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Specific Reading Assignment. The entire chapter will be covered. Crucial Figures are 3, 5, 6 22b. Important Figures are 4, 7, 8, 9, 10, 11, 12, 13, 15, 17, 18, 20, 24, 25, 28 and all Tables. 

THE NATURE OF DNA (pp. 24-32)

Objectives. To learn: 1) How DNA was found to be the genetic material by a set of experiments; 2) the molecular structure of DNA and its subunits, the nucleotides; 3) how the molecular organization of DNA explains some aspects of its function.

1a) Explain in your own words (you can use diagrams) how it was first demonstrated that DNA is the genetic material (Genetics in Process 2-1).

2a) Diagram the four deoxynucleotides. Where the book indicates P-O- (ionized phosphate, Fig. 2-3), draw instead P-OH (non-ionized); on the phosphate there should be two -OH, and note that the =O can point up, as in the book, sideways, or down.

2b) Draw two deoxynucleotides, one on top of the other (similar to the way they are in the left-hand side of Fig. 2-3) but in such a way that the phosphate group of the bottom deoxynucleotide is between the two sugars and one of the phosphate -OH is pointing up (toward the sugar of the top deoxynucleotide).

Next, imagine a condensation reaction in which a water molecule (H2O) is formed from the phosphate -OH and the H in sugar position 3'; this leaves the phosphate of the bottom deoxynucleotide attached to the 3' O of the sugar. Two deoxynucleotides attached in this way is called a dinucleotide. Draw this dinucleotide.

2c) After studying the structure of DNA, close the book and draw a diagram of a four-nucleotide- pairs long segment of DNA in the style of Fig. 2-5a. Label all its components, the phosphodiester, hydrogen and glycosidic (sugar to base) bonds. The width of the chain is 20 (2nm) and the distance between nucleotides is 3.4. Indicate the polarity (5' and 3' ends) of the two chains. Note that each phosphate group still has a -OH group left; these groups can ionize, releasing a H+, and leaving the DNA negatively charged at physiological pH; this is the reason DNA is called an "acid". Although we draw the molecule flat on a plane, in reality the two nucleotide chains rotate around each other as a tight spiral staircase going into an attic with 10 steps (nucleotides) per turn.

2d) Consider Fig. 2.4a or c, and imagine that there is a break of the phosphodiester bond on one strand between the first and second nucleotide from the top, and also a similar break in the other strand between the last and the second to the last nucleotides. What would happen to the DNA molecule?

2e) Write a very short assay (5-10 lines) on what James Watson and Francis Crick did about DNA. Did they invent it, discover it, describe it?

3a) How did the proposed structure of DNA explain its three basic functions?

Answers to THE NATURE OF DNA

1a) Check whether you made the following points.

The difference between non-pathogenic/non-encapsulated Pneumococcus and pathogenic/encapsulated bacteria is hereditary.

Bacteria are "transformed", converted from one to the other when they are treated with DNA but not other cellular components.

That the active compound is in fact DNA is shown because its transforming capacity is destroyed by the enzyme DNase, which breaks down DNA, but not by enzymes that break down other compunds (polysaccharides, etc.).

2a) See Answer 2b. The diagram shown here is to highlight the three components of the nucleotide and how they are connected to one another. The green outlines show how the condensation reaction takes place to for the ester bond (between the sugar C5’ and the phosphate) and the glycosidic bond (between the sugar C1’ and the base). H directly connected to C are usually left implicit (Each C has enough H to fill its four valences, 1’, 3’ and 4’have one H and 2’and 5’ have 2 H).

2b) When drawing nucleotides, the carbons are usually left implicit at the corners of the cyclic molecules (sugar and bases).

2c) Diagram of a short segment of DNA.

2d) Breaks such as those are called staggered cuts. If the conditions that the DNA is in are mild (neutral pH, not very hot), the DNA molecule will remain together held by the hydrogen bonds that connect the two strands between the cuts. If the temperature or pH are raised, the hydrogen bonds break and the molecule would separate in two. Each half of the molecule would terminate in a sticky end, a single stranded piece of DNA that is complementary to the one in the other molecule. The length of the sticky ends will depend on the distance between the cuts, and the shorter they are the more likely it is that the DNA molecule will separate into two. The figure below illustrates this phenomenon; a few bases are indicated to show the complementarity between the two strands.

