TLS Online TPP Program

#Question id: 24563


Which of the following is NOT true for classical pathway of  complement activation?

#Unit 4. Cell Communication and Cell Signaling
  1. For activation of the classical pathway by binding Ig Classical pathway Binding to apoptotic blebs and initiating phagocytosis of apoptotic cells
  2. Classical pathway creates C3 and C5 convertase for hydrolyzing the molecules
  3. Classical pathway includes serine protease, cleaving C1r and C1s
  4. Classical pathway does not require antigen-antibody interactions
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TLS Online TPP Program

#Question id: 15654

#Unit 13. Methods in Biology

You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium. Next you isolate a mutation, designated MelB–, which gives uninducible melibiase activity. Mapping experiments show that MelB– is linked to Mel1. Using an F' factor that carries the chromosomal region surrounding Mel1, you perform the following genetic tests:

                      
Describe the basic genetic properties of the MelB– mutation?

TLS Online TPP Program

#Question id: 15655

#Unit 13. Methods in Biology

You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium. Next you isolate a mutation, designated MelB–, which gives uninducible melibiase activity. Mapping experiments show that MelB– is linked to Mel1. Using an F' factor that carries the chromosomal region surrounding Mel1, you perform the following genetic tests:

 
Describe the proposal for the type of regulatory functions affected by the MelB– mutation.

TLS Online TPP Program

#Question id: 15656

#Unit 13. Methods in Biology

You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium. Explain the two regulatory pathways for Mel1 that can explain the behavior of the MelA– and MelB– mutations. For each regulatory pathways what is a role for the inducer melibiose?

TLS Online TPP Program

#Question id: 15657

#Unit 13. Methods in Biology

You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium.
You next construct a MelA– MelB– double mutant, which gives the following behavior:
 
Which of your two models is consistent with this new data?
a) the MelA- MelB- double mutant is identical to the MelB- mutant
b) MelA- is epistatic to MelB-
c) MelB- is epistatic to MelA-
d) the MelA- MelB- double mutant is identical to the MelA- mutant
Which of the following is the correct prediction about the double mutants?

TLS Online TPP Program

#Question id: 15658

#Unit 13. Methods in Biology

You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium.
Next, you isolate a third mutant, MelC–, which gives constitutive melibiase expression. The MelC– mutation is closely linked to Mel1 and MelB–. Genetic tests of the MelC– mutation yield the following:
 
As above, classify the MelC– mutation in terms of its basic genetic properties and explain how you arrived at your conclusions.

TLS Online TPP Program

#Question id: 15659

#Unit 8. Inheritance Biology

You calculate map distances between genes A, B, and C based on all pairwise dihybrid crosses. When you perform the trihybrid cross to verify your results, you discover that, despite analysis of a very large number of progeny, you have only about half as many double crossover progeny as you expect. Which explanation is the most reasonable?