Exam 3

Question 1

Chloroplast DNA (cpDNA) Heredity

Answer 1

maternally inherited

Green (Normal cpDNA) – In healthy plants, chloroplasts contain functional genes for chlorophyll production, resulting in green leaves due to efficient photosynthesis.

White or Yellow (Mutant cpDNA) – Some mutations disrupt chlorophyll biosynthesis, leading to chloroplasts that lack pigmentation, causing white or yellow sectors in leaves.

Variegation (Mixed cpDNA Populations - Heteroplasmy) – Some plants inherit a mix of normal and mutant chloroplasts, leading to patchy green and white/yellow patterns on leaves. This can occur due to random segregation of cpDNA during cell division.

Question 2

Mitochondrial DNA (mtDNA)

Answer 2

inherited maternally

double-stranded circular D N A

A cell can contain a mix of normal and mutated mtDNA

Threshold Effect – Symptoms of mtDNA mutations appear when the percentage of mutated mitochondria crosses a certain threshold

Question 3

Endosymbiotic theory (Lynn Margulis et al.)

Answer 3

Mitochondria and chloroplasts (organelles) arose independently 2 billion years ago from free-living bacteria
Organelles possessed attributes of aerobic respiration and photosynthesis, respectively

Question 4

m t D N A susceptible to mutations

Answer 4

No structural protection from histones

D N A repair mechanism limited

High concentrations of R O S (reactive oxygen species) generated by cell respiration

Question 5

The Particle Theory of Inheritance

Answer 5

states that hereditary traits act like particles, units, or factors as they are passed from generation to generation.

Question 6

Law of Segregation

Answer 6

states that his hereditary factors do not blend but remain distinct during breeding—thus, disproving the blending theory.

Question 7

Law of Independent Assortment

Answer 7

which states that character traits are not connected but are inherited independent of one another

Question 8

Map distance math

Answer 8

(recombinant #)/total X 100

Question 9

Chi square math

Answer 9

(Expected-observed)^2/observed

Add together

Degree of freedom: # of freedom - 1

Question 10

Three-point

Answer 10

Cross two true-breeding strains that differ with regard to three alleles

Perform a testcross by mating F1 female heterozygotes to male flies that are homozygous recessive for all three alleles

Collect and analyze data from F2 offspring (propose a hypothesis, apply the chi square formula and interpret the chi square value)

Calculate the map distance between pairs of genes

Construct the map

Question 11

Aneuploidy

Answer 11

involves a condition where one or a few chromosomes are added or deleted from the normal chromosome number

Question 12

Euploidy

Answer 12

a state where the chromosome number is an exact multiple of the basic chromosome set,

Question 13

Nondisjunction

Answer 13

Paired homologs fail to separate during segregation

Question 14

monosomy

Answer 14

a chromosomal abnormality that occurs when a cell has only one chromosome from a pair, instead of the usual two

Question 15

trisomy

Answer 15

a genetic condition where an individual has an extra copy of a specific chromosome

Question 16

Chromosomal Deletion

Answer 16

a genetic mutation where a portion of a chromosome is missing

Question 17

Duplications

Answer 17

usually caused by abnormal events during recombination

Repetitive sequences can cause misalignment between homologous chromosomes.

If a crossover occurs, nonallelic homologous recombination results

Question 18

Chromosomal Inversion

Answer 18

A chromosome with an inversion has a segment that has been flipped to the opposite orientation

Question 19

Paracentric inversion heterozygote results in gametes

Answer 19

One gamete with normal chromosome

One gamete dicentric (two centromeres; duplication and deletion)

One gamete with inversion

One gamete acentric (no centromere; duplication and deletion)

Question 20

Chromosomal Paracentric inversion

Answer 20

a chromosomal rearrangement that occurs when a chromosome breaks in two places on the same arm. The broken segment is then reinserted after a 180° rotation.

Question 21

Pericentric inversion heterozygote results

Answer 21

One gamete with normal chromosome
One gamete with inversion
Two gametes with duplications and deletions

Question 22

Chromosomal Pericentric inversion

Answer 22

a chromosomal rearrangement where a segment of a chromosome, including the centromere, is inverted, meaning it’s flipped 180 degrees and reinserted into the chromosome.

