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Biology Bio 93 Final Exam Review

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Lecture 1-9 Lecture 10-18 Lecture 19-22 Lecture 23-26 100 200 300 400 500 600

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Lecture 1-9 for 100 Describe a Structural Difference between DNA and RNA Answer

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Lecture 1-9 100 Answer: What is C) DNA nucleotides contain a different sugar than RNA nucleotides. DNA has deoxyribose, and RNA has the ribose

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Lecture 1-9 for 200 One liter of a solution of pH 2 has how many more hydrogen ions (H ) than 1 L of a solution of pH 6? Answer

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Lecture 1-9 200 Answer What is 10 4 = 10,000 times more

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Lecture 1-9 for 300 Each organism is uniquely adapted to its environment even at the molecular level. Human fats are about 3% polyunsaturated while that of cod, a fish living in cold ocean waters, is approximately 50% polyunsaturated. Knowing this, why can the cod survive colder temperatures than a human? A) Unsaturated fats act as hormones activating genes for proteins that act as insulators. B) Unsaturated fats dissolve in the aqueous portions of cells acting like antifreeze and slow the formation of ice crystals. C) Unsaturated fats remain fluid at lower temperatures than saturated fats due to their shape. D) Unsaturated fats can be assembled into thicker cell membranes providing additional insulation against the cold Answer

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Lecture 1-9 for 300 Answer What is C) Unsaturated fats remain fluid at lower temperatures than saturated fats due to their shape

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Lecture 1-9 for 400 There are 20 different amino acids. What makes one amino acid different from another? A) different carboxyl groups attached to an alpha carbon B) different amino groups attached to an alpha carbon C) different alpha carbons D) different side chains (R groups) attached to an alpha carbon Answer

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Lecture 1-9 for 400 Answer What is D) different side chains (R groups) attached to an alpha carbon

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Lecture 1-9 for 500 What are the 3 phases of cell signaling? Answer

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Lecture 1-9 for 500 Answer 1)Reception 2)Transduction (Relay) 3)Response

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Lecture 1-9 for 600 Answer What are the 4 kinds of signal types?

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Lecture 1-9 for 600 Answer 1)Paracrine 2)Endocrine 3)Cell Surface 4)Intracellular

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Lecture 10-18 for 100 Answer What is the difference between an exergonic and endergonic reaction?

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Lecture 10-18 for 100 Answer Exergonic releases energy; products more stable than reaction Endergonic absorbs energy; nonspontaneous products less stable

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Lecture 10-18 for 200 Answer

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Lecture 10-18 for 200 Answer What is E) mitochrondrion

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Lecture 10-18 for 300 Answer What role do NAPH and FADH2 play in cellular respiration?

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Lecture 10-18 for 300 Answer Electron carriers

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Lecture 10-18 for 400 Answer What is the difference between a oncogene and a protooncogene?

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Lecture 10-18 for 400 Answer Proto-oncogene: normal gene involved in cell cyle regulation Oncogene: mutate protooncogene

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Lecture 10-18 for 500 Answer

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Lecture 10-18 for 500 Answer What is C) glucose is oxidized and oxygen is reduced Glucose loses electron and Oxygen gains electron (OIL RIG) Remember, this is not the same as oxidizing agent and reducing agent! Oxidizing agent gains e- and gets reduced Reducing agent loses e- and gets oxidized

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Lecture 10-18 for 600 Answer What phases of cell cycle do law of independent assortment and law of segregation affect?

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Lecture 10-18 for 600 Answer Independent Assortment: Metaphase I (depends on genes being on separate chromosome) Segregation Anaphase I depends on alleles segregating into different games (each sister chromatid has copy of allele)

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Lecture 19-22 for 100 Which would you expect of a eukaryotic cell lacking telomerase? A) a high probability of becoming cancerous B) production of Okazaki fragments C) inability to repair thymine dimers D) a reduction in chromosome length E) high sensitivity to sunlight Answer

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Lecture 19-22 for 100 Answer What is D) a reduction in chromosome length Telomerase is an enzyme that adds DNA sequence repeats to the 3 end of DNA strands in the telomere regions

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Lecture 19-22 for 200 Which of the following types of mutation, resulting in an error in the mRNA just after the AUG start of translation, is likely to have the most serious effect on the polypeptide product? A) a deletion of a codon B) a deletion of 2 nucleotides C) a substitution of the third nucleotide in an ACC codon D) a substitution of the first nucleotide of a GGG codon E) an insertion of a codon Answer

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Lecture 19-22 for 200 Answer What is B) a deletion of 2 nucleotides

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Lecture 19-22 for 300 Polytene chromosomes of Drosophila salivary glands each consist of multiple identical DNA strands that are aligned in parallel arrays. How could these arise? A) meiosis followed by mitosis B) fertilization by multiple sperm C) replication followed by mitosis D) replication without separation E) special association with histone proteins Answer

