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3-11 AM Thu Sep 20 a access1.lon-capa.uiuc.edu Science Science of Qirat Sun and Mo... for One Day Mens Pants... ton On Men cE 502 Bad Prelecture... 2-25 Flash Sign Up Quiz3 LON-C. Taymur Taymur Hussain Student-section: ALI Main Menu Contents Grades MCB 450 Fall 2018 | Messages Courses Help Logout Course Contents. Homework 3 s (4 pts) Timer NotesEvaluate Feedback Print The peptide DAGADPHIVVLKSSGMTLRCNSTEQMDACDY will yield O3 fragments with chymotrypsin, three with cyanogen bromide, none with trypsin. O One fragment with chymotrypsin, none with trypsin, two with cyanogen bromide. O 2 fragments with chymotrypsin, two with trypsin, one with cyanogen bromide. O No fragments with cyanogen bromide, trypsin, or chymotrypsin. O 3 fragments with trypsin, none with chymotrypsin, three with cyanogen bromide. Submit AnsrAnswer Submitted: Your final submission will be graded after the due date. Tries 1/99 Previous Tries Post Discussion Send Feedback

Explanation / Answer

the given sequence is

DAGADPHIVVLKSSGMTLRCNSTEQMDACDY

Trypsin cleaves after - lys (K), arg (R)

Chymotrypsin cleaves after – phe (F), tyr (Y), trp (W)

Cyanogen bromide cleaves after – met (M)

Therefore, with cyanogen bromide we get three fragments

DAGADPHIVVLKSSGM TLRCNSTEQM DACDY

Then with trypsin we get three fragments

DAGADPHIVVLK SSGMTLR CNSTEQMDACDY

Since there is no F, W or Yin the middle, chymotrypsin will not make any fragments

Therefore the answer is option 5

3 fragments with trypsin, none with chymotrypsin and three with cyanogen bromide.

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