Question 23

Translocations

Answer 23

a chromosomal abnormality that occurs when parts of two chromosomes are exchanged or fused

Question 24

Mutation

Answer 24

a heritable change in the genetic material

Mutations provide allelic variations

Question 25

1. Chromosome mutations

Answer 25

Changes in chromosome structure Deletions, insertions, duplications, translocations

Question 26

2. Genome mutations

Answer 26

Changes in chromosome number Euploidy (multiple of all), aneuploidy (multiple of one)

Question 27

3. Gene mutations

Answer 27

Relatively small change in DNA structure that affects a single gene

Question 28

transition base

Answer 28

a change of a pyrimidine (C, T) to another pyrimidine or a purine (A, G) to another purine

Question 29

transversion base

Answer 29

a change of a pyrimidine to a purine or vice versa

Question 30

Spontaneous mutation:

Answer 30

Changes in nucleotide sequence that occur naturally Arise from normal biological or chemical processes that alter nitrogenous bases Spontaneous mutation rates vary, but are exceedingly low for all organisms

Question 31

Spontaneous Tautomeric Shifts

Answer 31

Enols may form due to tautomeric shifts in which a carbonyl group will be converted into a hydroxy group or an amine group may convert into an imino group. Changes available donor and acceptors with nitrogenous bases

Question 32

Induced Mutations

Answer 32

changes in the number or structure of chromosomes caused by environmental factors or mutagens like chemicals or radiation

Question 33

Alkylating agent mutagens

Answer 33

Adding alkyl groups (methyl, ethyl, propyl, etc) to an amino or keto group Alter base-pairing affinity

Question 34

Intercalating Agents

Answer 34

Chemicals with dimensions and shapes that wedge between D N A base pairs This causes base-pair distortions and D N A unwinding Forces bulges in DNA so it cannot be unbound

Question 35

Base analogs (mutagenic chemicals)

Answer 35

Can substitute for purines or pyrimidines during nucleic acid biosynthesis Increase tautomeric shifts Increase sensitivity to U V light—mutagenic Example: 5-Bromouracil behaves as thymine analog

Question 36

U V Light—Pyrimidine Dimers

Answer 36

Two identical pyrimidines that distort D N A conformation Errors can be introduced during D N A replication

Question 37

Nucleotide Excision Repair

Answer 37

This type of system can repair many types of DNA damage, including Thymine dimers and chemically modified bases missing bases, some types of crosslinks

Question 38

Two types of excision repair

Answer 38

Base excision repair (B E R) Nucleotide excision repair (N E R)

Question 39

Base excision repair (B E R)

Answer 39

Corrects D N A containing a damaged D N A base D N A glycosylase recognizes altered base

Question 40

Nucleotide excision repair (N E R)

Answer 40

Repairs bulky lesions that alter/distort double helix

Question 41

Base and nucleotide excision repair steps

Answer 41

Recognition of mutation (eg: deamination of cytosine to uracil which uracil DNA glycosylase can immediately detect uracil in DNA) Excision of mutated base (the strand is broken with AP endonuclease between polypeptide bonds) Repair (the other strand is used as a template to fill in the gap with Pol and ligase)

Question 42

D N A Repair associated with DNA replication

Answer 42

Proofreading and mismatch repair D N A polymerase “proofreads,” removes and replaces incorrectly inserted nucleotides Mismatch repair (if proofreading fails) becomes activated Mismatches are detected, cut, and removed (endonuclease and exonuclease). Correct nucleotide inserted by D N A polymerase

Question 43

Strand Discrimination

Answer 43

Based on D N A methylation Adenine methylase (enzyme in bacteria) recognizes D N A sequences and adds methyl group to adenine residues Newly synthesized strand of replication remains unmethylated Mismatch repair recognizes unmethylated strand and repairs

Question 44

Postreplication Repair

Answer 44

Responds after damaged D N A has escaped repair and has failed complete replication RecA protein directs recombination exchange with corresponding region on undamaged parental strand (donor D N A) Gap can be filled in by repair synthesis

Question 45

Double-Strand Break Repair - 2

Answer 45

Homologous recombination repair pathway Nonhomologous end joining pathway

Question 46

Homologous recombination repair pathway

Answer 46

Recognizes break, digests 5’ end, and leaves 3’ overhang 3’ end aligns with sequence complementary on sister chromatid Occurs during late S or early G2 phase of cell cycle

Question 47

Nonhomologous end joining pathway

Answer 47

Activated in G1 prior to replication Repairs double-strand breaks Complex of proteins is involved in end joining May include kinase and B R C A 1 Proteins bind free ends and ligate ends back together