DOC PREVIEW
MIT 7 014 - Solutions to Problem Set 1

This preview shows page 1-2 out of 6 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

MIT Department of Biology 7.014 Introductory Biology, Spring 2004 Solutions to 7.014 Problem Set 1 Question 1 a) Describe the conditions of the atmosphere on prebiotic earth and how these conditions differ from what is found today. Prebiotic earth had an atmosphere that lacked oxygen, but was rich in H2, N2, CO2, H2O, and sulfurous gases. 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 type of 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 prokaryotic organisms may resemble these early cells. However, these modern cells do not freely absorb the organic compounds 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 in atmospheric conditions influenced the course of evolution. At some point, photosynthetic organisms developed the ability to use the highly abundant H2O instead of H2S as a source of electrons. This released oxygen into the seas that reacted with dissolved ions and precipitated as oxides (predominantly iron oxide). As the ions were exhausted, the seas became saturated with O2, and finally, O2 escaped, began oxidizing the iron on land and finally began accumulating in the atmosphere. The change from a reducing environment to a more oxidizing one took hundreds of millions of years after the evolution of oxygenic photosynthesis. This change likely resulted in the extinction of many organisms adapted to the reducing conditions, and created an ozone shield that allowed the introduction of many aerobic life forms. Question 2 You are given four test tubes, each contains cells from a different organism. One tube contains bacterial cells, one contains yeast cells (eukaryotic), one contains human cells and the last contains dinosaur cells. Can you identify the cells from each tube if you are given a light microscope? Explain your answer. 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 would distinguish yeast cells from human and dinosaur cells. Distinguishing human and dinosaur cells with a simple microscope may not be possible.Question 3 Growth factor receptors (shown below) are transmembrane proteins found on the cell surface. H2C O P OO O H2C O P OO O – – – – CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CO O CC O O C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 H H H H 2C O PO O O CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CO O CC O O C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 H H H H 2C O PO O O – – – – extracellular O – – O – O O P O O H H H H2COPOOOH2COPOO O – ––– CHCHCHCHCHCHCHCHCHCHCO OC H CHCHCHCHCHCHCHCHCHCHCO OC H 2222222222H2C O PO O O 2222222222H2C O PO O O – –– – H H O P H H H H O C H C HHHH C C C C C H2C O OO O O H2C C C O O O O C C O O C C O O C O C O CC O O CO CH2 CH2 CH2 CH 2CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH 2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH 2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH 2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH 2 CH2 CH2 CH2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH 2CH2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH 2CH2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH 2CH2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH 2CH2 CH2 CH2 CH2 CH2 CH2 CH2CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH 2CH2 CH2 CH2 CH2 CH2 CH2 CH 2 C CH2 CH2 CH2 CH2 CH2 C O O C C O O C O C O C O OO O O C O O CO O O O O C C H C C C C H2C O H HH H H H2C C CHHC HH – H H H H O – P O O P OO O O – – – intracellular growth factor receptors a) The molecules that form the above membrane belong to what class of macromolecules? ___________________ Phospholipids Explain the important qualities/properties of these molecules that allow them to form membranes. The phospholipid molecules have hydrophilic heads that prefers to interact with aqueous environments and hydrophobic tails that cluster together to exclude water.Question 3, continued A smaller schematic of the growth factor receptor is shown here. ] transmembrane region b) 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 change occurs in the receptor. Growth factor binding causes dimerization of two adjacent receptors in the cell membrane. Upon dimerization, the intracellular domains of the receptors become activated. See schematic below. ligand ligand intracellular domain Receptor 1 Receptor 2 plasma membrane extracellullar ligand-binding domain inactive domains active domainsQuestion 3, continued c) 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. O 2 S 2 + CH C O CH3 S C O S Asp68 OH CH2 CH2 CH2 CH2 Ser53 Cys36 Ala45 -Phe50 Cys82 Cys75 CH2 2 H2N Gln12 CH2 CH2 CH2 CH2 H3N Lys65 H3C H3C Val98 Receptor 2Receptor 1 H CHCHCHInteracting Side chains Type of interaction i) Phe50 : Val98 van der Waals ii) Asp68 : Lys65 ionic iii) Cys75 : Cys82 covalent iv) Ser53 : Gln12 hydrogen d) 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 acids can be found in close proximity to one another.Question 4 You have discovered a new enzyme, enzyme E, which breaks down proteins by cleaving peptide bonds


View Full Document

MIT 7 014 - Solutions to Problem Set 1

Documents in this Course
Ecology

Ecology

21 pages

Quiz 2

Quiz 2

9 pages

Quiz II

Quiz II

13 pages

Quiz II

Quiz II

9 pages

Quiz 1

Quiz 1

9 pages

Quiz 3

Quiz 3

2 pages

Quiz 1

Quiz 1

16 pages

Quiz II

Quiz II

13 pages

Quiz III

Quiz III

10 pages

Quiz III

Quiz III

14 pages

Quiz 2

Quiz 2

14 pages

Quiz 2

Quiz 2

14 pages

S

S

4 pages

Load more
Download Solutions to Problem Set 1
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Solutions to Problem Set 1 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Solutions to Problem Set 1 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?