Fa08E1 - BICD 100 Exam 1A 2 1. Suppose Meselson and Stahl...

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Unformatted text preview: BICD 100 Exam 1A 2 1. Suppose Meselson and Stahl discovered that DNA replicated in a conservative fashion. When cells are grown in a heavy media, then shifted to a light media and allowed to replicate, how many bands would there be after 3 rounds of replication followed by density centrifugation, and would the bands in the centrifuge tube be heavy, light or hybrid? a. 1 band, hybrid d. 3 bands; heavy, light and hybrid b. 2 bands; heavy and hybrid e. None of the above c. 2 bands; light and hybrid Conservative replication would mean that only heavy and light bands would be produced. No intermediate bands would be formed, no matter how many rounds of replication. Samples of 7 newly discovered organisms were analyzed. Use the information in the table to answer the following questions: (Note: some information is missing from the table. This could either be due to the fact that the geneticist could not determine the content or that this particular organism does not contain that nucleotide.) Organism Adenosine Guanine Cytosine Thymine Uracil 1 22% 26% 22% 2 14% 14% 36% 3 23% 23% 4 16% 24% 24% 5 14% 33% 26% 6 24% 32% 12% 7 21% 29% 29% 21% 2. Which organism(s) could possibly be single stranded RNA? [Include ALL possibilities.] All possibilities means everything that could possibly be single stranded RNA, which means all organisms that DON'T have thymine. So the correct answer would be 1, 3, 4, and 6. This was not one of the choices. a. 4, 5, 6 c. 1 e. None of the b. 1, 4, 5 d. 6 above. 3. Which organisms could possibly be single stranded DNA? [Include ALL possibilities.] All possibilities means everything that could possibly be single stranded DNA, which means all organisms that DON'T have uracil. So the correct answer would be 2, 4, 5, and 7. This was not one of the choices. a. 4, 7 c. 4 e. None of the above b. 4, 5 d. 5 4. Mitosis produces ____ daughter cells that are ______ to the parent cell. ______ parent cells produce ____ daughter cells. Mitosis is simple cell division, so one cell produces 2 identical cells. If parent cell is diploid, then daughter cells are diploid. a. Four; different; diploid; haploid d. Four; identical; diploid; diploid b. Two; identical; diploid; diploid e. Two; different; diploid, haploid c. One; identical; haploid, diploid 5. Redgreen color blindness is an Xlinked recessive disorder. Charlie, a man with normal vision, marries Jessica, a normal female whose father and paternal BICD 100 Exam 1A 3 grandmother's father both suffered from redgreen color blindness. What is the probability that their first child, boy or girl, suffers from redgreen color blindness? (Disregard Charlie's family history) Jessica is a carrier (XCXc) while Charlie is a normal man (XCY). Their progeny could be carrier females, normal females, normal males, or colorblind males, in a 1:1:1:1 ratio. of the possibilities are colorblind. a. 1/2 c. 1/6 e. None of the above b. 1/4 d. 1/8 6. After having a normal boy, Jessica and Charlie decide to have another child. What is the probability that they will have an affected girl? They cannot have an affected girl, only carrier girls or normal girls. a. 0 c. 3/4 e. None of the above b. 1/8 d. 1/3 7. Jessica and Charlie's third child, a girl, has one eye with normal color vision and one eye with redgreen colorblindness. What is the best explanation? The only reasonable choice among the answers given is b. a. A nondisjunction event occurred which made this daughter XXY. b. The normal allele for color vision was on a chromosome that was inactivated in all the retina cells of the colorblind eye. c. The telomeres in her affected eye were progressively shorter than in the normal eye. d. A and C e. None of the above. 8. DNA is replicated in the _____ _____ direction on the leading strand and in the _____ _____ direction on the lagging strand DNA is always replicated in the 5'3' direction, regardless of which strand. a. 3'5', 3'5' c. 3'5', 5'3' e. None of the b. 5'3', 3'5' d. 5'3', 5'3' above. 9. If a double stranded DNA molecule contains 15% guanine, it must contain what percentage of thymine? If the DNA is 15% guanine, it must also be 15% cytosine, since G=G. If 30% of the DNA is G & C, then 70% must be A & T, and they must be present in equal numbers. of 70=35. a. 15% b. 85% c. 70% d. 35% e. 30% For questions 1014, use the following choices. Match each blank to the best possible answer. Use each answer once. a. DNA ligase d. DNA primase b. Okazaki fragments e. Telomerase c. DNA polymerase 10. The enzyme that attaches dNTPs to a growing strand of newly synthesized DNA is called____C______. The polymerase is responsible for adding nucleotides to new DNA. 11. The relatively short nucleic acid fragments that are synthesized during discontinuous DNA replication and joined together to form a continuous strand BICD 100 Exam 1A 4 are called___B______. The short pieces are called Okazaki fragments, and are only made on the lagging strand. 