IB Exam (Unit 3)

0.0(0)
Studied by 0 people
call kaiCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/28

flashcard set

Earn XP

Description and Tags

Last updated 12:11 PM on 4/26/24
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai

No analytics yet

Send a link to your students to track their progress

29 Terms

1
New cards

U1: What are DNA and RNA’s monomers?

nucleotides

2
New cards

U2: Compare the structure of RNA and DNA.

number of strands present

Base Composition

Type of Pentose

DNA

2

A, C, G, T

deoxyribose

RNA

1

A, C, G, U

ribose

3
New cards

U3: Explain the structure of DNA

  • double-helix

  • two antiparallel strands of nucleotides

  • linked by hydrogen bonding between complimentary base pairs

4
New cards

U4: Explain the structure of DNA

5
New cards

A1: Who clarified the structure of DNA with models?

Watson and Crick

6
New cards

S1: Drawing simple diagrams of the structure of single nucleotides of DNA and RNA, using circles, pentagons, and rectangles to represent phosphates, pentoses and bases

knowt flashcard image
7
New cards

U: State qualities of the replication of DNA

  • semi-conservative

  • depends on complimentary base pairing

8
New cards

U: Explain DNA Helicase’s Job

  • unwinds the double helix

  • separates the two strands by breaking hydrogen bonds

9
New cards

U: Explain DNA Polymerase’s Job

  • links nucleotides together to form a new strand

  • uses a pre-existing strand as a template

10
New cards

U: Define Transcription

the synthesis of mRNA copied from the DNA base sequences by RNA polymerase

11
New cards

U: Define Translation

the synthesis of polypeptides on ribosomes

12
New cards

U: Explain that the amino acid sequence of polypeptides is determined by…

mRNA according to the genetic code

13
New cards

U: Describe how Codons of three bases on mRNA correspond to…

one amino acid in a polypeptide

14
New cards

U: What does Translation depend on?

complimentary base-pairing between codons on mRNA and anticodons on tRNA

balls

15
New cards

A: Use of Taq DNA polymerase to produce multiple copies of DNA rapidly by the Polymerase Chain Reaction (PCR)

  • PCR allows for strands of DNA to be replicated into the millions

  • To speed up PCR, it is carried out at high temperatures

  • special type of heat-stable DNA polymerase: Taq DNA polymerase

    • obtained from Thermus aquaticus, a bacterium that is adapted to living in hotsprings,

  • Millions of copies of the DNA can be produced by PCR in a few hours because of the high temperatures used

16
New cards

A: Taq DNA Steps

  1. DNA heated to 95°C to separate strands

  2. DNA cooled to 53°C to allow primer to bind

  3. DNA reheated to 73°C to separate strands

17
New cards

A:

  • The gene that codes for insulin has been transferred from humans to the bacterium E. coli and to other organisms, to produce insulin

  • aa sequence for insulin in bacteria = aa sequence in humans

    • because of universality of genetic code

    • codons in bacteria = codons in human mRNA

  • there are some exceptions

<ul><li><p>The gene that codes for insulin has been transferred from humans to the bacterium <em>E</em>. coli and to other organisms, to produce insulin</p></li><li><p>aa sequence for insulin in bacteria = aa sequence in humans</p><ul><li><p>because of <strong><span style="color: purple">universality of genetic code</span></strong></p></li><li><p>codons in bacteria = codons in human mRNA</p></li></ul></li><li><p>there are some exceptions</p></li></ul>
18
New cards

S: Use a table of the genetic code to deduce which codons corresponds to which amino acids

  1. Identify if the given nitrogenous base is the

    1. 5’ strand - match complementary letters, turn into codons, and chart

    2. 3’ strand - match same letter for codon, chart

    3. mRNA strand -order codon, chart

    4. Anti-codon - find opposite letter, chart

  2. ALWAYS DOUBLE CHECK

<ol><li><p>Identify if the given nitrogenous base is the </p><ol><li><p><strong><span style="color: purple">5’ strand</span></strong> - match complementary letters, turn into codons, and chart</p></li><li><p><strong><span style="color: green">3’ strand</span></strong> - match same letter for codon, chart </p></li><li><p><strong><span style="color: yellow">mRNA strand</span></strong> -order codon, chart</p></li><li><p><strong><span style="color: blue">Anti-codon</span></strong> - find opposite letter, chart</p></li></ol></li><li><p><strong><span style="color: red">ALWAYS DOUBLE CHECK</span></strong></p></li></ol>
19
New cards

S: Analyze of Messelson and Stahl’s results to obtain support for the theory of semi-conservative replication of DNA

  • E. coli cultures in N15 environments

  • they were transferred to N14 environments

  • After 1 generation, DNA was intermediate in density between N14 and N15

  • After 2 generations, N14+N15 DNA was present along side solid N14 DNA

  • simply: it kept a strand of the old DNA and created a new strand with the light nitrogen

<ul><li><p><em>E. </em>coli cultures in <strong><span style="color: yellow">N<sup>15</sup></span><span> </span></strong>environments </p></li><li><p>they were transferred to <strong><span style="color: blue">N<sup>14</sup></span><span><sup> </sup></span></strong>environments</p></li><li><p>After <mark data-color="red">1 generation</mark>, DNA was intermediate in density between <strong><span style="color: blue">N<sup>14</sup></span></strong> and <strong><span style="color: yellow">N<sup>15</sup></span></strong></p></li><li><p>After <mark data-color="purple">2 generations</mark>, <strong><span style="color: blue">N<sup>14</sup></span></strong>+<strong><span style="color: yellow">N<sup>15</sup></span><sup> </sup></strong>DNA was present along side solid <strong><span style="color: blue">N<sup>14</sup></span></strong> DNA</p></li><li><p>simply: it kept a strand of the old DNA and created a new strand with the light nitrogen</p></li></ul>
20
New cards

S: Deduce the DNA base sequence for an mRNA strand

  1. Template strand of DNA is usually 3’-XXX-5’

  2. mRNA is always 5’-XXX-3’

<ol><li><p>Template strand of DNA is usually 3’-XXX-5’</p></li><li><p>mRNA is <strong><span style="color: yellow">always</span></strong> 5’-XXX-3’</p></li></ol>
21
New cards

U: Define Cellular Respiration

the controlled release of energy from organic compounds to produce ATP

22
New cards

U: True of False: ATP from cell respiration is not immediately available as a source of energy in the cell

false, it can be used right away

23
New cards

U: True of False: ATP from cell respiration is not immediately available as a source of energy in the cell

24
New cards

U: State that Anaerobic Cell Respiration gives a ______ yield of ATP from glucose

small!

25
New cards

U: Explain that Aerobic Cell Respiration requires ______ and gives a _____ yield of ATP from glucose.

  1. Oxygen (O2)

  2. large

26
New cards
27
New cards
28
New cards
29
New cards