Set 1

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Alanine, an amino acid, has a molar mass of eighty nine grams per mole. Join three alanines into one chain. Three times eighty nine is two hundred and sixty seven, so the chain is two hundred and sixty seven grams per mole, right? Hmm, that is not quite right, because something leaves every time two units join. We settle it in Example two. This is Set one: fifteen mixed questions from the whole Biomolecules unit, in the JEE Main pattern, with twenty minutes suggested for the set. We build three quick tools from its questions, then solve two together. So, let's begin.

0:40Tool 1: sort the sugars

So, tool one: sort the sugars, with question one in the box. How many of these seven are disaccharides? Sort them by what hydrolysis gives. Monosaccharides cannot be hydrolysed to a simpler sugar, and glucose, fructose and ribose are monosaccharides. Okay? Disaccharides give two monosaccharide units on hydrolysis: sucrose, maltose and lactose. All three are C twelve H twenty two O eleven, where the small numbers written low count the atoms. Right? Now, cellulose is also in the list, and it is made of glucose units. So, is cellulose a disaccharide, or something else? Yes, exactly, it is something else, a polysaccharide, because it gives a large number of glucose units on hydrolysis, not two. On screen, Q one stands for question one, and the double arrow means so. The disaccharides are maltose, lactose and sucrose, so the answer is three. With me? Now question five of the set. One mole of lactose is hydrolysed. How many moles of reducing monosaccharide do we get? Lactose gives one galactose and one glucose, and the single arrow means gives. All monosaccharides are reducing sugars, galactose too, so both count, and the answer is two moles. Counting only the glucose is the trap here. Clear?

1:59Tool 2: count the atoms in glucose reactions

Now tool two: the reactions of glucose. Glucose, C six H twelve O six, reacts with HCN, hydrogen cyanide, to give a cyanohydrin. This shows a carbonyl group, C double bond O, where the equals sign is the double bond. Okay? Question two asks for the molecular formula of this cyanohydrin. What do you get? Very good if you simply added the atoms, because one mole of glucose takes up one mole of HCN, and nothing leaves. Seven carbons, thirteen hydrogens, a nitrogen and six oxygens give C seven H thirteen N O six. Right? Question three reduces glucose to sorbitol: the CHO group at the top of the chain becomes a second C H two O H group. So sorbitol is a chain of six carbons, a C H two O H group at each end and four CHOH groups in between, shown by the bracket with a small four. With me? Now, every one of the six carbons carries an O H group. So sorbitol has six asymmetric carbons, right? Oh no, careful, that is the trap, because each end carbon is a C H two O H group, with two hydrogens on the same carbon, so it is not asymmetric. Only carbon two to carbon five are asymmetric, written C dash two to C dash five, so the answer is four.

3:17Tool 3: amino acids and dipeptides

Got it? Tool three: amino acids, with question eight in the box. In water, an amino acid is a zwitterion, a dipolar ion: a raised plus on the N H three group, a raised minus on the COO group, and R, the side chain, with dashes for single bonds. The isoelectric point, p I, is the p H at which it has no net charge. Okay? For a neutral amino acid, like alanine, p I is half of p K a one plus p K a two, where the small one and two count the ionisations. So, what is the p I of alanine? Two point three four plus nine point six nine gives twelve point zero three, and half of that is six point zero one five. The wavy equals sign means approximately, so the answer is six point zero two. With me? Now question nine. Gly, Ala and Val are the codes for glycine, alanine and valine. How many dipeptides with two different amino acids can we make? Three pairs, so three dipeptides, right? No, no, wait, it is not three, because order matters: Gly Ala, with glycine on the left, is a different compound from Ala Gly, and the crossed equals sign means not equal. Three choices for the first unit, two for the second, so three times two gives six dipeptides. Clear?

4:37Example 1: percent adenine in DNA with 28% guanine

Now Example one, question fourteen of this set. Here, a sample of DNA contains twenty eight percent guanine. Find the percentage of adenine. Pause the video and try it first. Done? Okay, let's check. The figure shows the two strands of DNA, whose backbones run in opposite directions, marked five prime and three prime. The dashed lines are hydrogen bonds: adenine faces thymine with two, and guanine faces cytosine with three. So, by Chargaff's rule, A equals T and G equals C, okay? Now, guanine is twenty eight percent. So adenine is twenty eight percent too, right? Be careful here, because guanine pairs with cytosine, not with adenine. So cytosine is also twenty eight percent, and G plus C is fifty six percent. Got it? The rest of the bases are adenine and thymine, so A plus T is one hundred minus fifty six, which gives forty four percent. Right? And A equals T, so adenine is half of forty four, which is twenty two percent. The figure shows why: every adenine faces a thymine. With me?

5:48Example 2: molar mass of the tripeptide Ala-Ala-Ala

And now Example two, question ten, which is our puzzle from the start. Find the molar mass of the tripeptide Ala Ala Ala, three alanine units in one chain. Pause the video and try it first. Done? Okay, let's check. Alanine is C three H seven N O two. Its molar mass, capital M, is thirty six for three carbons, plus seven for the hydrogens, plus fourteen for nitrogen, plus thirty two for two oxygens. That makes eighty nine grams per mole, written g mol to the minus one. Okay? Now look at the figure: glycine and alanine join into Gly Ala, with the N terminal residue on the left. The O H and the H in the box leave as water, minus H two O on the arrow, and the shaded CONH is the peptide bond. So each peptide bond releases one water, eighteen grams per mole. Right? So, three alanines join. How many molecules of water leave? Is it three, one for each alanine? No, only two of them leave, not three, because three units in a line need just two links. So three residues in a chain have two peptide bonds. Clear? So M is three times eighty nine, minus two times eighteen. Two hundred and sixty seven minus thirty six gives two hundred and thirty one grams per mole. And there is our puzzle solved: the missing thirty six left as two molecules of water. Got it?

7:17Quick revision: disaccharides, pI, Chargaff, peptide mass

So, quick revision. First, disaccharides give two units on hydrolysis, and every monosaccharide is a reducing sugar. Second, for a neutral amino acid, p I is half the sum of the two p K a values, and in DNA, A equals T and G equals C, okay? Third, a linear peptide of n residues has molar mass M equal to sigma M, the sum of the molar masses of the amino acids, minus eighteen times n minus one. Now try all fifteen problems of Set one on this page in twenty minutes, then check the solutions. All the best!