Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle from Set three. Here is a short strand of DNA, with bases G, A, T, T, A, C, A, just seven of them. So how many hydrogen bonds hold it to its partner strand? Is it seven, or fourteen, or something else? Questions that mix the whole unit like this come up often in JEE Main. So, a quick fire round first, and then two full problems.

0:30Quick fire: Tollens' test and epimers

First, a question on Tollens' test: how many moles of silver are deposited per mole of glucose? On the right are Fischer projections of four hexoses: the vertical line is the carbon chain, the D or L in front gives the series, and the plus or minus sign in brackets gives the direction of rotation. At the top of D-glucose sits its only aldehyde group, the CHO group. So one aldehyde group means one mole of silver, right? That is the trap, so be careful here. Tollens' reagent oxidises the CHO group to the carboxylate ion, COO minus; the arrow means is turned into, and the raised minus is the charge. In this change, that carbon goes from oxidation state plus one to plus three. With me? So one glucose molecule loses two electrons, and each silver ion, A g plus, takes only one electron, written e minus. So two moles of silver, A g, are deposited per mole of glucose; mol on the screen is short for mole. Okay? Now compare D-glucose and D-galactose in the figure, row by row. Each horizontal line marks one carbon, with H on one side and O H on the other, and the label C two marks carbon two, just below the CHO group. At which carbon do these two sugars differ? If you said carbon four, that is exactly right: it is the third row, where glucose has its O H on the right and galactose has it on the left. Every other carbon is the same. Clear? And D-mannose, which is not in the figure, differs from D-glucose only at carbon two. Pairs like these, differing only in the configuration at one carbon, are called epimers. Epimers are not enantiomers, so do not call them mirror images. Got it?

2:17Quick fire: invert sugar and an enzyme

Next, the hydrolysis of which common disaccharide changes the sign of the optical rotation of the solution? Maltose hydrolyses to glucose, and its rotation changes from plus one hundred and forty degrees to plus fifty two point five; the small raised circle means degrees. So the answer is maltose, right? Oh no, careful, read the question again, because it asks about the sign. For maltose, the rotation falls, but it stays positive, so there is no change of sign. So maltose is not the answer. With me? Now sucrose, which is dextrorotatory, plus sixty six point five degrees. On hydrolysis it gives D-glucose, plus fifty two point five, and D-fructose, minus ninety two point four; the values in brackets are their rotations. Okay? They form in equal moles, and fructose rotates more strongly. So fifty two point five minus ninety two point four, over two, gives about minus twenty degrees; the wavy equals sign means approximately. The sign has flipped to minus, laevorotatory, so the answer is sucrose, and this mixture is called invert sugar. Clear? Now the last quick question: an enzyme contains zero point three two percent of copper by mass, and the atomic mass of copper is sixty four. What is the minimum molar mass of the enzyme? The key idea is that each enzyme molecule must contain at least one copper atom, C u. So the minimum molar mass, M with a small min written low, is one hundred times m, over p. Here m is the molar mass of copper, and p is its percentage by mass. Right? So one hundred times sixty four, over zero point three two, gives twenty thousand; g mol to the power minus one on the screen means grams per mole. With two copper atoms per molecule it would be double, so twenty thousand is the least value. Got it?

4:09Example 1: mass of glucose from hydrolysis of 100 g starch

Now, Example one, a medium problem from Set three. Find the mass of glucose obtained by the complete hydrolysis of one hundred grams of starch; g stands for grams. Pause the video and try it first. Done? Okay, let's check. Look at the figure: starch is made of chains of glucose units, and each hexagon is one unit. So one hundred grams of starch gives one hundred grams of glucose, right? That is not right, because water joins in. Starch is C six H ten O five in brackets, with a small n outside, for n units in the chain; the small numbers written low count the atoms. On hydrolysis, each unit takes up one water molecule, so n molecules of water, H two O, join in, and n units give n molecules of glucose, C six H twelve O six. With me? Now we need the molar masses. One unit, C six H ten O five, is one hundred and sixty two, and glucose is one hundred and eighty. So every one hundred and sixty two grams of starch gives one hundred and eighty grams of glucose; the extra eighteen grams are the water. Okay? So one hundred times one hundred and eighty, over one hundred and sixty two, gives one hundred and eleven grams of glucose, a little more than the starch we started with. Clear?

5:28Example 2: hydrogen bonds between GATTACA and its complement

And now Example two, also from Set three, and it is our opening puzzle. Write the strand complementary to G, A, T, T, A, C, A, and find the number of hydrogen bonds between the two strands. Pause the video and try it first. Done? Okay, let's check. In the figure, the dashed lines are hydrogen bonds, and each strand runs from five prime to three prime, but the two run in opposite directions; the small dash is read as prime. Pairing is specific: adenine, A, pairs only with thymine, T, and guanine, G, only with cytosine, C, right? So go base by base, in the same left to right order. G gives C, A gives T, T gives A, T gives A, A gives T, C gives G, and A gives T, so the complementary strand is C, T, A, A, T, G, T, written in the same order. With me? Now let us count the hydrogen bonds. There are seven base pairs, with two hydrogen bonds in each, so the answer is fourteen, right? Be careful, because not every pair has two. In the figure, a pair of G and C has three dashed lines, so three hydrogen bonds; the short dash on the screen just joins the pair. Our strand has one G and one C, so there are two such pairs, with three bonds each. Okay? The other five bases, A, T, T, A and A, make five pairs of A and T, with only two hydrogen bonds each. Got it? So two times three, plus five times two, is six plus ten, which gives sixteen hydrogen bonds. Puzzle solved: not seven, not fourteen, but sixteen, because each pair of G and C holds three bonds. Clear?

7:10Quick revision: Tollens, invert sugar, enzyme mass, DNA

So, quick revision. With Tollens' reagent, glucose gives two moles of silver per mole, and epimers differ in configuration at only one carbon. Of sucrose, maltose and lactose, only sucrose inverts, from plus sixty six point five to about minus twenty degrees. Minimum molar mass is one hundred times m, over p, and hydrolysis adds water, so one hundred and sixty two grams of starch become one hundred and eighty. And a pair of G and C holds three hydrogen bonds, a pair of A and T only two, so count each pair. Now try the rest of Set three on this page. All the best!