Pshs Upcat Review Bio (Part 2) Answer Guide

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Pshs Upcat Review Bio (Part 2) Answer Guide

  1. 1. Note: The discussion here is a very general overview of topics Genetics that MIGHT come out in the UPCAT. Your discussion of these topics in Bio3 will definitely be more formal and more Molecular Dogma detailed. Also, the more complicated problems here will most Chemical Biology probably NOT come out in the UPCAT, but the basic topics from which they stem might. Cell Division
  2. 2. G1 What determines the sex of an unborn child? The chromosome content of the sperm The chromosome content of the ovum Diet of the father before sexual intercourse The number of days between ovulation and fertilization
  3. 3. • 2 varieties of sex chromosomes in humans: X & Y – XX = female, XY = male • The two sex chromosomes segregate during meiosis, and each gamete receives one – In the ovaries: In the testes: X X X X X Y X Y • At the moment of conception: X X X X X Y X Y Female Male
  4. 4. G2 In pea plants, spherical seeds (S) are dominant to dented seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape, what fraction of the offspring would be spherical seeds? 1/4 1/2 3/4 1
  5. 5. In pea plants, spherical seeds (S) are dominant to dented seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape, what fraction of the offspring would be spherical seeds? Alleles (alternative form) of the gene • Heterozygous: has one of (segment of DNA) for seed shape each kind of allele (Ss) • Dominant: trait is expressed all the • Homozygous: has two of the time same kind of allele o SS or Ss  spherical o Homozygous dominant: SS • Recessive: trait is expressed only if the o Homozygous recessive: ss organism has 2 copies of the allele o ss  dented
  6. 6. In pea plants, spherical seeds (S) are dominant to dented seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape, what fraction of the offspring would be spherical seeds? the parents both have the genotype (genetic makeup) Ss and are both spherical (the phenotype, or physical and observable expression of the genotype) during meiosis, the alleles for seed shape are segregated as follows, for both male and female parents: S S s s
  7. 7. In pea plants, spherical seeds (S) are dominant to dented seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape, what fraction of the offspring would be spherical seeds? • During fertilization, the possible combinations of gametes displayed in the table below, called a Punnett square: S s Gametes from parent 1 S SS Ss Possible genotypes of offspring s Ss ss Gametes from parent 2
  8. 8. In pea plants, spherical seeds (S) are dominant to dented seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape, what fraction of the offspring would be spherical seeds? • The resulting ratios of genotypes and phenotypes are: Genotype Ratio Phenotype Ratio SS 1/4 Spherical seed (SS or Ss) 3/4 Ss 2/4 ss 1/4 Dented seed 1/4 4/4 4/4 • Take note that these are always the resulting ratios of a monohybrid cross mono: dealing w/ a single trait hybrid: parents are heterozygous
  9. 9. G3 A man with type A blood marries a woman with type B blood. Their child has type O blood. The genotypes of the father and the mother, respectively, are AB; BB AA; BB AO; BO AO; BB
  10. 10. • The ABO blood type is controlled by a single gene with three alleles: IA, IB, and i*. IA and IB are codominant; IA and IB are both dominant over i. • The following are the genotypes of the different ABO blood group phenotypes: Phenotype Genotype Type A IAIA or IAi Type B IBIB or IBi Type AB IAIB Type O ii * For simplicity, IA is sometimes represented as A, IB as B, and i as O
  11. 11. A man with type A blood marries a woman with type B blood. Their child has type O blood. The genotypes of the father and the mother, respectively, are… Individual Phenotype Genotype Father Type A IAIA or IAi Mother Type B IBIB or IBi Child Type O ii You can construct a Punnett square using the different combinations of parental genotypes to see which combination will yield a type O child.
