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MIT 7 014 - Solutions to 7.014 Problem Set 1

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Solutions to 7.014 Problem Set 1Question 1a) Describe the conditions of the atmosphere on prebiotic earth and how these conditionsdiffer from what is found today.Prebiotic earth had an atmosphere that lacked oxygen, but was rich in H2, N2, CO2, H2O, and sulfurousgases. In the absence of O2 and O3, the UV light penetrated in far greater strength than it does now.Today, we have an atmosphere rich in oxygen and an ozone that protects us from the sun’s UV light.b) Describe the proposed first organism as it compares to modern organisms; include the typeof genetic material and the source of energy it may have used.The first cells were anaerobic and likely absorbed free organic compounds from the primordial seas.They are thought to have used RNA as their genetic material. Some modern anaerobic prokaryoticorganisms may resemble these early cells. However, these modern cells do not freely absorb the organiccompounds needed, they make them. They also use DNA as their genetic material.c) Describe the process that led to an O2 atmosphere on earth and how this change inatmospheric conditions influenced the course of evolution.At some point, photosynthetic organisms developed the ability to use the highly abundant H2O insteadof H2S as a source of electrons. This released oxygen into the seas that reacted with dissolved ions andprecipitated as oxides (predominantly iron oxide). As the ions were exhausted, the seas becamesaturated with O2, and finally, O2 escaped, began oxidizing the iron on land and finally beganaccumulating in the atmosphere. The change from a reducing environment to a more oxidizing one tookhundreds of millions of years after the evolution of oxygenic photosynthesis. This change likely resultedin the extinction of many organisms adapted to the reducing conditions, and created an ozone shield thatallowed the introduction of many aerobic life forms.Question 2You are given four test tubes, each contains cells from a different organism. One tube containsbacterial cells, one contains yeast cells (eukaryotic), one contains human cells and the lastcontains dinosaur cells. Can you identify the cells from each tube if you are given a light microscope? Explain youranswer.With a light microscope you could easily distinguish the prokaryotic bacteria from the other cell types.The prokaryotic bacteria would not have a nucleus, the other cell types would. The yeast cell wall woulddistinguish yeast cells from human and dinosaur cells. Distinguishing human and dinosaur cells with asimple microscope may not be possible.Question 3Growth factor receptors (shown below) are transmembrane proteins found on the cell surface.CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2C OOC COOCCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HHHH2COPOOOCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2C OOC COOCCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HHHH2COPOOOCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOCCOO CCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HH HH2COPOOOCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOCCOO CCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HH HH2COPOOO––––––––CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2C OOC COOCCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HHHH2COPOOOCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2C OOC COOCCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HHHH2COPOOO––––CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2C OOC COOCCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HHHH2COPOOOCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2C OOC COOCCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HHHH2COPOOOCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOCCOO CCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HH HH2COPOOOCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2COOCCOO CCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HH HH2COPOOO––––––––CHCHCHCHCHCHCHCHCHCHCOOCCOO CCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HH HCHCHCHCHCHCHCHCHCHCHCOOCCOO CCH2CH2CH2CH2CH2CH2CH2CH2CH2CH2HH H2222222222H2COPOOO2222222222H2COPOOO––––intracellularextracellulargrowth factor receptorsa) The molecules that form the above membrane belong to what class of macromolecules?___________________ PhospholipidsExplain the important qualities/properties of these molecules that allow them to formmembranes.The phospholipid molecules have hydrophilic heads that prefers to interact with aqueous environmentsand hydrophobic tails that cluster together to exclude water.Question 3, continuedA smaller schematic of the growth factor receptor is shown here.] transmembrane regionb) Which amino acids would you expect to find in the transmembrane region of the receptor?Alanine, cysteine, glycine, isoleucine, leucine, methionine, phenylalanine, proline, tryptophan, tyrosine,valine.When growth factor binds to the extracellular domain of the receptor, a conformational changeoccurs in the receptor. Growth factor binding causes dimerization of two adjacent receptors inthe cell membrane. Upon dimerization, the intracellular domains of the receptors becomeactivated. See schematic below.active domainsintracellular domainligand Receptor 1 Receptor 2inactive domains plasma membraneextracellullar ligand- binding domainligandQuestion 3, continuedc) Regions of the two receptors that interact upon dimerization shown below. In parts (i - iv)below, name the strongest type of interaction (choose from; hydrogen bond, ionic, covalent,van der Waals) that occurs between the side chains of the amino acids indicated.OCH2SCH2+CHCOCH3SCOSAsp68OHCH2CH2CH2CH2Ser53Cys36Ala45-Phe50Cys82Cys75CH2CH2H2NGln12CH2CH2CH2CH2H3NLys65H3CH3CVal98Receptor 2Receptor 1HInteracting Side chains Type of interactioni) Phe50 : Val98 van der Waalsii) Asp68 : Lys65 ioniciii) Cys75 : Cys82 covalentiv) Ser53 : Gln12 hydrogend) Explain how Gln12 and Val98, which are far apart in the primary sequence of the protein,can be close to each other in the region of the protein diagrammed above.When the linear chain of amino acids folds into its three dimensional shape, non-adjacent amino acidscan be found in close proximity to one another.Question 4You have discovered a new enzyme, enzyme E, which breaks down proteins by cleavingpeptide bonds after tyrosine or phenylalanine.a) Enzyme E is the product of gene G that encodes a protein with the molecular weight of 50kilodaltons (50 kD). When purify enzyme E, only the expected polypeptide is present.However, active enzyme E has a molecular weight of 250 kilodaltons (250 kD), not 50 kD.i) Why might active purified enzyme E be larger than the product encoded by gene G?The active enzyme must have multiple subunits. Likely it is a pentamer of the 50 kD polypeptideencoded by gene G.ii) Define primary, tertiary, and quaternary structure.The primary structure is the linear sequence of amino


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MIT 7 014 - Solutions to 7.014 Problem Set 1

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