2e) Watson and Crick proposed a model describing the three-dimensional organization of the DNA molecule. (The characteristics of the model they proposed are those described by the diagram for question 2c)). They did this based on the experimental evidence available at the time, mainly 1) X-ray diffraction observations that suggested the existence of a helix, and 2) chemical studies that showed that the proportion of adenine in DNA is approximately the same as that of thymine, and the proportion of guanine is similar to that of cytosine.

3a) The proposed structure allowed for three basic functions needed of the genetic material: 1) contain information, 2) contain all the information needed to replicate, and 3) possibility of variation (mutation) allowing for evolution. The first function depends on the total freedom of nucleotide sequence along the chain, the second on the invariant pairing of A:T and G:C, which allows one strand to serve as template for the other, and the third on the possibility for the nucleotide sequence to change.

THE STRUCTURE OF GENES (pp. 32-35)

Objectives. To learn 1) The main parts of a gene and what they do. 2) The differences between prokaryotic and eukaryotic genes. This is a first introduction to these topics, we will see them later in the course in more detail.

4a) Diagram a prokaryotic gene with all the elements in Fig. 2-6, top, but using the following symbols: a pencil line to indicate DNA that is not represented in the final messenger RNA (mRNA) and a long rectangle on its side to indicate DNA that is represented in the mRNA.

4b) The same as 4a, but for a eukaryotic gene.

4c) Table 2-1 indicates that mRNA molecules are 2-3 times longer in higher eukaryotes compared to bacteria, but genes can be 14-17 times longer (you do not need to remember all the values in the Table). How do you explain this discrepancy between the size of genes and the size of mRNAs?

Answers to THE NATURE OF GENES

4a) & 4b)

4c) Introns make eukaryotic genes much longer than prokaryotic genes even when mRNA lengths are comparable.

THE NATURE OF GENOMES (pp. 35-41)

Objectives. To learn: 1) The concept of genome. 2) The general characteristics of different types of genomes in terms of a) their size, b) number and shape of molecules, c) number of copies per cell, and d) number of genes per genome. (Learn numerical values for just a few examples as indicated in the questions below).

5a) What is the range of genome sizes in cellular genomes (Fig. 2-10)? (Do not learn all the specific sizes).

5b) List the various types of genomes.

5c) List the shape (linear or circular), number of molecules, size (in kb), and number of genes of the following genomes: F plasmid of Escherichia coli, herpesvirus, E. coli, and humans (From Table 2-2; inspect the remaining values, but you do not need to remember them).

5d) Before E. coli divides its genome replicates, so that for a short period of time a bacterial cell has two chromosomes. Would you call this cell diplod? Why? or why not?

5e) What are the differences between bacterial and eukaryotic genomes?

Answers to THE NATURE OF GENOMES

5a) 10³ to 10⁸ kilobases, or 1 to 10⁵ megabases.

5b) Plasmid genome, organellar genomes (mitochondrial and chloroplast), viral genome, cellular genome (prokaryotic or eukaryotic nuclear).

5c) F plasmid: C- 1- 100kb- 29 genes; Herpesvirus: L-1- 229kb- 200 genes; E. coli: C- 1- 4,700kb- 4,000 genes; Human (all mammals have approximately the same size genome): L- 23- 3,000,000kb- 100,000 genes.

5d) I would not call it diploid. The two genomes in E. coli originate by replication and are, therefore, identical (except for new mutations). In diploid organisms, on the other hand, the two genomes come one from one parent and one from the other, and differ from one another at many sites (they have different alleles for the same gene).

5e) 1) Bacteria are always haploid, while eukaryotes may be haploid or diploid. 2) The bacterial genome is usually contained in a single molecule or chromosome (There are bacteria with two chromosomes), while eukaryotic genomes are divided into many chromosomes (at least two pairs, but usually more). 3) Bacterial chromosomes are circular molecules of DNA while eukaryotic chromosomes are linear.

Objectives. To learn: 1) What the main features of chromosomes (the subunits of eukaryoric genomes) are. 2) What the main features of polytene chromosomes are. 3) How proteins (histones) and DNA are associated to constitute chromosomes. 4) The concept of chromatin.

6a) What are the two main reasons eukaryotic chromosomes can be studied at the light microscope (and we can speak of ‘visible landmarks’), but prokaryotic chromosomes cannot?.

6b) Compare the size of the E.coli genome with the size of the smallest human chromosome (All the information you need is in this chapter).