  12. 12. • (Additional info) Immunological aspects of the ABO blood group: Recipient phenotype Donor phenotype(s) Type A A or O • has A antigens • produces antibodies vs. B antigens Type B B or O • has B antigens • produces antibodies vs. A antigens Type AB (universal recipient) A, B, AB, or O • has both A and B antigens • does not produce antibodies to either A or B antigens (because they have both) Type O (universal donor) O • does not have either A or B antigens; • produces antibodies vs. both A and B antigens
  13. 13. G4 The diagram below shows the family tree of a family with phenylketonuria (PKU). PKU is a disease that is expressed in homozygous recessive individuals. Which of the following correctly describes the genotype of individuals in the family tree? A. P and Q are heterozygous B. P and Q are homozygous dominant C. R and S are homozygous dominant D. R is homozygous dominant, S is heterozygous dominant
  14. 14. • A pedigree is a family tree describing the interrelationships of parents and children across the generations • Legend: –  is a ;  is a –   indicates a mating, with offspring listed below in their order of birth –  shaded symbols stand for individuals with the trait being traced
  15. 15. pp pp pp pp If we let P = dominant allele of the gene associated with the occurrence of PKU, and p = recessive allele, this means that all shaded symbols have a genotype of pp. In determining the unknown genotypes in a pedigree, it is usually helpful to focus first on matings that produce homozygous recessive offspring, since this can narrow down your choices for the parents’ genotypes.
  16. 16. pp pp pp pp Let’s focus first on the mating of R and S. Since they are both unaffected, and they have an affected offspring (which means one or both of them must carry but not express the recessive allele) their possible genotypes are P_ x P_. By constructing Punnett squares of the possible genotypes of the parents, we can get these results: Parents Offspring PP x Pp 2 PP: 2 Pp These results indicate that R and S Pp x Pp 1 PP: 2 Pp: 1 pp are both heterozygous.
  17. 17. pp pp pp Pp Pp pp Next, let’s focus on individual P (unknown genotype) and its mate (genotype pp). Individual P’s possible genotypes are PP or Pp (can you figure out why?). Again, constructing Punnett squares for the possible mating combinations would yield the following results: Parents Offspring At this point, you can already PP x pp All Pp answer the question, though I encourage you to try to figure out Pp x pp 2 Pp: 2 pp the genotype of individual Q.
  18. 18. MD1 The diagram below shows the steps in protein synthesis. If the DNA sequence used to produce X is 5’-ATGTGGAAT-3’, what will be the sequence of X? 3'-UACACCUUA-5' 3'-TACACCTTA-5' 5'-UACACCUUA-3' 5'-TACACCTTA-3' DNA X Protein
  19. 19. • The flow of genetic information, summarized in the diagram below, is usually referred to as the central dogma of molecular biology. Notice that this diagram is very much similar to the diagram in the question.
  20. 20. • DNA (a polymer) carries genetic information via the sequence of its nucleotides (its monomers). • Nucleotides are made of building blocks themselves: Nucleoside a nitrogenous base, Nitrogenous a pentose (5-carbon) sugar, base and a phosphate group. • Nitrogenous bases: O 5’C – Adenine (A) O P O CH2 O – Thymine (T) O (or uracil (U), in RNA) Phosphate 3’C – Cytosine (C) group Pentose sugar – Guanine (G)
  21. 21. • The DNA molecule consists of two nucleotide chains that spiral around an imaginary axis, forming a double helix. • The nitrogenous bases of one chain are linked to those of the other chain through hydrogen bonds. The two chains are said to be complementary. The base-pairing rules are as follows: – adenine with thymine (or uracil, in RNA) – cytosine with guanine
  22. 22. • The central dogma illustrates how information in DNA is used to make proteins. • The nucleotide sequence in DNA (or simply, the DNA sequence) is used as a basis to make an intermediate molecule, RNA. This is the process of transcription. • The RNA sequence is then used to come up with a sequence of amino acids, which eventually make up a protein. This is the process of translation.
  23. 23. • Going back to the question: If the DNA sequence used to produce X is 5’-ATGTGGAAT-3’, what will be the sequence of X? • We are thus being asked to make an RNA sequence given this DNA sequence. • Before we start, though, we must know that when an RNA molecule is made, it elongates from the 5’ end to the 3’ end (5’3’). Thus its DNA template must be the strand that goes from 3’5’, which means that we need to know the complementary strand of the DNA sequence given above. Given strand: 5’-ATGTGGAAT-3’ Complementary strand: 3’-TACACCTTA-5’ This will be the template for the RNA sequence we are being asked of.
