 2 haploid cells. Lecture 13: Meiosis and Mendel 3-37. Describe the movement of homologous chromosomes and twin chromatids during Mitosis, Meiosis I, and Meiosis II. Compare Meiosis with Mitosis. Mitosis: DNA replicated and then twin chromatids separated Meiosis 1: DNA replicated and then homologous pairs separated Meiosis 2: twin chromatids separated Meiosis- creates 4 haploid daughter cells that aren’t genetically identical ○ Mitosis- creates 2 diploid daughter cells that are genetically identical Lecture 13: Meiosis and Mendel 3-38. Given the 2n number for a particular cell, tell the number of cells and the number of chromosomes in each as a result of (a) mitosis and (b) meiosis. Mitosis: 2 daughter cells with 2n chromosomes Meiosis: 4 daughter cells with n chromosomes Ex: If 2n=16, then mitosis produces 2 daughter cells with 16 chromosomes and meiosis produces 4 daughter cells with 8 chromosomes Lecture 13: Meiosis and Mendel 3-39. Why is meiosis called “reduction division”? Meiosis is called reduction division because it produces cells with half the DNA of a diploid cell. One diploid cell (2n) is divided into four haploid cells (n). Lecture 13: Meiosis and Mendel 3-40. Name something that happens to the two members of a homologous pair in meiosis that does not happen to them during mitosis. Only in Meiosis (I) do the two members of homologous pairs zipper together and participate in crossing over before lining up to be separated ○ In mitosis, the homologous pairs aren’t separated. Instead, every chromosome is duplicated and the sister chromatids are separated. Lecture 13: Meiosis and Mendel 3-41. A cell with 40 chromosomes (20 homologous pairs) undergoes meiosis: what are the products? (give the number of cells and number of chromosomes in each) 4 daughter cells, each with 20 chromosomes ○ n = number of homologous pairs Lecture 13: Meiosis and Mendel 3-42. Distinguish and relate these terms: chromosome, DNA double-helix molecule, gene locus (location), allele (specific form of a gene). Chromosome- 1 DNA molecule worth of information DNA double-helix molecule- chromosomes are made of 1 DNA molecule, but replicated chromosomes (sister chromatids) are made of 2 DNA molecules Gene locus- location of genes on a chromosome; same locus for the same genes on homologous chromosomes Allele- different versions of the same gene; if the gene codes for flower color, there could be a purple allele and white allele ○ Homologous chromosomes will have alleles that code for the same gene at the same loci but could have different alleles for said gene; this is why homologous chromosomes aren’t identical Lecture 14: Mendelian Genetics 3-43. For the following pairs of terms, define and distinguish between them: gene & allele; genotype & phenotype; homozygous & heterozygous; dominant & recessive alleles Gene & allele- Genes are sections of DNA that code for a specific trait while alleles are different versions of the same gene Genotype & phenotype- Genotypes are the genetic makeup of an organism (alleles) while phenotype is the observable traits of said organism (appearance) Homozygous & heterozygous- homozygous means 2 of the same allele (AA or aa) while heterozygous means different alleles (Aa) Dominant & recessive alleles- Dominant alleles are expressed in the phenotype no matter what while recessive alleles are only expressed in homozygous recessive genotypes (aa). Lecture 14: Mendelian Genetics 3-44. Distinguish Mendel’s principle of segregation from the principle of independent assortment in terms of alleles moving into gametes. State the events of meiosis that explain each of those. Principle of segregation- while making gametes, the 2 alleles for a gene separate into different gametes ○ Doesn’t contribute to genetic diversity Principle of independent assortment- the separation of 2 alleles for one trait is completely independent of the separation of 2 alleles for a different trait ○ Inheritance of one trait doesn’t influence the inheritance of another trait ○ Chromosomes randomly line up during metaphase during meiosis 1 ○ Contributes greatly to genetic variation because of the different combinations of genotypes created by independent assortment Produces 2n combinations of daughter cells, where n= # homologous pairs Lecture 14: Mendelian Genetics 3-45. When given the parental genotypes for one trait, e.g.—Aa mating with AA—set up a Punnett Square for the cross, showing the possible gametes and offspring genotypes. (a) From the Punnett Square, predict the ratio of offspring genotypes. (b) When provided dominance information for those alleles (simple=complete vs incomplete), predict the ratio of offspring phenotypes. Genotypes ratios: 1:1, AA:Aa A A Phenotype ratios: All dominant phenotypes A AA AA a Aa Aa

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