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Homeostasis : JUNE 2012 / VARIANT 41 / Q10 B

HOMEOSTASIS JUNE 2012 / VARIANT 41 / Q10 B   Describe the mechanisms involved in reabsorption in the proximal convoluted tubule and describe how the epithelial cells of the proximal convoluted tubule are adapted to carry out this process. [9] This questions demands a great detail to be written. A good approach of writing the answer would be: The proximal convoluted tubule is the main site of selective reabsorption in the kidney. It reabsorbs approximately 100 % small proteins, glucose and amino acids. The mechanism of reabsorption of glucose is as follows. The Na+- K+ pump on the basolateral membrane actively pumps 3Na+ outside and 2K+ into the cells against the concentration gradient. This creates a concentration gradient to allow Na+ to move into the cell from the luminal fluid.  Na+ are co-transported into the PCT cells together with glucose via the Na+- glucose co-transporter. This mechanism of glucose reabsorption into the PCT cells is known as secondary active transpo...
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  NOVEMBER 2019 PAPER 4 Q10a Describe the features of ATP that make it suitable for its role as the universal energy currency of cells. (6) For this part of the question you need to list the features of ATP in a comprehensive manner. You should begin writing by saying the ATP as a molecule is small and water soluble and thus have a rapid rate of intracellular diffusion. It serves as the link between the energy yielding and energy- requiring reactions. It possesses energy rich bonds which can be easily hydrolysed to release about 30.5 kJ of energy per mole of bonds hydrolysed. Its rapid turnover allows it to be easily reformed when needed.
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MARCH 2020 P12 15 Which description of cell surface membrane permeability is correct? A An increase in the concentration of cholesterol molecules in the cell surface membrane can increase its permeability to hydrophilic substances.  B Cell surface membrane permeability to large hydrophilic molecules is high and can be increased by membrane transport proteins involved in facilitated diffusion.  C The permeability of the cell surface membrane to ions increases as the proportion of saturated fatty acid chains in the phospholipids increases.  D Without the presence of carrier and channel membrane proteins, the cell surface membrane has a low permeability to large polar molecules.  ANSWER D   In this question, we need to keep in mind that cell surface membrane has a low permeability to hydrophilic substances if transport proteins are not present since the phospholipid bilateral itself is hydrophobic in nature. Therefore, in A an increase in cholesterol molecules...
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MARCH 2020 PAPER 12 : Which cell structures contain RNA?  1 centrioles  2 mitochondria  3 nucleus  4 ribosomes  A 1, 2, 3 and 4  B 1 and 2 only  C 2, 3 and 4 only  D 3 and 4 only ANSWER C The centrioles do not contain RNA. Mitochondria has a prokaryotic background thus it contains features similar to those of prokaryotes. Since prokaryotes contain RNA formed by transcription and translation( because prokaryotes also carry out the processes of protein synthesis), mitochondria which are evolved from prokaryotes, also contain RNA. The nucleus is basically the site of transcription where gene is transcribed to form mRNA. The mRNA is a form of RNA. Therefore, the nucleus contains RNA. Ribosome is basically a complex structure made up of rRNA and proteins. rRNA is a form of RNA. Thus, ribosome is also the correct option available. 5 It is possible for a bacterium to synthesise a eukaryotic protein.  This involves introducing a eukaryotic gene ...
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BIOLOGY MARCH 2020 PAPER 12 Q1: The diagram shows an eyepiece graticule and cell viewed through a microscope. When the eyepiece graticule was calibrated at this magnification, the whole length of the graticule shown covered 12 divisions of a stage micrometer scale. There were 100 divisions in 10 mm of the stage micrometer.  What is the actual length of the cell?  A 2.5µm  B 3.6µm  C 360µm  D 3 mm ANSWER C In this question, you need to make use of the data given in the question. To begin with, find the magnitude of one division of the stage micrometer. If 100 divisions occupy 10mm, then 1 division of the stage micrometer would be ((1*10))/100) = 0.1mm. Since 100 divisions of a graticule covered 12 divisions of a stage micrometer, then 100 divisions would be equal to 12*0.1mm = 1.2mm. Therefore, 1 division of an eyepiece graticule would be equal to 0.012mm. In the picture shown above, the cell covers 29 divisions of a graticule, so the length of the cell is 29*0.0...
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WINTER 2019 VARIANT 12 Q1: A student calibrated the scale on an eyepiece graticule in the eyepiece lens of a light microscope. The student was given a stage micrometer scale to use. The divisions on the stage micrometer scale were 0.1 mm apart. Which data must the student collect in order to calibrate the eyepiece graticule?  1 magnification of the eyepiece lens of the microscope  2 number of divisions of the stage micrometer scale seen in one field of view of the microscope  3 number of divisions of the eyepiece graticule scale equivalent to each division of the stage micrometer      scale  A 1 and 3  B 2 and 3  C 2 only  D 3 only In this question, magnification of the eyepiece lens of the microscope is of no help to the calibration of the eyepiece graticule since it is the objective lens for which the eyepiece graticule has to be calibrated seperately. The eyepiece magnification is usually fixed Also, it is the eyepiece divisions that...
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Hola Folks! This is a platform where you can post your doubts related to your O-level and Alevel past papers. The board is bound to be GCSE or IGCSE.
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