  24. 24. • Going back to the question: If the DNA sequence used to produce X is 5’-ATGTGGAAT-3’, what will be the sequence of X? • Now that we know what the sequence of the template DNA is, we can put together the RNA that will result during transcription: Template strand: 3’-TACACCTTA-5’ RNA after transcription: 5’-AUGUGGAAU-3’ • Note that the resulting RNA sequence is the same as the DNA sequence initially given to us, except that the T’s were replaced by U’s. • Note, also, that the correct answer is not among the choices given in the problem. Sorry about that :P
  25. 25. MD2 Which component is NOT DIRECTLY involved in the process known as translation? mRNA DNA RNA ribosomes
  26. 26. • Translation is the RNA-directed synthesis of a polypeptide. This means that RNA (in particular, Proteins consist of one or more messenger RNA (mRNA), which was polypeptides, which are chains of produced by transcription) is the amino acids. template for making protein. • Recall from your lessons on cell structure and function the functions of ribosomes. • Given these information, you will then be able to answer the previous question.
  27. 27. MD3 Based on the genetic code below, which amino acid sequence is specified by the mRNA sequence UAUCGCACCUCAUAG when the first triplet (UAU) is used as the start of the reading frame?* Tyr-Gly-Ser-Leu Tyr-Arg-Thr-Ser Tyr-Arg-Thr-Asp Tyr-Arg-Thr-Ser * The PDF version of this reviewer only has tripeptides as choices for A and B. Please use the choices as listed here. Sorry about that.
  28. 28. • The genetic code is made of triplets of nucleotides and the DNA Gene 2 molecule amino acids for which they Gene 1 code. Gene 3 • During translation, our cells DNA strand 3 5 (ribosomes, in particular) (template) A C C A A A C C G A G T “read” the mRNA produced TRANSCRIPTION earlier by translation in sets of three nucleotides (codons). As mRNA 5 U G G U U U G G C U C A 3 the ribosome encounters a Codon certain codon, the TRANSLATION corresponding amino acid is Gly Protein Trp Phe Ser added to the growing Amino acid polypeptide.
  29. 29. • Given the mRNA sequence, separated into codons: UAU CGC ACC UCA UAG • The first codon: UAU – Locate the 1st nucleotide, U, on the leftmost column of the genetic code to the right – Then locate the 2nd nucleotide, A, on the top row – Then locate the 3rd nucleotide, U, on the rightmost column – The amino acid found at the intersection of those three nucleotides is the amino acid corresponding to that codon.
  30. 30. • Given the mRNA sequence, separated into codons: UAU CGC ACC UCA UAG • If you work out the rest of the sequence, you should come up with: Tyr-Arg-Thr-Ser • The stop codons – UAA, UGA, and UAG – are chemical signals to the cell that indicate the end of translation
  31. 31. MD4 Based on the genetic code below, identify a possible sequence of nucleotides in the DNA template for an mRNA coding for the polypeptide sequence Phe-Pro-Lys. AAA-GGG-UUU TTC-CCC-AAG TTT-CCA-AAA UUU-CCC-AAA
  32. 32. • Following the previous discussion, we now work our way back with this problem. • First, let us list the codons that code for the given amino acids: Amino acid Codons Phe UUU, UUC Pro CCU, CCC, CCA, CCG Lys AAA, AAG
  33. 33. Amino acid Codons • Since we have many choices for the middle amino acid (aa), Phe UUU, UUC let us focus first on the 1st and Pro CCU, CCC, CCA, CCG 3rd aa’s. Lys AAA, AAG • A Phe-Pro-Lys tripeptide will result from an RNA sequence that – starts with either 5’-UUU or 5’-UUC – ends with either AAA-3’ or AAG-3’ • This means that its DNA template would have a sequence such as - 3’-AAA or 3’-AAG - TTT-5’ or TTC-5’
  34. 34. • With these options for the end sequences: - 3’-AAA or 3’-AAG - TTT-5’ or TTC-5’ We can eliminate the obviously wrong answers from the options given to us. Notice, too, that the eliminated choices below could have also been discarded at the start because of the uracils in their sequence (DNA does not have uracil). A. AAA-GGG-UUU C. TTT-CCA-AAA B. TTC-CCC-AAG D. UUU-CCC-AAA • Now that you’re left with just options B and C, you can now look at the possible sequences for the 2nd aa in order to choose your final answer. Can you figure out why the BEST answer is C?
