CARBOHYDRATE BOOKLET ANSWERS

Transcription

1 CARBOHYDRATE BOOKLET ANSWERS FOR: QER QUESTION APPLICATION AND EXTENSION QUESTIONS. FEEDBACK CAN BE VIEWED AT:

2 LONG ANSWER QUESTIONS WITH QER The long answer question requires you to write at length about a topic. The quality of your extended responses (QER) will also be assessed. The mark scheme will have the following areas: 1. Indicative Content this is the biological concepts and terminology you are expected to use in your answer marks to gain these top marks your answer will show: The candidate constructs an articulate, integrated account correctly linking relevant points, such as those in the indicative content, which shows sequential reasoning. The answer fully address the question with no irrelevant inclusions or significant omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately marks to gain these marks your answer will show: The candidate constructs an account correctly linking some relevant points, such as those in the indicative content showing some reasoning. The answer addresses the question with some omissions. The candidate uses scientific conventions and vocabulary appropriately and accurately marks to gain these marks your answer will show: The candidate makes some relevant points, such as those in the indicative content, showing limited reasoning. The answer addresses the question with significant omissions. The candidate has limited use of scientific conventions and vocabulary.

3 Complete the QER questions number 4 below. 4. Carbohydrates are a large group of biological molecules with diverse structures and functions. Plants and animals can use different carbohydrates to perform a common function. In addition to this, plants and animals can use carbohydrates that are unique to them and for unique functions. The microscope images below show carbohydrates that fit the description stated in this question. Image A Image B State the names of the carbohydrates shown in image A and Image B. Describe the structure of these carbohydrates and explain how this structure permits their function. [9] Image A is a plant cell and contains starch grains and cellulose cell wall. Image B is of an animal cell which contains glycogen granules. Starch consists of two polysaccharides called amylose and amylopectin which are both polymers of α-glucose. Amylose is a coiled structure with only α-1,4 glycosidic bonds but amylopectin is branched and has both α-1,4 and α-1,6 glycosidic bonds. Starch is a storage of energy in plant cells. It can perform this function as it is insoluble and so cannot alter the water potential of the cell, is compact so large amount of starch can be stored as starch grains in the cell. The starch grains also prevent the starch form leaving the cell. In animal cells glycogen acts as the storage of energy. It s structure is identical to amylopectin in being a polymer of α-glucose and having both α-1,4 and α-1,6 glycosidic bonds.

4 However, glycogen is highly branched unlike amylopectin. Glycogen is a good storage polysaccharide for the same reasons as those for starch stated above, but glycogen forms glycogen granules. Cellulose is a structural polysaccharides and forms part of the cell wall of plants. It is a polymer of β-glucose which form long straight chains joined by β-1,4 glycosidic bonds. The β-glucose is rotated by 180 o so causing the glycosidic bonds to alternate up and down. Hydrogen bonds can form between the chains of β-glucose to form microfibrils. Microfibrils join together to form macrofibrils. The large number of hydrogen bonds in cellulose gives it a high tensile strength and makes it ridged and inelastic. These properties provide strength and support to plant cells. The inelastic nature of cellulose also causes plant cells to become turgid as the cytoplasm swells with water and pushes against the cell wall.

5 APPLICATION AND EXTENSION 1. Rare sugar syrup containing D -Allulose but not high-fructose corn syrup maintains glucose tolerance and insulin sensitivity partly via hepatic glucokinase translocation in Wistar rats. The above title is from a scientific paper published in the journal of agricultural and food chemistry (2017, 64, , Shintani et al). Below are extracts taken from this paper (some slightly modified). Using information in these extracts and your own knowledge answer the questions that follow. High-fructose corn syrup (HFCS) is commercially produced by isomerizing glucose to fructose. The taste of HFCS is similar to that of sucrose, and its price is affordable. Thus, HFCS has been used in various foods, such as beverages, desserts and confectionary and bakery products. However, the consumption of HFCS has been reported to be one of the risk factors for developing diabetes and obesity. From a preventative point of view it is advisable to reduce sugar intake or replace sugar with other low-calorie sweeteners. Thus, the development of new types of sweeteners is awaiting. D-Allulose is a C 3 epimer of D -fructose and is one of the rare sugars that are present in a limited quantity in nature. Studies have shown that D -allulose has various functions, such as reducing postprandial glucose elevation this has been confirmed in a number of studies and there have been two mechanisms proposed for the glucose lowering effect of D -allulose. These are: 1. The inhibition of α-glucosidase resulting in the decreased absorption of glucose. 2. The effect on the glycaemic response in the liver. D -Allulose promotes the conversion of blood glucose to glycogen in the liver via the use of

6 glucokinase. When glucose enters the hepatocytes, it is phosphorylated to glucose-phosphate by glucokinase which is a key step to forming glycogen. Thus, glucokinase increases the amount of glycogen in the liver, suppressing postprandial blood glucose elevation. D -Allulose has the effect of activating the glucokinase, causing its transport out of the nucleus and into the cytoplasm where it will then increase the amount of glycogen and thus lower the blood glucose level. Rare sugar syrup (RSS) is a mixture of 45% glucose, 29% fructose, 5% D - allulose and a number of other carbohydrates. Therefore, RSS containing D-allulose may activate glucokinase and lead to improved glycaemic control. The anti-obesity effect of RSS has been confirmed in both humans and animals in a clinical trial the results of which showed that the intake of 40g of RSS for 3 months lead to a 1.8Kg reduction in body weight compared to that observed after HFCS intake. It is of interest to assess whether RSS, a new sweetener derived from HFCS, maintains glucose tolerance and insulin sensitivity and, if so, whether anti-hyperglycaemic action of RSS is mediated by exporting glucokinase from the nucleus to the cytoplasm. To this end, glucose tolerance and the degree of glucokinase translocation in rats administered water, HFCS or RSS in drinking water were investigated. (a). State three controlled variables for this investigation. Same gender of Wistar rat, same age of Wistar rat, same concentration of RSS.

