Chapter 16/17 - DNA & Genetic Material

Question 1

What are the 2 forms Streptococcus pneumoniae?

Answer 1

normal virulent nonvirulent

Question 2

What form of Streptococcus pneumoniae causes pneumonia?

Answer 2

normal virulent

Question 3

What form of Streptococcus pneumoniae does not cause pneumonia?

Answer 3

nonvirulent

Question 4

What is the mutant form of Streptococcus pneumoniae?

Answer 4

nonvirulent

Question 5

Streptococcus pneumoniae: What form am I? Referred to as the S form

Answer 5

normal virulent

Question 6

Streptococcus pneumoniae: What form am I? Forms smooth colonies on a culture dish

Answer 6

normal virulent

Question 7

Streptococcus pneumoniae: What form am I? lacks an enzyme needed to manufacture the polysaccharide coat

Answer 7

mutant, nonvirulent form

Question 8

Streptococcus pneumoniae: What form am I? forms rough colonies.

Answer 8

mutant, nonvirulent form

Question 9

Performed a series of simple experiments in which mice were infected with bacteria

Answer 9

Griffith

Question 10

Question 11

Mice infected with the ___________ form died from pneumonia

Answer 11

virulent S

Question 12

Infection with the _________ form had no effect.

Answer 12

nonvirulent R

Question 13

If the virulent S form is first killed with heat, what is the result?

Answer 13

infection does not harm the mice

Question 14

Infecting mice with a mixture of heat-killed S form with live R form caused what?

Answer 14

pneumonia and death in the mice.

Question 15

High levels of live S form bacteria were found in the ____ of ____

Answer 15

lungs of the dead mice.

Question 16

When polysaccharide coat passes from the dead, virulent S bacteria to the live, coatless R bacteria in the mixture, permanently altering the coatless R bacteria into the virulent S variety this process is called?

Answer 16

transformation

Question 17

Transfer of virulence from one cell to another

Answer 17

Transformation

Question 18

Transfer of genetic material between cells

Answer 18

Transformation

Question 19

Identified the substance responsible for transformation in Griffith’s experiment.

Answer 19

Colin MacLeod and Maclyn McCarty

Question 20

What steps were used to identify the substance responsible for transformation :

Answer 20

1. prepared the mixture of dead S Streptococcus and live R Streptococcus
2. removed as much of the protein as they could from their preparation, eventually achieving 99.98% purity
3. transforming activity was not reduced.

Question 21

Transformation Principle: Oswald Avery, Colin MacLeod and Maclyn McCarty

After nearly all protein removal, transfromation activity was not ___________-

Answer 21

reduced

Question 22

Transformation Principle: Oswald Avery, Colin MacLeod and Maclyn McCarty

The substance responsible for transformation resembled what?

Answer 22

dna

Question 23

Extracting lipids and proteins did what to transforming activity?

Answer 23

did not reduce

Question 24

Protein-digesting enzymes and RNA-digesting enzymes did what to transforming activity?

Answer 24

did not affect

Question 25

DNA-digesting enzymes did what to transforming activity?

Answer 25

destroyed

Question 26

Who demonstrated that phage genetic material is DNA?

Answer 26

Alfred Hershey and Martha Chase

Question 27

Who labeled either protein capsule or DNA of bacteriophages with radioactive isotopes (32P and 35S)

Answer 27

Hershey and Chase (1952)

Question 28

Hershey and Chase (1952): Experiments provided several lines of evidence that genetic material is actually stored in what?

Answer 28

DNA and not protein

Question 29

**Hershey and Chase** deduced that \_\_\_\_\_, and not \_\_\_\_\_\_, constituted the genetic information that viruses inject into bacteria.

Answer 29

DNA PROTEIN

Question 30

Although the **three-dimensional structure** of the DNA molecule was not elucidated until Watson and Crick, it was known that it contained three main components which are :

Answer 30

**a 5-carbon sugar** **a phosphate (PO4) group** **a nitrogen-containing (nitrogenous) base**. The base may be a purine (adenine, A, or guanine, G), a two-ringed structure; or a pyrimidine (thymine, T, or cytosine, C), a single-ringed structure. RNA contains the pyrimidine uracil (U) in place of thymine.