  35. 35. MD5 Which of the following is NOT true of a codon? It consists of three nucleotides. It may code for the same amino acid as another codon does. It never codes for more than one amino acid. It is the basic unit of the genetic code. You can easily answer this by studying the genetic code from the previous problems.
  36. 36. MD6 Corn plants have been genetically modified to contain a gene which can produce toxins to kill pests of the plant. Which of the following arguments is NOT a valid reason for opposing the widespread use of this genetically modified crop? Beneficial insects may inadvertently be killed. The inserted gene could cause mutation to occur on other parts of the plant chromosome. The inserted gene could become incorporated into the genome of weed species growing nearby. The modified chromosomes could cause the chromosomes in the cell in the body of the consumer to mutate. Statements A and C describe possible scenarios. Thus your choices are down to B or D, both of which have little or no scientific evidence of happening (at least not yet).
  37. 37. CB1 Some substances were isolated from a chimpanzee that directly played roles in body structure, immunology, and oxygen transport. What is the classification of these substances? lipids carbohydrates proteins nucleic acids Carbohydrates and lipids are mainly sources of energy, while nucleic acids store genetic information. Proteins are called the workhorses of the body, since they perform various functions such as the ones mentioned in this problem.
  38. 38. CB2 Given the table below, what are substances Y and Z? fats; amino acids carbohydrates; amino acids fats; fatty acids and glycerol proteins; monosaccharides Digestive enzyme Substance digested Resulting simple molecules Amylase Polysaccharides Disaccharides (W) Proteins (X) Lipase (Y) (Z)
  39. 39. • The name of the enzyme – lipase – indicates that the enzyme acts on lipids. Thus the question is: what are the building blocks of lipids? • The following table lists the important molecules found in living organisms, as well as their building blocks. Note that only carbohydrates, proteins, and nucleic acids are considered polymers (long molecules consisting of many similar building blocks called monomers). Lipids consist of two types of smaller molecules. Biomolecule Building blocks Carbohydrates Monosaccharides Proteins Amino acids Nucleic acids Nucleotides Lipids 1 glycerol + 3 fatty acids
  40. 40. CB3 A cell may use lipids to speed up the chemical reaction. form a barrier between the cell and its environment. provide structural support. hydrolyze food molecules.
  41. 41. • Let’s analyze which biomolecules are being referred to by each option: A. speed up the chemical reaction.  these are enzymes, which are proteins B. form a barrier between the cell and its environment.  the barrier refers to the cell membrane, which consists of a lipid bilayer C. provide structural support.  proteins such as collagen and elastin are known to provide structural support in animals D. hydrolyze food molecules.  again, these refer to proteins
  42. 42. CB4 Which of the following best describes the difference between the structure of carbohydrates and fats? Carbohydrates Fats A. Do not contain the element Contain the element nitrogen nitrogen B. Ratio of H atoms to O atoms is No fixed ratio of H and O 2:1 atoms C. Made of glucose molecules Made of fatty acids and joined together glycerol joined together D. No fixed ration of H to O Ratio of H atoms to O atoms is atoms 2:1
  43. 43. • All you need to answer the question is to recall the structure of carbohydrates and fats: – Carbohydrates: C(H2O)n – Fats or lipids: 1 glycerol + 3 fatty acids
  44. 44. CB5 Which of the following is NOT a property of ALL enzymes? Enzymes catalyze catabolic reactions only. Enzymes have active sites of specific shapes. Temperature and pH affect the rate of enzyme reactions. Enzymes are required in very small quantities. You can answer this easily 
  45. 45. CB6 Scientists can create new genes and place them in another organism’s cells. What is this technique? DNA transfer DNA mutation cloning recombinant DNA technology You can answer this easily  The definitions of these terms are in Campbell.
  46. 46. CB7 Which is NOT true about the structure of eukaryotic DNA? It has double helical structure. It is made up of two complementary strands. It contains nucleotides as its basic structural unit. It is composed of phosphates, ribose, and nitrogenous bases. You can answer this easily 
  47. 47. CB8 Which of the following statements about nucleic acids is correct? In a given sample of DNA the amount of thymine is equal to the amount of cytosine. Both DNA and RNA are assembled from nucleoside triphosphates. The bases and sugar molecules found in DNA and RNA are the same. All of the above are correct.
