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*** This review should be used as supplement to your studies and is not all-encompassing. ***

PCB 3063 Exam 3 Review

Chapter 25

1. What assumptions must be met for a population to be in Hardy–Weinberg equilibrium? Solution:Large population, random mating, and not affected by migration, selection, or mutation

2. Jean Manning, Charles Kerfoot, and Edward Berger studied genotypic frequencies at the phosphoglucose isomerase (GPI) locus in the cladoceran Bosmina longirostris (a small crustacean known as a water flea). At one location, they collected 176 of the animals from Union Bay in Seattle, Washington, and determined their GPI genotypes by using electrophoresis

Genotype Number

S1S1 4

S1S2 38

S2S2 134

Determine the genotypic and allelic frequencies for this population.

Solution:f(S1S1) = 4/176 = 0.023

f(S1S2) = 38/176 = 0.216

f(S2S2) = 134/176 = 0.761

f(S1) = (8+38)/352 = 0.13

f(S2) = (268 + 38)/352 = 0.87


3. Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some populations along the coast of British Columbia. Kermit Ritland and his colleagues determined that white coat color in these bears results from a recessive mutation (G) caused by a single nucleotide replacement in which guanine substitutes for adenine at the melanocortin-1 receptor locus (mcr1), the same locus responsible for red hair in humans. The wild-type allele at this locus (A) encodes black or brown color. Ritland and his colleagues collected samples from bears on three islands and determined their genotypes at the mcr1 locus.

Genotype Number

AA 42

AG 24

GG 21

a. What are the frequencies of the A and G alleles in these bears?

Solution:f(A) = (2*42 + 24)/174 = 0.62 f(G) = (2*21 + 24)/174 = 0.38

b. Give the genotypic frequencies expected if the population is in Hardy–Weinberg equilibrium.

Solution:Expected genotype frequencies and numbers of each in the population:

f(AA) = (0.62)(0.62) = 0.384; 0.384*87 = 33

f(AG) = 2(0.62)(0.38) = 0.471; 0.471*87 = 41

f(GG) = (0.38)(0.38) = 0.144; 0.144*87 = 13

4. Full color (D) in domestic cats is dominant over dilute color (d).


Of 325 cats observed, 194 have full color and 131 have dilute color.

a. If these cats are in Hardy–Weinberg equilibrium for the dilution locus, what is the frequency of the dilute allele?

Solution: f(dilute color) = f(dd) = q2 = 131/325 = 0.403; q = 0.635

b. How many of the 194 cats with full color are likely to be heterozygous?

Solution:If q = f(d) = 0.635, then p = f(D) = 1 – q = 0.365.

f(Dd) = 2pq = 2(0.365)(0.635) = 0.464; 0.464(325) = 151 heterozygous cats

5. Tay–Sachs disease is an autosomal recessive disorder. Among Ashkenazi Jews, the frequency of Tay–Sachs disease is 1 in 3600. If the Ashkenazi population is mating randomly for the Tay–Sachs gene, what proportion of the population consists of heterozygous carriers of the Tay–Sachs allele?

Solution: If q = the frequency of the Tay–Sachs allele, then q2 = 1/3600; q = 0.017

The frequency of the normal allele = p = 1 – q = 0.983.The frequency of heterozygous carriers = 2pq = 2(0.983)(0.017) = 0.033; approximately 1 in 30 are carriers.

Chapter 10


6. What four general characteristics must the genetic material possess?

Solution:

(1) The genetic material must contain complex information.

(2) The genetic material must replicate or be replicated faithfully.

(3) The genetic material must have the capacity to vary or mutate to generate diversity.

(4) The genetic material must encode the phenotype or have the ability to code for traits.

7. How does an RNA nucleotide differ from a DNA nucleotide? Solution:DNA nucleotides, or deoxyribonucleotides, have a deoxyribose sugar that lacks an oxygen molecule at the 2′ carbon of the sugar molecule. Ribonucleotides, or RNA nucleotides, have a ribose sugar with an oxygen linked to the 2′ carbon of the sugar molecule. Ribonucleotides may contain the nitrogenous base uracil, but not thymine. DNA nucleotides contain thymine, but not uracil.

8. How does a purine differ from a pyrimidine? What purines and pyrimidines are found in DNA and RNA?

Solution:A purine consists of a six-sided ring attached to a five-sided ring. A pyrimidine consists of only a six-sided ring. In both DNA and RNA, the purines found are adenine and guanine. DNA and RNA differ in their pyrimidine content. The pyrimidine cytosine is found in both RNA and DNA. However, DNA contains the pyrimidine thymine, whereas RNA contains the pyrimidine uracil but not


thymine.

9. What different types of chemical bonds are found in DNA and where are they found?

Solution:The deoxyribonucleotides in a single chain or strand of DNA are held by covalent bonds called phosphodiester linkages between the 3′ end of the deoxyribose sugar of a nucleotide and the 5′ end of the deoxyribose sugar of the next nucleotide in the chain. Two chains of deoxyribonucleotides are held together by hydrogen bonds between the complementary nitrogenous bases of the nucleotides in each chain.

10. DNA molecules of different sizes are often separated with the use of a technique called electrophoresis. With this technique, DNA molecules are placed in a gel, an electrical current is applied to the gel, and the DNA molecules migrate toward the positive (+) pole of the current. What aspect of its structure causes a DNA molecule to migrate toward to the positive pole?