12. The enzyme that has a builtin RNA strand template that initiates DNA synthesis at the end of chromosomes is called_____E___. Telomerase is responsible for synthesizing the ends of chromosomes, but is only active in germ cells and cancer cells. Other cells progressively lose small parts of their telomeres at each division. 13. Short pieces of RNA are synthesized during DNA replication by an enzyme called___D_____. Primase (or DNA Primase) is the enzyme that makes the small RNA primers needed for DNA polymerase to extend the chain. 14. The enzyme that forms a phosphodiester bond between adjacent nucleotides, but does not extend the strand is called___A____. DNA Ligase is the enzyme that forms the phosphodiester bond between two discontinuous pieces of DNA, sealing the backbone and making one complete molecule. Dragons have 38 pairs of homologous chromosomes. How many chromosomes and chromatids are present in cells at the following stages of the cell cycle, and are the cells haploid or diploid? Use the following choices to answer questions 2224. 15. Metaphase I E At metaphase I, the cell has just recently entered meiosis, and so it still has the chromosome complement that it had after G2. This means that there are 76 chromosomes, each of which is doubled, so 152 chromatids. Since there are two of each of the pairs of chromosomes, the cell is diploid. 16. Anaphase II B At anaphase II, the cell has already had the homologous pairs of chromosomes split during meiosis I, plus the remaining 38 chromosomes have just had their sister chromatids split into their two daughter chromosomes. So there are 76 chromosomes, 76 chromatids, and the cell is haploid since the chromosomes that are present only represent one half of the original chromosomal content of the progenitor cell. 17. Horn cell in G1 D In G1, the cells have just completed cell division and mitosis, so each chromosome is comprised of a single chromatid. The cells are diploid, though, so there are two homologous chromosomes for each of the 38, which means 76 chromosomes, 76 chromatids, and diploid. a. 38 chromosomes, 38 chromatids, haploid b. 76 chromosomes, 76 chromatids, haploid c. 38 chromosomes, 76 chromatids, diploid d. 76 chromosomes, 76 chromatids, diploid e. 76 chromosomes, 152 chromatids, diploid 18. Parts of genes called ______________ are transcribed into mRNA but are later removed. Introns are part of the sequence that is transcribed into RNA, but it is spliced out to form the mature mRNA. a. Prions d. Introns b. Exons e. None of the above c. Caps 19. If the genetic code contained only 3 different bases and these bases made up 4 base code words (codons), how many different codons would be possible? BICD 100 Exam 1A 5 If the genetic code was one base, then there would be three possible codons. If the genetic code was two bases, then there would be 3x3=9 possible codons. If three bases, 3x3x3=27. For a code of 4 bases, each with three possible nucleotides, there are 3x3x3x3=81 possibilities. a. 64 b. 256 c. 27 d. 81 e. 12 20. A 1 B 2 C 3 D 4 E Given above is a prespliced sequence of introns (numbers 14) and exons (letters AE) within an mRNA. Of the following spliced sequences of mature mRNA, which cannot be made? The mature mRNA must contain the exons in the order in which they are in the newly transcribed RNA. Exons can be cut out by alternative splicing, but they don't get reshuffled. a. ABCDE c. ABDC e. None of the b. ABCE d. ACE above. 21. Colinearity implies that there is a direct correspondence between the nucleotide sequence of DNA and the amino acid sequence of protein. Although this is generally true of prokaryotes, it is not the case for eukaryotes because: a. Deletions and insertions of nucleotides by RNA polymerase are more likely to occur in eukaryotes than in prokaryotes. Not true. b. After transcription, a 5' cap and 3' polyA tail are added to mRNA, skewing the direct correspondence of the DNA nucleotide sequence to the protein amino acid sequence in eukaryotes. Cap and tail do not affect amino acid sequence. c. During translation, not every mRNA codon is converted to an amino acid via tRNA in eukaryotes. Not true. d. During transcription, introns are transcribed but are spliced out during mRNA processing in eukaryotes. YES e. Due to the complexity of eukaryotes, it is necessary for more RNA to be transcribed to form proteins than is necessary for prokaryotes. Not true. 22. During what phase of the cell cycle