Print Options

Card layout: ?

← Back to notecard set|Easy Notecards home page

Instructions for Side by Side Printing
  1. Print the notecards
  2. Fold each page in half along the solid vertical line
  3. Cut out the notecards by cutting along each horizontal dotted line
  4. Optional: Glue, tape or staple the ends of each notecard together
  1. Verify Front of pages is selected for Viewing and print the front of the notecards
  2. Select Back of pages for Viewing and print the back of the notecards
    NOTE: Since the back of the pages are printed in reverse order (last page is printed first), keep the pages in the same order as they were after Step 1. Also, be sure to feed the pages in the same direction as you did in Step 1.
  3. Cut out the notecards by cutting along each horizontal and vertical dotted line
To print: Ctrl+PPrint as a list

40 notecards = 10 pages (4 cards per page)

Viewing:

Genetics Exam 2

front 1

Frederick Griffith’s transformation experiment showed that

back 1

A chemical component of cells could introduce a new trait to a cell

front 2

In Avery, MacLeod and McCarty's experiment, what enzyme was present in the mixture of R & S Strain that when injected into the mice did not result in death?

back 2

DNase

front 3

If Hershey and Chase phage experiment, what finding led to the conclusion that DNA is the genetic material?

back 3

The presence of radioactivity in the bacteria when 32 P was used

front 4

What would be the outcome of Hershey-Chase experiment if protein (and not DNA) was the carrier of genetic information?

back 4

The infected bacteria would be positive for 35 S radioactive labeling, and not 32 P

front 5

This molecule is most likely found on ___. It has a(n) ___ ring

back 5

RNA; purine

front 6

Which of the following is not included in the Watson-Crick model of D N A structure?

back 6

The strands run parallel

front 7

Which of the following is not consistent with Erwin Chargaff’s findings?

back 7

(C + G) = (A + T)

front 8

The virus phi X 174 infects Escherichia coli. Its base composition is as follows:
A = 24.0 %
G = 23.3 %
C = 21.4 %
T = 31.3 %
What conclusions can you draw about phi X 174

back 8

It has a single-stranded D N A genome.

front 9

If you heated a sample of D N A from 30°Ceto 100°C while you measured the A 260 , which of the following graphs would you expect to see?

back 9

(A)

front 10

In the Meselson–Stahl experiment, if DNA had been replicated conservatively, after two rounds of replication, there would be _____

back 10

one old, heavy band and one new, light band

front 11

All known bacterial D N A polymerases _____.

back 11

have 5′ to 3′ polymerization activity

front 12

What activity provides D N A pol III the ability to proofread?

back 12

D N A pol III has 3′→5′ exonuclease activity

front 13

What protein is functioning at the point marked with the large arrow?

back 13

D N A polymerase III

front 14

What enzyme will replace the R N A primers found in the newly synthesized strand

back 14

D N A pol I

front 15

Okasaki fragments are a consequence of

back 15

the inability of the DNA polymerase to polymerize in the 3′ to 5′ direction

front 16

Is there any role for an R N A polymerase in D N A replication?

back 16

Yes, to generate primers.

front 17

What does the term “processive” mean in the context of D N A replication?

back 17

that length of D N A that is replicated by the core enzyme before it detaches from the template

front 18

Telomerase _____.

back 18

a. contains its own R N A template
b. is a reverse transcriptase
c. extends one strand of the telomere
d. is responsible for helping to maintain chromosome size
e. all of the above

front 19

How do bacteria package their D N A

back 19

by supercoiling the D N A

front 20

If there were six bases in D N A and still only 20 amino acids, how long would a codon need to be in order to specify an amino acid?

back 20

Two bases

front 21

The wobble hypothesis predicts that codons coding for the same amino acid _____

back 21

may differ at the third position

front 22

Nobel Prize in Physiology or Medicine 2024

back 22

Victor Ambros and Gary Ruvkun

front 23

In humans, the phenylalanine hydroxylase gene is 90,000 (90 k b) bases long, yet the mRNA is only 2,400 (2.4 k b). What explains this difference?

back 23

Presence of introns in DNA

front 24

The sigma subunit of bacterial R N A polymerase _____.

back 24

binds to a bacterial gene’s promoter

front 25

An aminoacyl tRNA synthase Valine is mutated so that it now attaches the amino acid glycine to the tRNA Valine instead of valine. What will happen at translation?

back 25

There will be glycines at all glycine positions and at all valine positions

front 26

During the elongation phase of translation, charged tRNAs enter the ribosome at the ______.

back 26

A site

front 27

Which of the following will most likely result in frameshift mutations

back 27

Acridine orange dye

front 28

Which of the following will more likely result in double stranded breaks? Select all that apply

back 28

Ionizing radiation, transposable elements, and oxidizing damage

front 29

Which of the following will likely result in thymine dimer?

back 29

UV

front 30

Constitutive genes ____

back 30

are expressed continuously

front 31

The lac operon is optimally expressed _____

back 31

in the presence of lactose and the absence of glucose

front 32

Which of the following is a trans-acting element?

back 32

a. Lac O
b. T r p R protein
c. Lac I protein
d. C A P
e. b, c, and d

front 33

Attenuation in the E. coli t r p operon _____.

back 33

results from the formation of secondary structures in the t r p m R N A

front 34

Eukaryotic regulation of gene expression occurs at the level of _____

back 34

a. transcription
b. splicing and processing
c. m R N A degradation
d. translation
e. all of the above

front 35

What is an interchromatin compartment?

back 35

D N A-free area of the nucleus located between chromosomes

front 36

Where might you most likely find the histone variant H2A.Z?

back 36

In promoters and enhancers of expressed genes

front 37

A protein causes the acetylation of histones. What effect do you think this will have on a gene in that region?

back 37

It will cause an increase in expression

front 38

What is the role of D N A methylation in eukaryotes?

back 38

To aid in silencing transcription

front 39

What is the difference between the Initiator element (I n r) and the TATA box?

back 39

R N A P binds to the TATA sequence and then initiates transcription downstream at I n r

front 40

An enhancer sequence _____.

back 40

can be inverted without altering its functional ability