Solution:The phosphate backbone of DNA molecules typically carries a negative charge, thus making the DNA molecules attractive to the positive pole of the current.

11. One nucleotide strand of DNA molecule has the base sequence illustrated below. 5′—ATTGCTACGG—3′

Give the base sequence and label the 5′ and 3′ends of the complementary DNA nucleotide strand.

Solution:Answer: 3′—TAACGATGCC—5′

12. If a double-stranded DNA molecule is 15% thymine, what are the percentages of all the other bases?


Solution:The percentage of thymine (15%) should be approximately equal to the percentage of adenine (15%). The remaining percentage of DNA bases will consist of cytosine and guanine bases (100% – 15% – 15% = 70%); these should be in equal amounts (70%/2 = 35%). Therefore, the percentages of each of the other bases if the thymine content is 15% are adenine = 15%; guanine = 35%; and cytosine = 35%.

13. Heinz Shuster collected the following data on the base composition of ribgrass virus (H. Shuster, in The Nucleic Acids: Chemistry and Biology, vol. 3, E. Chargaff and J. N. Davidson, Eds. New York: Academic Press, 1955). On the basis of this information, is the hereditary information of the ribgrass virus RNA or DNA? Is it likely to be single stranded or double stranded?

Percent A G C T U

Ribgrass virus 29.3 25.8 18.0 0.0 27.0

Solution:Most likely, the ribgrass viral genome is a single-stranded RNA. The presence of uracil indicates that the viral genome is RNA. For the molecule to be double- stranded RNA, we would predict equal percentages of adenine and uracil bases and equal percentages of guanine and cytosine bases. Neither the percentages of adenine and uracil bases nor the percentages of guanine and cytosine bases are equal, indicating that the viral genome is likely single stranded.

Chapter 11

14. Describe the composition and structure of the nucleosome.


Solution:The nucleosome core particle contains two molecules each of histones H2A, H2B, H3, and H4, which form a protein core with 145–147 bp of DNA wound around the core. For each nucleosome, one molecule of a fifth histone, H1, binds to DNA entering and exiting the nucleosome core to clamp the DNA around the nucleosome.

15. What is a chromosomal puff? Solution: localized swelling where chromosome is relaxed

16. What are epigenetic changes and how are they brought about?

Solution:Epigenetic changes are changes in gene expression that are passed on to cells or future generations but do not involve alteration of the nucleotide sequence. Epigenetic changes are brought about by altering DNA structure, such as methylation of the DNA, or altering chromatin structure by modifying histones.

17. What is the difference between euchromatin and heterochromatin?

Solution:Euchromatin undergoes regular cycles of condensation during mitosis and decondensation during interphase, whereas heterochromatin remains highly condensed throughout the cell cycle, except transiently during replication. Nearly all transcription takes place in euchromatic regions with little or no transcription within heterochromatin.

Chapter 12


18. List the different proteins and enzymes taking part in bacterial replication. Give the function of each in the replication process.

Solution:

(1) DNA polymerase III is the primary replication polymerase. It elongates a new nucleotide strand from the 3′–OH of the primer.

(2) DNA polymerase I removes the RNA nucleotides of the primers and replaces

them with DNA nucleotides.

(3) DNA ligase connects Okazaki fragments by sealing nicks in the sugar phosphate

backbone.


(4) DNA primase synthesizes the RNA primers that provide the 3′–OH group needed for DNA polymerase III to initiate DNA synthesis.

(5) DNA helicase unwinds the double helix by breaking the hydrogen bonding between the two strands at the replication fork.

(6) DNA gyrase reduces DNA supercoiling and torsional strain that is created ahead of the replication fork by making double-stranded breaks in the DNA and passing another segment of the helix through the break before resealing it. Gyrase is also called topoisomerase II.

(7) Initiator proteins bind to the replication origin and unwind short regions of DNA.

(8) Single-stranded binding protein (SSB protein) stabilizes single-stranded DNA prior to replication by binding to it, thus preventing the DNA from pairing with complementary sequences.

19. Why is DNA gyrase necessary for replication?

Solution:DNA synthesis relies on a single-stranded template; thus, double-stranded DNA molecules must be unwound prior to replication. During DNA unwinding by DNA helicase, tension builds up ahead of the separation (supercoiling). DNA gyrase (also referred to as topoisomerase) reduces supercoiling (relaxes tension) which builds up during DNA unwinding, preventing DNA breakage.

20. What similarities and differences exist in the enzymatic activities of DNA polymerases I and III? What is the function of each DNA polymerase in bacterial cells?

Solution:Each of the five DNA polymerases has a 5′ to 3′ polymerase activity. They differ in their exonuclease activities. DNA polymerase I has a 3′ to 5′ as well as a 5′ to 3′ exonuclease activity. DNA polymerase III has only a 3' to 5' exonuclease activity.

(1) DNA polymerase I carries out proofreading. It also removes and


replaces the RNA primers used to initiate DNA synthesis.

(2) DNA polymerase III is the primary replication enzyme and also has a proofreading function in replication.

21. Why is primase required for replication?

Solution:Primase is a DNA-dependent RNA polymerase. Primase synthesizes the short RNA molecules, or primers, that have a free 3'–OH to which DNA polymerase can attach deoxyribonucleotides in replication initiation. The DNA polymerases require a free 3'–OH to which they add nucleotides, and therefore they cannot themselves initiate replication. Primase does not have this requirement.

22.

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