do the chromosomes arrive at the spindle poles? You know this one! a. Interphase c. Metaphase e. Telophase b. Prophase d. Anaphase 23. Which statement is not true about mitosis and meiosis? a. Mitosis and meiosis are the same except that meiosis consists of two divisions while mitosis is only one. The two divisions are not the same, in meiosis one division is a reductional division and one division is an equational division. All of the other statements are true. b. Homologous chromosomes do not synapse in mitosis but do in meiosis. c. Mitosis produces two daughter cells whereas meiosis forms four daughter cells. d. Mitosis produces identical progeny cells whereas meiosis does not. e. In both mitosis and meiosis, chromosomes line up on the metaphase plate. BICD 100 Exam 1A 24. The human betaglobin protein is 255 amino acids long. How many mRNA nucleotides are required to encode human betaglobin protein? For a 255amino acid protein, there must be 255*3=765 nucleotides in the coding region of the mRNA, plus an additional 3 nucleotides to specify the stop codon, for a total of 768. Without that last stop codon, the protein would have continued to translate the mRNA into proteins, and the protein would have been longer than 255 amino acids, so all 768 nucleotides are required for the proper formation of the protein. a. 85 c. 258 a. 768 b. 255 d. 765 Questions 2529 refer to the following diagrams: 6 I II III IV V VI VI I VIII 25. What is number of DNA molecules in II? 6 chromosomes with 2 chromatids each, so 12 DNA molecules. a. 6 c. 18 e. Unable to b. 12 d. 24 determine 26. Which diagram represents anaphase I of meiosis? Anaphase I is when the homologous chromosomes separate and migrate to opposite poles. a. VII b. V c. IV d. II e. VIII BICD 100 Exam 1A 7 27. How many chromosomes are present at anaphase I? Since no cell division has occurred at anaphase I, the same number of chromosomes are present that were there at the beginning of meiosis, which is 6. a. 3 b. 6 c. 12 d. 18 e. 24 28. Which diagram shows an end product of mitosis? Meiosis? At the end of mitosis, there will still be 6 chromosomes (1 chromatid each) per cell; at the end of meiosis, there will be only 3 chromosomes per cell. a. I, VII c. II, III e. VII, VIII b. V, VII d. I, V 29. What is the diploid number (2n) of chromosomes in this organism? The diploid number refers to the total number of chromosomes, which is 6 in this cell. a. 2 c. 6 e. Unable to b. 3 d. 12 determine 30. If a couple has 4 children, what is the probability that 3 are boys and 1 a girl? There are several ways to solve this problem. You can use binomial theorem, where P=[n!/s!t!](ps)(qt), where n=4, s=3, t=1, p=q=1/2, so P=[4*3*2*1/3*2*1*1](1/2)3(1/2) = 4/16 = . OR you could have drawn out a branch diagram with 4 kids, which would have given a total of 16 branches. 4 of those branches would have had 3 B and 1 G (G could be 1st, 2nd, 3rd, or 4th). a. 1/4 b. 3/8 c. 1/3 d. 1/16 e. 4/7 31. If a couple already has two girls and one boy, what is the chance that their next child is a boy? The previous children are irrelevant; the chance is 50% for any child. a. 0% b. 25% c. 50% d. 75% e. 100% 32. Nondisjunction... a. ...is related to the failure of centrosomes in chromosome pair separation. Centrosomes are the organelles from which the microtubules originate. b. ...is the failure of homologous chromosomes to separate in meiosis II. Homologous chromosomes separate during meiosis I, not meiosis II. c. ...is the failure of sister chromatids to separate in meiosis I. Sister chromatids separate during meiosis II, not meiosis I. d. ...is the failure of sister chromatids to separate in mitosis. This is the only true statement, since NDJ can occur in mitosis as well as meiosis. e. All of the above are true statements about nondisjunction. 33. What is the best explanation for the 3:1 ratio that Mendel obtained in the F2 generation when he crossed pea plants to observe seven different traits? a. There was a linkage of the pea color and each of the other traits. Linkage is unrelated to Mendel's observed traits. b. The phenotype is not directly linked with the genotype. Mendel's traits were examples of phenotype being directly correlated with genotype. c. The phenotype of the F1 generation was a blend of the two parents. F1s were like one or the other parent, not a blend at all. d. Each of the traits was governed by dominant and recessive alleles. This is the only true statement. BICD 100 Exam 1A e. A test cross is needed for every experiment that is involved in the study of genes. Test crosses are only useful for distinguishing between homozygous dominant and recessive individuals. 