  48. 48. To evaluate the correctness of each statement, you need to recall a few facts about nucleic acids. A. In a given sample of DNA the amount of thymine is equal to the amount of cytosine.  base-pairing rules: A-T and C-G B. Both DNA and RNA are assembled from nucleoside triphosphates.  Nucleoside triphosphates are the ingredients of polynucleotide synthesis C. The bases and sugar molecules found in DNA and RNA are the same.  Clue: DNA, RNA
  49. 49. CB9 A scientist analyzed several DNA samples to determine the relative proportions of purine (adenine and guanine) and pyrimidine (cytosine and thymine) bases. Her data are summarized in the table below: Percentages of Nitrogenous Bases in Three Samples Sample G C A T I 35 35 15 15 II 40 10 40 20 III 25 25 25 25 Which sample(s) supports the base pairing rules? A. Sample I only C. Sample I and II B. Sample II only D. Samples I, II, and III
  50. 50. • If a sample supports the rules on base pairing (already mentioned earlier), that would mean that the amounts of G and C will be equal, and that the amounts of A and T will be equal. Note that none of the choices is valid. Again, sorry for that :P
  51. 51. CB 10 A scientist analyzed several DNA samples to determine the relative proportions of purine (adenine and guanine) and pyrimidine (cytosine and thymine) bases. Her data are summarized in the table below: Percentages of Nitrogenous Bases in Three Samples Sample G C A T I 35 35 15 15 II 40 10 40 20 III 25 25 25 25 If the scientist had analyzed mRNA rather than DNA, what percentage of uracil would you expect to find in Sample II? A. 10 C. 35 B. 25 D. 40 For this question, you just need to remember that in RNA uracil is present rather than thymine.
  52. 52. CD1 Nerve cells and cardiac (heart) cells have long life spans and do not undergo mitosis once they are formed. Smooth muscles cells also have long life spans but are capable of undergoing mitosis. Which of the following statements is FALSE regarding consequences of injuries to the spinal cord, heart, or smooth muscle? Damage caused by injuries to smooth muscle may be reversed because of its capability to regenerate. Injuries to the spinal cord and heart cause serious and permanent damage. Injuries to the spinal cord, heart, and smooth muscle cause serious damage and take a very long time to heal. All of the above statements are false. You can answer this easily 
  53. 53. CD2 Which of the following stages of mitosis can be viewed as the opposite of prophase if we consider only the changes in the nucleus? anaphase interphase metaphase telophase
  54. 54. Recall the stages of mitosis:
  55. 55. CD3 The following statements regarding mitosis and meiosis are true EXCEPT Replication of DNA occurs before nuclear division begins. Mitosis involves only one round of cell division, while meiosis involves two rounds. The second meiotic division is similar to mitosis in that the chromosome number is reduced. Mitosis results in two genetically identical diploid daughter cells, while meiosis results in four genetically nonidentical haploid daughter cells.
  56. 56. Recall the stages of meiosis (those of mitosis are on a previous slide already).
  57. 57. CD4 Mature cells no longer divide, so they do not replicate their DNA. A cell biologist found that there was X amount of DNA in a human nerve cell. The biologist then measured the amount of DNA in four other types of human cells; the results are recorded in the chart below. Complete the chart by filling in the type of cell from these choices: A. sperm cell B. bone marrow cell just beginning interphase of the cell cycle C. skin cell in the S phase of the cell cycle D. intestinal cell beginning mitosis. Cell Amount of DNA Type of Cell I 2X II 1.6X III 0.5X V X
  58. 58. Here are some clues: A. sperm cell  haploid B. bone marrow cell just beginning interphase of the cell cycle C. skin cell in the S phase of the cell cycle D. intestinal cell beginning mitosis.  Choices B, C, and D are all diploid cells. At each of the stages of the cell cycle mentioned, is there any DNA replication going on? Cell Amount of DNA Type of Cell I 2X II 1.6X III 0.5X V X Try to figure this out first.
  59. 59. “Chance favors the prepared mind.” - Louis Pasteur

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