6c) What are the main features by which chromosomes in dividing cell (i.e., exclude polytene chromosomes) can be distinguished at the microscope?

6d) How are polytene chromosomes constituted?

7a) Using the spacing between nucleotides from the Watson and Crick model of DNA (.34 nm), estimate the length in milimeters or meters of the E.coli chromosome and of the 23 human chromosomes (as if these were linked end to end in a single molecule).

7b) Compare the amount of DNA in Figs. 2-24 and 2-12.

7c) Describe the nucleosome. What is represented by the black and white balls in Fig. 2-22a?

7d) Study carefully Fig. 2-22 of the nucleosome and the 30 nm solenoid, and be prepared to apply all proper labels to such a figure.

7e) Ignore higher order coiling and the coiling of the red line in Fig. 2-26; let us just say that in most cases it is not entirely clear how the scaffold itself folds. SAR regions in the DNA attach to a scaffolding made of proteins in such a way that the 30 nm solenoid loops out not quite regularly from this scaffold (Figs. 2-25 and 2-26). In other words, ignore point 4 in the message in page 48. Note that the 30 nm solenoid (also called the chromatin fiber) is present both in condensed chromosomes and dispersed chromatin in interphase nuclei. The condensation of chromosomes during cell division is due to the folding of the scaffoldi

Ignore 7f

7f) Pay special attention to "Representation of chromosomes" (p. 47) and Fig. 2-24. We will use primarily the two bottom representations. Notice that although the cells are in interphase when the chromosomes cannot be distinguished because they are so long and thin, for our purposes we represent them as identifiable units. Using the "solid representation", diagram the nucleus of a diplod cell with 4 chromosomes: some of the chromosomes should be long metacentric and some short telocentric. Identify the chromosomes with numbers. Instead of drawing the chromosomes as little sausages, just draw a line, with a little circle to indicate the centromere.

Answers to THE NATURE OF EUKARYOTIC NUCLEAR CHROMOSOMES

6a) The much greater DNA content of human chromosomes and the fact that they "condense" during cell division.

6b) The smallest human chromosome is about 2% of the total genome size (Table 2-4); 2% of 3,000,000 (Table 2-2) kb is 60,000 kb. The E. coli genome is 4,700 kb (Table 2-2). Thus the smallest human chromosome is approximately 12 larger than the bacterial genome.

6c) 1) Size, 2) centromere position or relative arm length (telocentric, acrocentric, metacentric), 3) position of nucleolar organizer, 4) chromomere pattern, 5) heterochromatin patterns, and 6) banding patterns.

6d) Polytene are chromosomes in interphase rather than cell division so the DNA is greatly extended. The chromosomes have replicated many times (There may be up to 2,000 copies of each chromosome) but all the copies remain associated in a bundle; this increases the diameter of the chromosmes sufficiently to make them visible in the microscope. Each chromosome has a particular distribution of chromomeres; because all the strands of a chromosome are in register the chromomeres appear as bands across the diameter of each chromosome. The pattern of distribution of these bands (spacing, thickness, darkness) allows the identification of each chromosome region (Fig. 2-18c).

7a) The E. coli chromosome is 4.7 x 10⁶ np (nucleotide pairs) (Table 2-2). If the length of one np is .34 nm, the length of 4.7 x 10⁶ np would be 0.34 x 4.7 x 10⁶ = 1.6 x 106 nm = 1.6 mm (1 nm = 10-9 m, 1 mm = 1 x 10-3 m).

The human genome, 3 x 10⁹ np (Table 2-2), would measure, 0.34 x 3 x 10⁹ nm = 1 m (approx).

7b) Let’s assume that Fig. 2-24 corresponds to a medium size human chromosome, which would be 2x larger than the smallest human chromosome (Table 2-2). From Question 6b) we can expect the DNA in this chromosome to be 20 times longer than the DNA in E. coli. It may appear that not all the DNA has come out of the cell in Fig. 2-12.

7c) A 10 nm complex of histones (which are positively charged) and DNA (which is negatively charged). Two molecules each of histones H2A, H2b, H3 and H4 form a "core" around which is wound a length of DNA in two turns. In the presence of histone H1 (one molecule per nucleosome), the nucleosomes aggregate into a coil with six nucleosomes per turn (the 30 nm solenoid) (Fig. 2-22).

7f)

Problems 6, 7, 11, 14, 15, 16, 17, 18, 19, 28, 21, 23, 34.