test 3 hw probs

Which positions in the purine ring of a purine nucleotide in DNA have the potential to form hydrogen bonds but are not involved in Watson-Crick base pairing?

All purine ring nitrogens (N-1, N-3, N-7, and N-9) have the potential to form hydrogen bonds (see Figs 8-1, 8-11, and 2-3). However, N-1 is involved in Watson-Crick hydrogen bonding with a pyrimidine, and N-9 is involved in the N-glycosyl linkage with deoxyribose and has very limited hydrogen-bonding capacity. Thus, N-3 and N-7 are available to form further hydrogen bonds.

When cloning a foreign DNA fragment into a plasmid, it is often useful to insert the fragment at a site that interrupts a selectable marker (such as the tetracycline-resistance gene of pBR322). The loss of function of the interrupted gene can be used to identify clones containing recombinant plasmids with foreign DNA. With a yeast artificial chromosome (YAC) vector, it is not necessary to do this; the researcher can still distinguish vectors that incorporate large foreign DNA fragments from those that do not. How are these recombinant vectors identified?

Bacteriophage λ DNA can be packaged into infectious phage particles only if it is between 40,000 and 53,000 bp long. The two essential pieces of the bacteriophage λ vector have about 30,000 bp in all, so the vector is not packaged into phage particles unless the additional, foreign DNA is of sufficient length: 10,000 to 23,000 bp.

One critical function of chondroitin sulfate is to act as a lubricant in skeletal joints by creating a gel-like medium that is resilient to friction and shock. This function seems to be related to a distinctive property of chondroitin sulfate: the volume occupied by the molecule is much greater in solution than in the dehydrated solid. Why is the volume so much larger in solution?

In solution, the negative charges on chondroitin sulfate repel each other and force the molecule into an extended conformation. The polar molecule also attracts many water molecules (water of hydration), further increasing the molecular volume. In the dehydrated solid, each negative charge is counterbalanced by a counterion, such as Na+, and the molecule collapses into its condensed form.

What property of the waxy cuticles that cover plant leaves makes the cuticles impermeable to water?

Long, saturated acyl chains, nearly solid at air temperature, form a hydrophobic layer in which a polar compound such as H2O cannot dissolve or diffuse.

The manufacture of chocolates containing a liquid center is an interesting application of enzyme engineering. The flavored liquid center consists largely of an aqueous solution of sugars rich in fructose to provide sweetness. The technical dilemma is the following: the chocolate coating must be prepared by pouring hot melted chocolate over a solid (or almost solid) core, yet the final product must have a liquid, fructose-rich center. Suggest a way to solve this problem. (Hint: Sucrose is much less soluble than a mixture of glucose and fructose.)

Prepare the core as a semisolid slurry of sucrose and water. Add a small amount of sucrase (invertase), and quickly coat the semisolid mixture with chocolate. After the chocolate coat has cooled and hardened, the sucrase hydrolyzes enough of the sucrose to form a more liquid center: a mixture of fructose, glucose, and sucrose. .

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RNA duplexes are considerably more stable than DNA duplexes of comparable sequence. Although the physical basis for this thermal stability is not understood, it means that even though the A=U base pairs of double-stranded RNA are weaker than the A=T base pairs of double-stranded DNA, the melting point of the RNA duplex will be higher than that of the DNA duplex.

Smith and Wilcox found that sample A had 136 counts/min of 32P; sample B had 3,740 counts/min. Did the nuclease cleavage leave the phosphate on the 5′ or the 3′ end of the DNA fragments? Explain your reasoning.

The 5′ end. If the phosphate were left on the 3′ end, the kinase would incorporate significant 32P as it added phosphate to the 5′ end; treatment with the phosphatase would have no effect on this. In this case, samples A and B would incorporate significant amounts of 32P. When the phosphate is left on the 5′ end, the kinase does not incorporate any 32P: it cannot add a phosphate if one is already present. Treatment with the phosphatase removes 5′ phosphate, and the kinase then incorporates significant amounts of 32P. Sample A will have little or no 32P, and B will show substantial 32P incorporation—as was observed.

Explain why the absorption of UV light by double-stranded DNA increases (the hyperchromic effect) when the DNA is denatured.

The double-helical structure is stabilized by hydrogen bonding between complementary bases on opposite strands and by base stacking between adjacent bases on the same strand. Base stacking in nucleic acids causes a decrease in the absorption of UV light (relative to the non-stacked structure). On denaturation of DNA, the base stacking is lost and UV absorption increases.

8 2

This sequence has a palindrome, an inverted repeat with twofold symmetry: (5′)GCGCAATATTTCTCAAAATATTGCGC(3′) (3′)CGCGTTATAAAGAGTTTTATAACGCG(5′) Because this sequence is self-complementary, the individual strands have the potential to form hairpin structures. The two strands together may also form a cruciform.

Which bond(s) in α-D-glucose must be broken to change its configuration to β-D-glucose? Which bond(s) to convert D-glucose to D-mannose? Which bond(s) to convert one "chair" form of D-glucose to the other?

To convert α-D-glucose to β-D-glucose, the bond between C-1 and the hydroxyl on C-5 must be broken and reformed in the opposite configuration. To convert D-glucose to D-mannose, either the —H or the —OH on C-2 must be broken and reformed in the opposite configuration. Conversion between chair conformations does not require bond breakage; this is the critical distinction between configuration and conformation.

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Treat a suspension of the bacteria as follows: Add lactose at a concentration well above the Kt, so that virtually every molecule of galactoside transporter binds lactose. Next, add nonradiolabeled NEM and allow it to react with all available —SH groups on the cell surface. Remove excess lactose by centrifuging and resuspending the cells, then add radiolabeled NEM. The only Cys residues now available to react with NEM are those in the transporter protein. Dissolve the membrane proteins in sodium dodecylsulfate (SDS), and separate them on the basis of size by SDS gel electrophoresis. The Mr of the labeled band should represent that of the galactoside transporter.

Explain the difference between a hemiacetal and a glycoside.

hemiacetal is formed when an aldose or ketose condenses with an alcohol; a glyco-side is formed when a hemiacetal condenses with an alcohol

Calculate the weight in grams of a double-helical DNA molecule stretching from Earth to the moon (~320,000 km). The DNA double helix weighs about 1 × 10−18 g per 1,000 nucleotide pairs; each base pair extends 3.4 Å. For an interesting comparison, your body contains about 0.5 g of DNA.

km to anstram (1nm to A) angstram to bp (i bp is 3.4 A) bp to g (1x10^-18 per 1000 bp)