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QUESTION NO. 21

QUESTION NO. 21. In a heterozygous monohybrid cross, the dominant trait can be expressed in the phenotype of the F1 ________ of the time. 0 percent 25 percent 33 percent 75 percent 100 percent. The correct answer is (D).

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QUESTION NO. 21

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  1. QUESTION NO. 21 In a heterozygous monohybrid cross, the dominant trait can be expressed in the phenotype of the F1 ________ of the time. • 0 percent • 25 percent • 33 percent • 75 percent • 100 percent www.sirjamesdeverajr.weebly.com

  2. The correct answer is (D). • In a heterozygous cross – height, for example, when T represents the dominant, tall and t represents the recessive, short, trait-the following genotype would represent both parents: Tt. Following Mendelian procedures and using a Punnett square to ensure accuracy, the following genotypes would result: TT, Tt, Tt, and tt. This would produce an occurrence of the dominant trait 75 percent of the time. www.sirjamesdeverajr.weebly.com

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  4. QUESTION NO. 22 Which of the following blood types are possible if the parents are A and O blood types? • A and O • B and O • AB only • O only • A, B, and O www.sirjamesdeverajr.weebly.com

  5. The correct answer is (A) • This is a case of multiple alleles controlling a trait (not to be confused with polygenesis). The alleles for blood types A and B (represented by IAand IB , respectively) are separately dominant, when one is present but the other is not. Taken together, they are incompletely dominant as in AB (represented only by IA IB) blood type. Blood type O (represented by i) is recessive. If two alleles are IAIA or IA I, then the blood type will be A. if the two alleles are IBIB or IBi, then the blood type will be B. Type O blood can only be ii. www.sirjamesdeverajr.weebly.com

  6. QUESTION NO. 23 Which of the bonding examples below is NOT possible? • a DNA adenine to a DNA thymine • a DNA adenine to an RNA thymine • a DNA cytosine to an RNA cytosine • a DNA adenine to an RNA uracil • a DNA guanine to a DNA cytosine www.sirjamesdeverajr.weebly.com

  7. The correct answer is (B). • In RNA, thymine is replaced by uracil, so the correct choice is (B), where DNA adenine is said to bond to an RNA thymine. All the other choices are incorrect. www.sirjamesdeverajr.weebly.com

  8. Remember the Complementary Bases On DNA: A-T C-G On RNA: A-U C-G www.sirjamesdeverajr.weebly.com

  9. Name the Amino Acids • GGG? • UCA? • CAU? • GCA? • AAA? www.sirjamesdeverajr.weebly.com

  10. Protein Synthesis • The production or synthesis of polypeptide chains (proteins) • Two phases: Transcription & Translation • mRNA must be processed before it leaves the nucleus of eukaryotic cells www.sirjamesdeverajr.weebly.com

  11. Nuclear membrane DNA Transcription Pre-mRNA RNA Processing mRNA Ribosome Translation Protein DNA RNAProtein Eukaryotic Cell www.sirjamesdeverajr.weebly.com

  12. QUESTION NO. 24 For the DNA strand 5’-ACC-GTG-ACA-TTG-3’, the correct compliment DNA would be • 3’-TGG-CAC-TGT-AAC-5’ • 5’-TGG-CAC-TGT-AAC-3’ • 3’-UGG-CAG-UGU-AAC-5’ • 5’-ACC-GUG-ACA-UUG-3’ • 3’-TCC-GAG-TGT-AAC-5’ www.sirjamesdeverajr.weebly.com

  13. The correct answer is (A). • The complimentary strand for the indicated DNA strand would begin at the 3’ end, not at the 5’ end. In addition, it would also have thymine – T. finally, opposite A would be T, and vice-versa, and opposite G would be C, also vice-versa. www.sirjamesdeverajr.weebly.com

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  16. QUESTION NO. 25 An mRNA is 429 nucleotides long. The number of amino acids in the polypeptide chain formed from this mRNA is • 143 • 142 • 141 • 429 • 428 www.sirjamesdeverajr.weebly.com

  17. The correct answer is (B). • tRNA, which contains 3 anti-codon bases for every amino acid it carries, matches up opposite the mRNA codons that are also three bases long. This means that the mRNA in question potentially codes for 429/3 amino acids. This gives us a potential of 143 amino acids or choice (A). However, the stop sequence does not code for an amino acid, so this gives us 142 amino acids coded for. Care should be exercised when considering the start sequence. While not a part of the resulting protein, that site does not code for the amino acid methionine in the translation process. www.sirjamesdeverajr.weebly.com

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  20. Question no. 26 is about a breeding experiment. The researcher’s goal was to develop white mice with short tails. P brown mice with short tails x white mice with long tails F1 all offspring are brown and have long tails F2 292 mice are brown with long tails 97 mice are brown with short tails 103 mice are white with long tails 36 mice are white with short tails www.sirjamesdeverajr.weebly.com

  21. The results of the above cross indicate that among the original parents (P-generation) ________ • both were heterozygous for coat color and tail • one was homozygous dominant for coat color and tail length, whereas the other was homozygous recessive for both traits • one was homozygous dominant for coat color and homozygous recessive for tail length, whereas the other was homozygous recessive for coat color and homozygous dominant for tail length • one was homozygous dominant for both traits, whereas the other was heterozygous for both traits • one was homozygous recessive for both traits, whereas the other was heterozygous for both traits www.sirjamesdeverajr.weebly.com