8 34. Spotted coat (p) in Lamas is recessive to solid colored coat (P). If a solidcoated Lama mated with a spotted coated Lama, what ratio of the F1 generation would be spotted? You are not told anything about the solidcoated llama (sorry about the typo in the question). There's no way to predict what the F1 will look like without doing a test cross to determine the genotype of the solid parent. a. None of the offspring would be spotted since the solid trait is dominant b. All of the offspring would be spotted since the solid trait is a deleterious trait c. It would be impossible to tell without a test cross of the solid parent. d. The F1 generation would be 25% spotted and 75% solid colored. e. The F1 generation would be 50% spotted and 50% solid colored. 35. In eggplant, large seeds (L) are dominant over small seeds (l) and long body shape (G) is dominant over short curved shape (g). If a homozygous long eggplant with large seeds is crossed with a short, curved eggplant with small seeds, what is the phenotypic ratio of the F1 generation? GGLL x ggll all GgLl, (long, large) a. all short, curved with small seeds b. all long with large seeds c. 1 short, curved with small seeds: 1 long with large seeds d. 1 long with large seeds: 2 long with small seeds: 2 short with large seeds: 1 short with small seeds e. 9 long with large seeds: 3 long with small seeds: 3 short with large seeds: 1 short with small seeds 36. If a researcher crosses a short eggplant with small seeds to a heterozygous (GgLl) eggplant, what is the probability that the offspring will either have a long body or small seeds (but not both)? GgLl x ggll GgLl; Ggll; ggLl; ggll OR long, large; long, small; short, large; short, small. Only the first and last classes satisfy the requirement, so or 50%. a. 100% b. 50% c. 25% d. 12.5% 75% 37. What is the correct order of mitosis? a. Prophase, Metaphase, Anaphase, Telophase b. G2, S, Metaphase, Telophase, G1 c. S, Prophase, G1, Metaphase, Anaphase, Telophase d. Telophase, S, G1, G2, Anaphase, Prophase e. None of the above Questions 3839. For the following modes of inheritance, how many of the children will be affected if the mother is affected with a disorder, and the father is not affected and does not have any family history of the disorder? Assume that the disorder is extremely rare in the general population. BICD 100 Exam 1A 9 The last sentence meant that you can assume that the father is not a carrier, plus that the mother will be heterozygous in the case of a dominant disorder. 38. Autosomal recessive. Mother must be aa, father AA, so all kids are Aa, none are affected. a. None of the children b. All of the sons and none of the daughters c. All of the daughters and half of the sons d. All of the daughters and none of the sons e. All of the children 39. Xlinked dominant. Mother must be XAXa; father must be XaY, so kids will be XAXa (affected daughter); XaXa (unaffected daughter); XAY (affected son); XaY (unaffected son). a. None of the children b. All of the sons and none of the daughters c. Half of the daughters and half of the sons d. All of the daughters and half of the sons e. All of the sons and half of the daughters 40. The patchy distribution of color on a calico cat results from what? Only females show the patchy distribution of color, which tells you that it must be related to the X chromosome somehow. If one X chromosome carries the allele for orange color, and one for black color, the patches of orange occur where the black containing X was inactivated, and vice versa. a. A genotype resulting in that coloration. b. The fact that one parent was black, the other brown, and blending occurred. c. Random inactivation of X chromosomes. d. Environmental factors. e. A recessive, sexlinked trait. 41. In humans, sex is determined by what? The Y chromosome must contain the SRY locus (which contains the TDF gene) in order to get a normal male. The SRY locus COULD be located on another chromosome, and the individual will be male even if they lack the Y chromosome. a. The presence of a Y chromosome in males. b. The absence of an X chromosome in females. c. The absence of a second X chromosome in males. d. The presence of a 1:1 ratio of X to autosomal sets in females. e. The presence of the SRY locus in males. 42. After meiosis, the resulting gametes have _______ the number of chromosomes as the parental cell. Meiosis is a process of dividing the cells so that each of the four resulting daughter cells has of the genetic content of the mother cell. a. Half c. One fourth e. Quadruple b. Equal d. Double 43. What modifications do eukaryotes make on RNA after it has been transcribed from DNA and before it is translated into a polypeptide strand? Mature eukaryotic mRNA must have all of the modifications listed. BICD 100 Exam 1A a. 5' Cap b. 3' PolyA Tail c. Splicing of introns 10 d. All of the above e. None of the above 44. What are telomeres? Telomeres are both the ends of the chromosomes (choice A) and the parts of the chromosome that may shorten over time (choice D). a. Ends of chromosomes. b. Responsible for the origin of replication. c. The location on the chromosome where microtubules attach in order to separate the chromosomes during mitosis and meiosis. d. Parts of the chromosome that may shorten over time. e. Two of the above answers are correct. 