7 (b). Which of the following molecules are mentioned in the passage? A B C D E F A, B,C,D, E (c). What are the ethical issues of this investigation. The investigation uses rats. Rats may have been harmed in the investigation, e.g. been in pain. Results would not be usefull as the effects may be different in humans. (d). Explain why the statement glucokinase increases the amount of glycogen in the liver is scientifically inaccurate. Amount is the term meaning moles. The term concentration should have been used.

8 (e). Below is some selective data from this investigation (the data has been significantly modified for the purposes of this question). Figure 1. Effects of water, HFCS and RSS on glucose levels expressed as a percentage ± range. Figure 2. Effects of water, HFCS and RSS on Plasma glucose levels ± range.

9 Figure 3. Effects of water, HFCS and RSS on Plasma insulin levels ± range. Figure 4. Effects of water, HFCS and RSS on cytoplasmic glucokinase levels ± range. (a) Using figure 1 describe the trend for the effect of HFCS on glucose levels. At time 0 minutes the glucose concentration was 100%. The percentage of glucose gradually reduced to approximately 90% at 60 minutes. After 60 minutes the glucose percentage raised to approximately 80% at 180 minutes.

10 (ii) Explain which set of data: water, RSS or HFCS from figure 1 has the greatest consistency. RSS shows the greatest consistency because the range bars are very short and do not overlap. This means that the replicates results were close together. Due to this we have confidence in the trend and confidence that there is a true difference in the mean values. (iii) Insulin sensitivity is a measure of how sensitive the body is to the effects of insulin. Glucose tolerance is a measure of glucose levels in the blood. Using both figure 2 and figure 3 explain whether water, HFCS or RSS shows the greatest glucose tolerance and greatest insulin sensitivity. The blood glucose levels for RSS treatment are the lowest compared to water and HFCS in figure 2. In figure 3 RSS treatment caused the lowest levels of blood insulin. This lowest level of insulin caused the greatest reduction in blood glucose with RSS treatment indicating high insulin sensitivity and the greatest glucose tolerance was caused by RSS. (iv) It has been suggested that the polymerisation of glucose to glycogen is achieved by the action of glucokinase in the cytoplasm of hepatic cells and that the glucokinase levels in the cytoplasm are greatest with d -allulose present. Does the data presented in this question fully confirm this suggestion? Explain your answer. The data in figure 3 shows that the greatest percentage of cytoplasmic glucokinase is achieved by RSS which contains d -allulose. However, there is no data to support that its glucokinase that converts glucose to glycogen.

11 (v) Compare the structure of glycogen and starch. Starch is a polymer of α-glucose and so is glycogen. Glycogen is branched and so is amylopectin. Starch is made up of two different polysaccharides, but glycogen is made up of just one polysaccharide. (vi) Does the data disprove the prediction stated at the start of this question? Explain your answer. No, it does not disprove it. The data proves the prediction in that RSS with D-Allulose maintains glucose tolerance and insulin sensitivity due to the data in figure 2 and 3 as well as the data in figure 4 that shows the highest cytoplasmic kinase concentration.

12 Q2. Below is the structure of a carbohydrate called raffinose. Raffinose can be found in beans, cabbage, Brussel sprouts and other vegetables. Study this structure carefully and answer the following questions. (a) (i) What type of carbohydrate is raffinose? Trisaccharide (ii) State the names of the two bonds in raffinose including carbon numbers. α-1,6 and α 1,6 glycosidic bonds. (b) The enzyme α-galactosidase can breakdown raffinose but is not found in the human digestive tract. (i) What is the name of the reaction catalysed by α-galactosidase? Hydrolysis (ii) Place an arrow on the diagram of raffinose to show the site of action of α-galactosidase and state the names of the products of this reaction. Sucrose and galactose.

13 (c) The molar mass for raffinose is g/mol. You need to make up 5 standard solutions of concentration: 0.60 moldm -3, 0.45 moldm -3, 0.25 moldm -3, 0.15 moldm -3, 0.08 moldm -3. (i) Describe and calculate how you would make up the above 5 standard solutions. Each standard solution must be in a total volume of 100cm g/mol x 0.6 = g dissolved in 1 dm 3 = 0.6moldm -3. This is the stock solution that will be used to make the other concentrations. Its volume is 1dm 3 As the question asked for a total volume of 100cm 3 the V1xC1 = V2xC2 equation can be used to calculate the volumes needed to do the dilution. V1 and C1 are the concentration and volume of your stock solution and V2 and C2 are the concentration and volume of the solution that you require. Worked example for 0.45 moldm -3 So V1 = this is what we need to calculate, C1 = 0.6moldm -3 V2 = 100cm 3, C2 = 0.45moldm -3 We need to make V1 the subject of the equation: V1 = V2 C2 C1 V1 = 100cm3 0.45moldm 3 0.6moldm 3 = 75cm 3 So, we need 7.5cm 3 of the 0.6moldm -3 stock solution and add this to a volume of water to make a total volume of 100cm 3. The volume of water is calculated by 100cm 3 75cm 3 = 25cm 3.

14 The above stages are done for the other 3 standard solutions. The answers are: 0.25 moldm -3 V1 = 41.6cm 3, volume of water = 58.3cm moldm -3, V1 = 25cm 3, volume of water = 75cm moldm -3 V1 = 13.3cm 3, volume of water = 86.7cm 3 (ii) Explain why you would need to make up standard solutions. To calibrate an instrument and plot a calibration curve.