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Lecture 19-22 for 300 Answer What is D) replication without separation

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Lecture 19-22 for 400 Each of the following options is a modification of the sentence THECATATETHERAT. Which of the following is analogous to a frameshift mutation? A) THERATATETHECAT B) THETACATETHERAT C) THECATARETHERAT D) THECATATTHERAT E) EHHCATATETHERAT Answer

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Lecture 19-22 for 400 Answer What is D) THECATATTHERAT There is a single deletion of E, which would cause a shift in the reading frame

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Lecture 19-22 for 500 When DNA is compacted by histones into 10 nm and 30 nm fibers, the DNA is unable to interact with proteins required for gene expression. Therefore, to allow for these proteins to act, the chromatin must constantly alter its structure. Which processes contribute to this dynamic activity? A) DNA supercoiling at or around H1 B) hydrolysis of DNA molecules where they are wrapped around the nucleosome core C) accessibility of heterochromatin to phosphorylating enzymes D) nucleotide excision and reconstruction E) methylation and phosphorylation of histone tails Answer

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Lecture 19-22 for 500 Answer What is E) methylation and phosphorylation of histone tails Methylation and phosphorylation to histone tails alter chromatin structure

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Lecture 19-22 for 600 A mutant bacterial cell has a defective aminoacyl synthetase that attaches a lysine to tRNAs with the anticodon AAA instead of a phenylalanine. The consequence of this will be that A) none of the proteins in the cell will contain phenylalanine. B) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU. C) the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-specifying anticodons. D) the ribosome will skip a codon every time a UUU is encountered. E) None of the above will occur; the cell will recognize the error and destroy the tRNA. Answer

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Lecture 19-22 for 600 Answer What is B) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU

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Lecture 23-26 for 100 The bicoid gene product is normally localized to the anterior end of the embryo. If large amounts of the product were injected into the posterior end as well, which of the following would occur? A) The embryo would grow to an unusually large size. B) The embryo would grow extra wings and legs. C) The embryo would probably show no anterior development and die. D) Anterior structures would form in both sides of the embryo. E) The embryo would develop normally. Answer

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Lecture 23-26 100 Answer What is D) Anterior structures would form in both sides of the embryo

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Lecture 23-26 for 200 Answer During the early part of the cleavage stage in frog development, the rapidly developing cells A) skip the mitosis phase of the cell cycle. B) skip the S phase of the cell cycle. C) skip the G1 and G2 phases of the cell cycle. D) rapidly increase the volume and mass of the embryo. E) skip the cytokinesis phase of the cell cycle.

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Lecture 23-26 for 200 Answer What is C) skip the G1 and G2 phases of the cell cycle Frog cleavage cycles consists of S phase and M phase but lack G1 and G2 phases

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Lecture 23-26 for 300 If gastrulation was blocked by an environmental toxin, then A) embryonic germ layers would not form. B) cleavage would not occur in the zygote. C) fertilization would be blocked. D) the blastula would not be formed. E) the blastopore would form above the gray crescent in the animal pole. Answer

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Lecture 23-26 for 300 Answer What is A ) embryonic germ layers would not form

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Lecture 23-26 for 400 If a Drosophila female has a homozygous mutation for a maternal effect gene, A) she will not develop past the early embryonic stage. B) only her male offspring will show the mutant phenotype. C) her offspring will show the mutant phenotype only if they are also homozygous for the mutation. D) only her female offspring will show the mutant phenotype. E) all of her offspring will show the mutant phenotype, regardless of their genotype. Answer

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Lecture 23-26 for 400 Answer What is E) all of her offspring will show the mutant phenotype, regardless of their genotype. With maternal effect gene, an organism shows the phenotype expected from the genotype of the mother, irrespective of its own genotype

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Lecture 23-26 for 500 This segment of DNA has restriction sites I and II, which create restriction fragments A, B, and C. Which of the gels produced by electrophoresis shown below best represents the separation and identity of these fragments? A)C)E) B)D) Answer

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Lecture 23-26 for 500 Answer What is B) DNA is negatively charged due to the phosphate ions present in the ribose-phosphate backbone. It moves towards the positive pole during electrophoresis. The shorter the DNA fragment, the faster it moves towards the positive pole.

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Lecture 23-26 for 600 For a neuron with an initial membrane potential at - 70 mV, an increase in the movement of potassium ions out of that neuron's cytoplasm would result in A) depolarization of the neuron. B) hyperpolarization of the neuron. C) the replacement of potassium ions with sodium ions. D) the replacement of potassium ions with calcium ions. E) the neuron switching on its sodium-potassium pump to restore the initial conditions. Answer

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Lecture 23-26 for 600 Answer What is B) hyperpolarization of the neuron.