Concept explainers
The lipid portion of a typical bilayers is about
a. Calculate the minimum number of residues in an
b. Calculate the minimum number of residues in an
c. Explain why
d. The epidermal growth factor receptor has a single transmembrane helix. Find it in this partial sequence:
...RGPKIPSIATGMVGALLLLVVALGIGILFMRRRH…
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- What can you infer about the structure and cellular localization of the protein below based on its hydropathy index? 40 -20 -40 20 40 60 80 100 120 140 160 180 200 220 240 Residue number A. It is a cytosolic protein B. It is a lysosomal enzyme that works in the lumen of the lysosome. O C. It is a cytosolic protein with a single-pass transmembrane domain O D. It is a multi-pass transmembrane protein that will be ultimately found on the cell surface but may also be found In the ER, Golgi, or lysosomal membranes. Hydropathic indexarrow_forwardSome alpha helices might be amphipathic. Where exactly in the protein structure would you find this type of helices. Explain how the position of amino acids in the sequence defines the structure of such helices. a. b.arrow_forwardThe lipid portion of a typical bilayer is about 30 Å thick.(a) Calculate the minimum number of residues in an a-helix required tospan this distance.(b) Calculate the minimum number of residues in a b-strand required tospan this distance.(c) Explain why a-helices are most commonly observed in transmembraneprotein sequences when the distance from one side of a membrane to theother can be spanned by significantly fewer amino acids in a b-strandconformation.(d) The epidermal growth factor receptor has a single transmembrane helix.Find it in this partial sequence: . . . RGPKIPSIATGMVGALLLLVVALGIGILFMRRRH . . .arrow_forward
- Make a table with a scale of absorbance and the concentration of protein in Chromatin sample from the following data for excel graph Absorbance=660nm following data are of tubes with concern A =0 B=0.036 C=0.011 D=0.001 E=0.027 F=0.020 G=0.032 H1=0.176 H2=0.183 I1=0.150 I2=0.171 also plot the graph??arrow_forwardB) We also know from studies of protein structure that one turn of an a-helix includes 3.6 amino acid residues and extends the long axis of the helix by - 0.56 nm. Approximately how many amino acids must a helical transmembrane segment of an integral membrane protein have if the segment is to span the lipid bilayer defined by two stearate molecules laid end to end, plus the polar head groups to which the stearate molecules are attached (assume a polar head group has a diameter of 0.5 nm)?arrow_forwardAn unspecified protein contains a 15-residue long α-helix with the following sequence: WEANIKQRLSTMGEYKQCHANG a. How many full turns are there in this α-helix? b. What is the length of the helix (in Angstroms) in the direction of the helix axis? c. How many hydrogen bonding interactions would be there in this α-helix? d. If it was a 310 helix instead with a helical pitch of 6 Å, is it long enough to span the 30 Å bilayer? Calculate and explain.arrow_forward
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- I'm honestly confused on where to start with this question. I think it has something to do with the residues per turn but can you please explain and show detailed work on the the problem. Researchers have discovered a new membrane protein that has three transmembrane domain αhelixes that span the lipid bilayer. a. If the lipid bilayer is 30 Å thick, what is the smallest number of amino acid residues that can be in each α-helixes for them to be long enough to span the bilayer? b. If one of the three α-helixes is titled at a 32° angle relative to an axis perpendicular to the surface of the lipid bilayer, what is the smaller number of amino acid helixes required for that helix to still span the lipid bilayer? c. How many hydrogen bonding interactions would there be in the smallest possible tilted transmembrane α-helix? d. If the tilted α-helix was a 310 helix instead how many amino acid residues would be required to span the lipid bilayer? How many amino acid residues would be required if…arrow_forwardThe major reason that antiparallel beta-stranded protein structures are more stable than parallel beta-stranded structures is that: Select one: a. parallel ones are in a slightly less extended configuration than antiparallel strands. b. parallel ones do not have as many disulfide crosslinks between adjacent strands. c. parallel ones do not stack in sheets as well as antiparallel strands. X d. parallel ones have fewer hydrogen bonds than antiparallel strands. e. parallel ones have weaker hydrogen bonds between adjacent strands.arrow_forwardWhich of the following statements are correct? explain your answers.A. An individual ribosome can make only one type of protein.B. All mRNAs fold into particular three-dimensional structures that are required for their translation. C. the large and small subunits of an individual ribosome always stay together and never exchange partners.D. ribosomes are cytoplasmic organelles that are encapsulated by a single membrane.E. Because the two strands of DNA are complementary, the mRNA of a given gene can be synthesized using either strand as a template.F. An mRNA may contain the sequence ATTGACCCCGGTCAA.G. the amount of a protein present in a cell depends on its rate of synthesis, its catalytic activity, and its rate of degradation.arrow_forward
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