• wonderlic tests
  • EXAM REVIEW
  • NCCCO Examination
  • Summary
  • Class notes
  • QUESTIONS & ANSWERS
  • NCLEX EXAM
  • Exam (elaborations)
  • Study guide
  • Latest nclex materials
  • HESI EXAMS
  • EXAMS AND CERTIFICATIONS
  • HESI ENTRANCE EXAM
  • ATI EXAM
  • NR AND NUR Exams
  • Gizmos
  • PORTAGE LEARNING
  • Ihuman Case Study
  • LETRS
  • NURS EXAM
  • NSG Exam
  • Testbanks
  • Vsim
  • Latest WGU
  • AQA PAPERS AND MARK SCHEME
  • DMV
  • WGU EXAM
  • exam bundles
  • Study Material
  • Study Notes
  • Test Prep

Biochem Module 4 Exam Latest Update -

Exam (elaborations) Dec 14, 2025 ★★★★★ (5.0/5)
Loading...

Loading document viewer...

Page 0 of 0

Document Text

Biochem Module 4 Exam Latest Update - Questions and 100% Verified Correct Answers Guaranteed A+ Verified by Professor

3 homopolysaccharides of glucose - CORRECT ANSWER: starch, glycogen, and

cellulose

A hike is lost in the wilderness and without food. He runs across several beetles with hard chitin exoskeletons. Would he get nutrition in the form of glucose from eating these

beetles? Explain. - CORRECT ANSWER: The exoskeleton of beetles are composed of

chitin. Chitin is made of glucose residues with beta 1-->4 linkage, however, human enzymes cannot breakdown this type of linkage. Human enzymes are able to breakdown alpha 1-->4 linkages only. Therefore, the hiker will not intake glucose or get nutrition from the beetle and should not eat it.

  • How many monosaccharide units are furanoses, and how many are pyranoses?
  • What is the linkage between the monosaccharides?

C) Is this a reducing sugar? - CORRECT ANSWER: A) 0 furanoses and 2 pyranoses

  • beta 1-->4
  • Yes, this is a reducing sugar because a true anomeric carbon is present.
  • Which carbon in the following molecule determines if the molecule is D or L? Explain.

B) Is this molecule D or L? - CORRECT ANSWER: A) Carbon D, the penultimate

carbon, determines if the molecules is D or L. In the case of glucose, this would be carbon-5.

  • The monosaccharide is L because the -OH group is on the left side.

amylopectin has alpha 1-->6 branching about every ___ residues.

  • 1 to 10 1 / 2
  • 8 to 12
  • 10 to 30
  • 24 to 30

e. no branching - CORRECT ANSWER: d. 24 to 30

amylose and amylopectin are both polymers of:

  • alpha-D-glucose
  • beta-D-glucose
  • galactose
  • idose

e. maltose - CORRECT ANSWER: a. alpha-D-glucose

*cellulose and chitin = beta-D-glucose *starch, glycogen, amylose, and amylopectin = alpha-D-glucose

amylose folds into which of the following structures?

  • beta sheet
  • beta turn
  • alpha helix
  • D-configuration

e. alpha form - CORRECT ANSWER: c. alpha helix

based on the form of the cyclic sugar below in a Haworth projection, which Fischer projection formula could have formed this structure? - CORRECT ANSWER: *look at the Haworth projection and create Fischer projections for it

besides C, H, and O, what other element is found in the structure of chitin?

  • / 2

User Reviews

★★★★★ (5.0/5 based on 1 reviews)
Login to Review
S
Student
May 21, 2025
★★★★★

The detailed explanations offered by this document made learning easy. A outstanding purchase!

Download Document

Buy This Document

$1.00 One-time purchase
Buy Now
  • Full access to this document
  • Download anytime
  • No expiration

Document Information

Category: Exam (elaborations)
Added: Dec 14, 2025
Description:

Biochem Module 4 Exam Latest Update - Questions and 100% Verified Correct Answers Guaranteed A+ Verified by Professor 3 homopolysaccharides of glucose - CORRECT ANSWER: starch, glycogen, and cellul...

Unlock Now
$ 1.00