Question 1 3 points Fall 2014 7 012 Problem Set 2 20 Points total DUE WEDNESDAY 9 24 2014 AT 9 45AM ONLINE C6H12O6 6H2OBBBBBBB 6CO2 6H2O energy Name Overall the process of cellular respiration can be summarized in the equation below This process does not occur in one step but in many steps These steps can be grouped into the following metabolic pathways Glycolysis Pyruvate oxidation Citric acid cycle Electron transport chain a For each of these pathways briefly summarize the process by writing the primary reactants and products You can either use names or molecular compositions for these summaries Make sure to include the type and number of the energy carrying molecules as well as electron carriers produced in each pathway Glycolysis occurs in the cytoplasm INPUTS OUTPUTS Pyruvate oxidation occurs in the mitochondria INPUTS OUTPUTS Citric acid cycle occurs in the mitochondria INPUTS OUTPUTS Electron transportation occurs in the mitochondria INPUTS OUTPUTS b Which of the pathways would not occur in the absence of oxygen Does the absence of oxygen affect these processes directly or indirectly Explain You are working with a version of yeast which has a mitochondrial ion channel that allows for the H ions accumulated in the intermembrane space to freely flow into the matrix 1 Pyruvate oxidation is indirectly affected because pyruvate requires NAD which cannot be produced in the absence of oxygen because the electro transport chain cannot work without oxygen full explained below Citric acid cycle is indirectly affected because NADH is an electron receiver so if the electron transport chain cannot break down NADH because it does not have oxygen as stated below With a lack of oxygen there will be a subsequent lack of NADH so the citric acid cycle cannot run Electron Transportation is directly affected because oxygen is the proton accepter at the end of the electron transport chain thus without oxygen this step could not occur Question 2 3 5 points Without oxygen pyruvate oxidation citric acid cycle and electron transportation will not occur Pyruvate will be directed to the fermentation process in this mutant yeast To get normal yeast to behave like the mutant yeast in part c it would have to be placed in anaerobic conditions The amino acids would be hydrophobic non polar side chains are within the hydrophobic inner layers of the plasma membrane Anaerobic conditions same amounts of energy would be produced by the normal yeast and the mutant yeast Aerobic the mutant yeast will not be able to complete the electron transport chain The mutant yeast would produce less energy than the normal yeast Name a In general what can you say about the amino acid composition of the transmembrane domain of membrane spanning proteins b You grow and divide these variant yeast cells under both aerobic and anaerobic conditions and compare the energy produced to normal yeast that grow and divide under these same conditions Explain what you would expect to see c Suppose you identify another mutant yeast strain which cannot carry out pyruvate oxidation What process will pyruvate be directed to in this mutant yeast d How could you get normal yeast to behave like the mutant yeast in part c Explain Go to the MITx edX site https lms mitx mit edu courses MITx 7 012 2014 Fall about and log into your 7 012X account Navigate to Courseware Within week 2 you will find Problem Set 2 You must complete the problem titled Enzymes here Part a b c Your TA can check that you completed this through the 7 012x site so you do not need to submit any work here You are working in a lab with fruit flies and you cross two trueBbreeding fliesBone male with long wings and yellow colored body Parent 1 and one female with short wings and tan colored body Parent 2 You may assume that body color and wing size are regulated by two different genes a What does it mean for an organism to be trueBbreeding b As a result of the cross described in the question prompt all of your progeny F1 have tan bodies and long wings Give the genotype and phenotype of both parents and the progeny Use the letter A for the wing trait and the letter D for body color Use a capital letter to indicate the allele associated with a dominant phenotype and a lower case letter to indicate the allele associated with a recessive phenotype 2 Parent 1 AAdd Long wings yellow body Parent 2 aaDD Short wings tan body Offspring AaDd Long wings tan body True breeding means that the offspring will have the same phenotypic trait as their parents Question 4 5 points Question 3 1 5 points This is a test cross Fly phenotype Fly phenotype 250 250 250 250 Number of flies Possible gametes for parent 1 AD Ad aD ad Possible gametes for parent 2 ad Below are the numbers of the flies you actually observe Name c You cross a fly from the F1 progeny to a yellow fly with short wings You observe 1000 flies in the F2 Write all the possible gametes for both flies in this cross d What is another term used to describe the cross in part c e Write the number of flies that you would expect to see for each of the following fly categories resulting from the cross in part c assuming that the genes of interest are not linked Tan body long wings Tan body short wings Yellow body long wings Yellow body short wings Tan body long wings Tan body short wings Yellow body long wings Yellow body short wings f What does this information tell you g If the genes are linked calculate the distance between these two genes If they are not linked say they are unlinked You are studying two traits in a hypothetical plant petal color and leaf shape You know that purple color is dominant to pink color and that long leaves are dominant to wide leaves You also know that the genes regulating these traits are linked and located 10 centimorgans apart You cross a plant with purple petals and long leaves P1 to a pink plant with wide petals P2 As a result of your cross you get a mixture of the following types of plants Purple petals wide leaves Pink petals wide leaves Purple petals long leaves Pink petals long leaves They are linked Distance between these two genes 90 150 1000 0 24 cm Question 5 7 points This shows us that the two genes are linked Number of flies 90 400 360 150 3 plants plants plants plants P1 AaBb P2 aabb Plant phenotype Possible Outcome 2 Number of Possible Outcome 4 Number of Possible Outcome 1 Number of Possible Outcome 3 Number of 190 60 40 210 70 170 180 80 130 110 120 140 220 15 35 230 Name a What are the genotypes
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