
Concepts of Biology
1st Edition
ISBN: 9781938168116
Author: Samantha Fowler, Rebecca Roush, James Wise
Publisher: OpenStax College
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Chapter 17, Problem 20CTQ
How do B and T cells differ with respect to antigens that they bind?
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Chapter 17 Solutions
Concepts of Biology
Ch. 17 - Figure 17.5 Which of the following statements...Ch. 17 - Figure 17.6 Influenza virus is packaged in a viral...Ch. 17 - Figure 17.17 The Rh antigen is found on...Ch. 17 - Which statement is true? a. A virion contains DNA...Ch. 17 - The viral ________ plays a role in attaching a...Ch. 17 - Which statement is true of viral replication? a....Ch. 17 - Which of the following is a barrier against...Ch. 17 - Although interferons have several effects, they...Ch. 17 - Which innate immune system component uses MHC...Ch. 17 - The humoral immune response depends on which...
Ch. 17 - The fact that the body does not normally mount an...Ch. 17 - Foreign particles circulating in the blood are...Ch. 17 - Allergy to pollen is classified as _______. a. an...Ch. 17 - A potential cause of acquired autoimmunity is...Ch. 17 - Autoantibodies are probably involved in ________....Ch. 17 - Why can’t dogs catch the measles?Ch. 17 - Why is immunization after being bitten by a rabid...Ch. 17 - Different MHC class I molecules between donor and...Ch. 17 - If a series of genetic mutations prevented some,...Ch. 17 - How do B and T cells differ with respect to...Ch. 17 - Why is the immune response after reinfection much...Ch. 17 - Some photographers develop a sensitivity to...
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- From your previous experiment, you found that this enhancer activates stripe 2 of eve expression. When you sequence this enhancer you find several binding sites for the gap gene, Giant. To test how Giant interacts with eve, you decide to remove all of the Giant binding sites from the eve enhancer. What results do you expect to see with respect to eve expression?arrow_forwardWhat experiment could you do to see if the maternal gene, bicoid, is sufficient to form anterior fates?arrow_forwardYou’re curious about the effect that gap genes have on the pair-rule gene, evenskipped (eve), so you isolate and sequence each of the eve enhancers. You’re particularly interested in one of the enhancers, which is just upstream of the eve gene. Describe an experimental technique you would use to find out where this particular eve enhancer is active.arrow_forward
- For short answer questions, write your answers on the line provided. To the right is the mRNA codon table to use as needed throughout the exam. First letter U บบบ U CA UUCPhe UUA UCU Phe UCC UUG Leu CUU UAU. G U UAC TV UGCys UAA Stop UGA Stop A UAG Stop UGG Trp Ser UCA UCG CCU] 0 CUC CUA CCC CAC CAU His CGU CGC Leu Pro CCA CAA Gin CGA Arg CUG CCG CAG CGG AUU ACU AAU T AUC lle A 1 ACC Thr AUA ACA AUG Mot ACG AGG Arg GUU GCU GUC GCC G Val Ala GAC Asp GGU GGC GUA GUG GCA GCG GAA GGA Gly Glu GAGJ GGG AACASH AGU Ser AAA1 AAG Lys GAU AGA CAL CALUCAO CAO G Third letter 1. (+7) Use the table below to answer the questions; use the codon table above to assist you. The promoter sequence of DNA is on the LEFT. You do not need to fill in the entire table. Assume we are in the middle of a gene sequence (no need to find a start codon). DNA 1 DNA 2 mRNA tRNA Polypeptide C Val G C. T A C a. On which strand of DNA is the template strand (DNA 1 or 2)?_ b. On which side of the mRNA is the 5' end (left or…arrow_forward3. (6 pts) Fill in the boxes according to the directions on the right. Structure R-C R-COOH OH R-OH i R-CO-R' R R-PO4 R-CH3 C. 0 R' R-O-P-OH 1 OH H R-C-H R-N' I- H H R-NH₂ \H Name Propertiesarrow_forward4. (6 pts) Use the molecule below to answer these questions and identify the side chains and ends. Please use tidy boxes to indicate parts and write the letter labels within that box. a. How many monomer subunits are shown? b. Box a Polar but non-ionizable side chain and label P c. Box a Basic Polar side chain and label BP d. Box the carboxyl group at the end of the polypeptide and label with letter C (C-terminus) H H OHHO H H 0 HHO H-N-CC-N-C-C N-C-C-N-GC-OH I H-C-H CH2 CH2 CH2 H3C-C+H CH2 CH2 OH CH CH₂ C=O OH CH2 NH2arrow_forward
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