Overview

Transcript

0:00Introduction

Hello everyone! Let us warm up with a one-minute puzzle, in the JEE Main style. Glucose reduces Tollens' reagent, so it is a reducing sugar. Fructose is a ketose, yet it reduces Tollens' reagent as well. Now join one glucose unit to one fructose unit, and you get sucrose, our ordinary cane sugar. Two reducing sugars joined together, so sucrose must be a reducing sugar too, right? Hold that thought for a few minutes. This chapter gives you three sets of mixed problems, with questions like this one from every chapter of the unit. So let us see how to use them. Let's begin.

0:41How this chapter works

So, first, what is in this chapter? There are three timed sets, each of fifteen original problems, written in the style of recent JEE Main papers. Okay? Each set mixes all the chapters of the unit, from carbohydrates right up to hormones. So you never know which topic comes next, just as in a real paper. Right? Now, give yourself twenty minutes for each set. So how much time does one problem get? Twenty minutes is twelve hundred seconds, and twelve hundred over fifteen gives eighty. So about eighty seconds a problem, as the brackets say, and only then check the solutions. Clear? And the real past JEE questions on this unit are at the end of Chapters one to five. These sets are original, in the same style. Got it?

1:32Tool 1: which sugars are reducing?

Now the first tool, for the carbohydrate questions. All monosaccharides are reducing sugars, and that includes fructose, even though it is a ketose. With me? On the right are the three common disaccharides from the course, with their units named under each one. In maltose, two alpha glucose units are joined from carbon one to carbon four, written C one to C four. Alpha and beta, the Greek letters in brackets, name the two anomers. Lactose has the same carbon one to carbon four linkage, but beta. The small numbers in the figure mark the carbons joined through the oxygen. Okay? But in both of them, carbon one of the second unit is not used in the linkage. It is still a free hemiacetal, which can open to an aldehyde, so maltose and lactose are reducing sugars. Right? Now, back to our puzzle. In sucrose, the figure shows carbon one of glucose joined to carbon two of fructose, and the label under it already says non-reducing. So two reducing sugars join to give a non-reducing one. Can you tell me why? Here is the reason, and this is exactly the trap in our puzzle. Carbon one carries the reducing group of glucose, and carbon two carries the reducing group of fructose. This linkage uses up both of them, so sucrose is non-reducing, and that settles our opening puzzle. Clear?

3:00Tool 2: peptides, count the bonds

Next, the tool for peptides. The figure on the right shows glycine joining alanine. The boxed O H of glycine and H of alanine leave as water, and the arrow marked minus H two O shows that loss. The shaded CONH group is the peptide bond, and the dashes on screen are its bonds to the rest of the chain. Okay? Now a quick question. Each amino acid unit in a chain is called a residue. So a straight chain of five residues has how many peptide bonds? The answer is four, yes, exactly, one bond between each neighbouring pair. So a linear peptide of n residues has n minus one peptide bonds. But join the two ends into a ring, a cyclic peptide, and there are n bonds. Got it? And this gives the molar mass. For a linear peptide, capital M, its molar mass, equals the sum of the molar masses of the residues, minus eighteen times, in brackets, n minus one. The big Greek sigma means sum, M with a small i is the molar mass of each amino acid, and eighteen is the molar mass of water, lost once for each bond. With me?

4:14Tool 3: proteins, vitamins and DNA

Now proteins. When egg white is boiled, it sets solid, and that is denaturation. So tell me, does boiling break the peptide bonds? No, it does not, and many students slip here. Heat disturbs the hydrogen bonds, so the secondary and tertiary structures are lost; on screen, two and three with a small circle are short for secondary and tertiary. But the primary structure, one with a small circle, which is the sequence of amino acids, stays intact. Okay? Next, vitamins. Only four are fat-soluble, vitamins A, D, E and K, and the water-soluble ones are vitamin C and the B group. So learn the four fat-soluble ones by heart. Right? And nucleic acids. In DNA, adenine pairs with thymine through two hydrogen bonds, written H-bonds on screen, and guanine pairs with cytosine through three. And RNA has uracil, the letter U, in place of thymine. Clear?

5:14Example 1: molar mass of the tripeptide Ala-Ala-Ala

Now Example one, from Set one. Find the molar mass of the tripeptide alanine, alanine, alanine, written with the code ALA three times. The brackets give the molar mass of alanine, eighty nine grams per mole; g mol to the power minus one is read as grams per mole. Pause the video and try it first. Done? Okay, let's check. Three alanine units, so three times eighty nine. Each unit loses a water molecule, so subtract three times eighteen, giving two hundred and thirteen, right? Wait, that is exactly the trap, because three residues in a straight chain have only two peptide bonds. The double arrow means so: only two water molecules are lost, not three. Okay? So capital M equals three times eighty nine, minus two times eighteen. Right? Three times eighty nine is two hundred and sixty seven, and two times eighteen is thirty six. Two hundred and sixty seven minus thirty six gives two hundred and thirty one grams per mole. Count the bonds first, and this trap cannot catch you. Got it?

6:25Example 2: H-bonds between GATTACA and its partner strand

Now Example two, a harder one from Set three. Write the strand complementary to G, A, T, T, A, C, A, and find the number of hydrogen bonds between the two strands. The figure on the right is from the course: each dashed line is one hydrogen bond, and five prime and three prime, the numbers with a small dash, mark the ends of the strands. Pause the video and try it first. Done? Okay, let's check. Pair each base in turn: guanine with cytosine, adenine with thymine, thymine with adenine, and cytosine with guanine. So the arrow takes us to the partner strand, C, T, A, A, T, G, T, as on screen. With me? Now the hydrogen bonds. Seven base pairs, two hydrogen bonds each, so fourteen, right? Oh no, careful, that is the trap, because not every pair has two bonds. The given strand has one G and one C, so there are two G C pairs; the short dash on screen joins the partners. The other five are A T pairs, from its three adenines and two thymines. Okay? Each G C pair has three hydrogen bonds and each A T pair has two, as the figure shows. So two times three plus five times two, that is six plus ten, gives sixteen hydrogen bonds. See how counting the pairs first changed the answer?

7:54Quick revision: sucrose, peptide bonds, DNA base pairs

So, quick revision. Sucrose is joined from carbon one to carbon two, so it is non-reducing, while maltose and lactose reduce. For a linear peptide of n residues, count n minus one bonds, with one water lost for each. In DNA, the A T pairs have two hydrogen bonds and the G C pairs have three, and denaturation keeps the primary structure. Now take Set one, give yourself twenty minutes, and check the solutions only at the end. All the best!