45. A eukaryotic chromosome usually has only one: A chromosome will have many origins of replication, since linear chromosomes take too long to replicate starting at just one end. All chromosomes contain many genes, so will have many reading frames and promoters, and will each contain millions of nucleotides. Each has only one centromere, though. a. Origin of b. Reading frame d. Promoter replication c. Centromere e. Nucleotide Unwebbed feet and albino are two recessive traits found in the tree frog of the Amazon. The unwebbed feet (f) prevent the frogs from jumping as fast as the frogs with normal feet (F). Albinism (a) prevents the frogs from camouflaging to its environment, as frogs that are wildtype (A) for that trait can. When a researcher crossed two frogs that are wildtype and with normal feet (without knowing their genotypes), he found the following results for the F1 generation: wildtype, normal feet wildtype, unwebbed feet albino, normal feet albino, unwebbed feet 153 57 68 15 46. From the given information, what is the most likely genotype of the parents? The answer can't be choice B, C, or D, since none of them fit the description of the frogs in the problem (both need to be AF). Either A or E could work, but we can rule out A since we would expect equal number of the 4 phenotypic classes from cross A, and something closer to a 9:3:3:1 ratio from cross E, which is what we see. a. AAFf x AaFF c. AaFf x aaFF *** e. AaFf x AaFf b. Aaff x AaFf *** d. aaff x AAFF *** 47. If a wildtype, unwebbed feet frog mated with an albino, normal feet, what would be one possible phenotypic ratio of their offspring? Each frog has several different possibilities. The first one could be AAff or Aaff; the second one could be aaFF or aaFf. The cross between aaFf and AAFf would give a 1:1:1:1 ratio, while AAff x aaFF would give only one phenotypic class. Either Aaff x aaFF or AAff x aaFf would give a 1:1 ratio. a. 1:1 c. 3:1 e. None of these are b. 9:3:3:1 d. 2:3:1 possible. BICD 100 Exam 1A 11 48. Why would the Chisquared test be necessary even if an experiment were done with a large population? a. It wouldn't be necessary to do a Chisquared test with a large population. Why on earth not? Large populations still show variation. b. A large population doesn't assume that variation is due to chance. Yes! c. A Chisquared test can provide genetic information only. What does this even mean? d. The complex calculations of a Chisquared test don't allow it to be used for small populations. The test can be used on any size group. e. Population size does not have an effect on the outcome of a Chisquared test. The larger the population, the larger the expected variation. 49. Hemophilia is an Xlinked recessive disorder that results in the inability of an individual to control bleeding due to a defective bloodclotting factor. Maria, a phenotypically normal female with no family history of hemophilia, marries Ted, a phenotypically normal man whose mother's brother suffered from the disease. Maria and Ted are worried that if they decide to have children they might have an affected child. Therefore, they decide to visit a geneticist to talk about the probability of having a hemophiliac child. Knowing what you know about inheritance, what is the probability that Ted and Maria will have a child affected by hemophilia? Maria must have the genotype X+X+, and Ted must be X+Y. Neither is affected, and neither is a carrier. a. 0% c. 50% e. None of the b. 25% d. 35% above. 50. Suppose, that recent research has shown that hemophilia can also be Ylinked. Using the same information as in question 49, what is the probability that if Maria and Ted decide to have a child that it will be affected by hemophilia? Ted couldn't have gotten his Y chromosome from his mother, so having the disease on Ted's mother's side of the family won't matter for a Ylinked disorder. Plus, we still assume that Ted is normal (unaffected). So his genotype does not change, and none of the kids can be affected. a. 0% c. 50% e. None of the b. 25% d. 35% above. ...
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This note was uploaded on 01/03/2009 for the course BICD 100 taught by Professor Nehring during the Fall '08 term at UCSD.

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