  22. The correct answer is (C) • The question pertains to Mendelian inheritance. The only traits occurring in the F1 generation are brown coat color and long tails, whereas all four traits show up among the F2 progeny. This suggests that brown coat color and long tails are dominant over white coat and short tails respectively. The parents must have been homozygous for each trait, as only the dominant traits were present among their offspring. Thus, the parent with a brown coat and short tail must have been homozygous dominant for coat color and homozygous recessive for tail length, whereas the parent with white coat and long tail must have been homozygous recessive for coat color and homozygous dominant for tail length. www.sirjamesdeverajr.weebly.com

  23. QUESTION NO. 27 If a molecule of DNA is composed of approximately 16.2% adenine (A) and 33.4% guanine (G), the percentages of thymine (T) and cytosine (C) must be approximately • 16.3% T and 16.3% C • 34.1% T and 34.1% C • 34.1% T and 16.3%C • 16.3 T and 34.1% C • 33.4% T and 16.2 C www.sirjamesdeverajr.weebly.com

  24. The correct answer is (D) • In the DNA molecule, base pairing occurs between adenine and thymine, which are held together by two hydrogen bonds, and base pairing occurs between guanine and cytosine, which are held together by three hydrogen bonds. Therefore, the percentages of adenine and thymine would be similar, as would the percentages of guanine and cytosine. www.sirjamesdeverajr.weebly.com

  25. QUESTION NO. 28 IS ABOUT HARDY-WEINBERG EQUILIBRIUM Within the squirrel population at James City Park, 16% show the recessive phenotype of a curled tail (tt). What is the frequency of the dominant allele (T) in the population? • 0.40 • 0.16 • 0.26 • 0.60 • 0.32 www.sirjamesdeverajr.weebly.com

  26. The correct answer is (D). • If 16 percent (.16) of the population shows the recessive phenotype (q2), then the square root of .16 (0.40) would equal the frequency of the recessive allele, q. because p + q = 1, then p=1 – q = 0.60. www.sirjamesdeverajr.weebly.com

  27. QUESTION NO. 29 IS ABOUT HARDY-WEINBERG EQUILIBRIUM Within the squirrel population at James City Park, 16% show the recessive phenotype of a curled tail (tt). What is the frequency of the heterozygotes in the population? 0.08 • 0.24 • 0.36 • 0.48 • 0.64 www.sirjamesdeverajr.weebly.com

  28. The correct answer is (D). • The frequency of heterozygotes in the population is equal to two times the product of the frequencies of the recessive (q) and the dominant (p) alleles. Thus, the frequency of the heterozygotes is 2pq = 2(0.6)(0.4) = 0.48. www.sirjamesdeverajr.weebly.com

  29. QUESTION NO. 30 How many different kinds of gametes could the following individuals produce? 1.aaBb 2. CCDdee 3. AABbCcDD 4. MmNnOoPpQq 5. UUVVWWXXYYZz www.sirjamesdeverajr.weebly.com

  30. The answers are • Remember the formula 2n • Where n = # of heterozygous 1. aaBb = 2 2. CCDdee = 2 3. AABbCcDD = 4 4. MmNnOoPpQq = 32 5. UUVVWWXXYYZz = 2 www.sirjamesdeverajr.weebly.com

  31. Question no. 31 Sixteen percent of the human population is known to be able to wiggle their ears. • This trait is determined to be a recessive gene. • These is a population genetics question. • Use the following equation: 1 = p2 + 2pq + q2 www.sirjamesdeverajr.weebly.com

  32. Question #6A What of the population is homozygous dominant for this trait? • q2 = 16% or .16: q2 = .16 q = .4 • then use : 1 = p + q 1 = p + .4 1- .4 = p p = .6 • Now use p2 for answer: .62 = .36 or 36% www.sirjamesdeverajr.weebly.com

  33. Question 31.B What of the population is heterozygous for this trait? • We know that q = .4and p = .6 • Now use 2pq for answer: 2(.6)(.4) = .48 or 48% www.sirjamesdeverajr.weebly.com

  34. QUESTION NO. 32 (use the genetic codon table) Which of the following partial polypeptides is coded for by the series of codons below? -UUC –CCA –CAG –GGU –ACA – • Start-Thrp-Phe-Ala-stop • Met-Thr-Phe-Ala-stop • Leu-Lys-Ser-Arg-Val • Phe-Pro-Gln-Gly-Thr • Phe-Ser-Pro-Trp-stop www.sirjamesdeverajr.weebly.com

  35. The correct answer is (D) • More than one codon may code for the same amino acid. The codon AUG may code for methionine or it may code for a start site, signaling ribosomes to begin translating the mRNA at that site. Three “stop” codons (UAA, UAG, UGA), when present within a genetic message, signal the end of the message and termination of the translation in that region. www.sirjamesdeverajr.weebly.com

  36. QUESTIONNO. 33. Which of the following base substitution mutations in the mRNA above would have the least effect on the resulting polypeptides? • Substitution of UCC for UUG in Phe • Substitution of CAA for CCA in Pro • Substitution of CAC for CAG in Gln • Substitution of GGA for GGU in Gly • Substitution of CCA for ACA in the Thr www.sirjamesdeverajr.weebly.com

  37. QUESTION NO. 34 All of the arrows are associated with the process of ________ • carbon fixation • photochemical reactions • anaerobic respiration • aerobic respiration • oxygen fixation www.sirjamesdeverajr.weebly.com

  38. The correct answer is (D) The drawing denotes a mitochondrion. Cellular respiration culminates in oxidative phosphorylation, resulting in the formation of water when oxygen accepts the hydrogen ions and electrons at the end. www.sirjamesdeverajr.weebly.com

  39. End of Day 1 www.sirjamesdeverajr.weebly.com

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