AGR 3303 Exam 4

Effects of translocation

- Affects phenotype by physically linking genes that were formerly located on different chromosomes - Gene expression - Chromosomal breaks that bring about translocations may take place within a gene and disrupt its function

Ways aneuploidy arises

- Chromosome may be lost in course of mitosis or meiosis if its centromere is deleted - Small chromosome generated by Robertsonian translocation may be lost in mitosis or meiosis - Through nondisjunction - failure of homologous chromosomes or sister chromatids to separate in meiosis or mitosis

Telophase II

- Chromosomes arrive at spindle poles - Nuclear envelope reforms - Cytoplasm divides - Chromosomes relax and are no longer visible

Telophase I

- Chromosomes arrive at the spindle poles and cytoplasm divides

Prophase I

- Chromosomes condense and become more and more visible - Crossing over occurs here - Nuclear membrane breaks down - Spindle fiber forms

Prophase II

- Chromosomes recondense - Spindle reforms - Nuclear envelope breaks down again

How does nondisjunction in meiosis I create aneuploidy?

- Creates 2 diploid cells - 2 trisomic zygotes and 2 monosomic zygotes

How does nondisjunction in meiosis II create aneuploidy?

- Creates 2 haploid and 1 diploid cell - 1 trisomic, 1 monosomic, and 2 normal diploid zygotes

Process of nucleotide-excision repair

- Enzyme complex recognizes the distortion resulting from damage - DNA is separated and single-stranded binding proteins stabilize the single strands - Enzyme cleaves the strand on both sides of the damage - Part of the damaged strand is removed and the gap is filled in by DNA polymerase and sealed by DNA ligase

Sister chromatids have the following features EXCEPT that: a. both parts have telomeres. b. they are homologs to each other. c. they remain attached to each other at their centromere. d. they will separate during anaphase.

B

The primary difference between mismatch repair, base excision repair, and nucleotide excision repair is: a. which DNA polymerase is used. b. how the mutation is identified. c. the accuracy of the repair mechanism. d. the template strand. e. None of the above.

B

Two chromosomes have the following segments, where • represents the centromere: K L M • N O P Q R S T U V • W X Y Z What type of chromosome mutation would result in the following chromosomes: K L M • N O P O P Q R S T U V • W X Y Z a. Deletion b. Tandem duplication c. Displaced duplication d. Reverse duplication e. Pericentric inversion

B

Which type of chromosome mutation decreases the amount of genetic material for one entire chromosome? a. Translocation b. Aneuploidy c. Polyploidy d. Inversion

B

Why was the pea plant an ideal plant for Mendel to use? a. Generation time that is several years b. Simple traits that are easy to identify c. Low numbers of offspring produced d. Expensive and time-consuming to grow e. All of the above

B

Mismatch repair

Incorrectly paired bases are detected and corrected by mismatch-repair enzymes

Psuedodominance

Indicates one of the homologous chromosomes has a deletion

Base-excision repair

Modified base is excised and then entire nucleotide is replaced

Translocation

Movement of genetic material between nonhomologous chromosomes or within the same chromosome

Do homozygous individuals have problems with inversions?

No

Telophase

Nuclear membrane reforms around each set of chromosomes = 2 separate nuclei

Interkinesis

Nuclear membrane reforms around the chromosomes clustered at each pole, spindle breaks down, chromosomes relax

Down's syndrome

Polysomic individual (primarily trisomy 21) - nondisjunction of chromosome 21 during meiosis I

What does UV radiation cause?

Pyrimidine dimers

What happens to heterozygous individuals during reciprocal translocation?

Reduced fertility - 8 chromatids pairing - Three-way segregation - Gametes from adjacant-1 or -2 are nonviable due to missing or duplicating gene copies

Meiosis I is also known as ________.

Reduction division

Nucleotide-excision repair

Removes bulky DNA lesions (pyrimidine dimers) that distort the double helix

What is prone to strand slippage?

Stretches of repeated sequences

Transitions or transversions occur more frequently?

Transitions

Turner Syndrome

XO - female that is missing an X chromosome with 45 chromosomes (monosomy X)

Do heterozygous individuals have problems with inversions?

Yes, can cause inversion loops during prophase I - reduced recombination - unusual structure with crossing over

Cri-du-chat syndrome

deletion on the short arm of chromosome 5

Tandem duplication

duplicated region is immediately adjacent to the original segment

Displaced duplication

duplicated segment is located some distance from the original segment, either on the same chromosome or different

Reverse duplication

duplication is inverted

Segmental duplication

duplications greater than 1000 bp in length

Principle of segregation

each individual diploid organism possesses two alleles for any particular characteristic

Cystic Fibrosis

ex. of loss-of-function mutation - Mutations produce a nonfunctional form of cystic fibrosis transmembrane conductance regulator protein, which normally regulates the movement of chloride ions in and out of the cell

Reciprocal translocation

exchange of segments between chromosomes

Conditional mutations

expressed only under certain conditions

Interphase

extended period of growth and development between cell divisions - DNA synthesized, RNA and proteins produced, hundreds of biochemical reactions

What is generated when a TE inserts itself into DNA?

flanking direct repeats

Pyrimidine dimers

formation of chemical bonds between adjacent pyrimidine molecules on the same strand - Distorts configuration of DNA and often blocks replication so cell division is inhibited

Trisomy

gain of a single chromosome (2n+1)

Tetrasomy

gain of two homologous chromosomes (2n+2)

Locus

gene location

Gene

genetic factor (region of DNA) that helps determine a characteristic

Nonreciprocal translocation

genetic material moves from one chromosome to another without any reciprocal exchange

Position effect

if positions are altered by an inversion and are expressed at inappropriate times/tissues

Testcross

individual of unknown genotype crossed with another individual with a homozygous recessive genotype for a trait in question - If homozygous (DD), then all progeny are tall - If heterozygous (Dd), then 1/2 are tall, 1/2 are short

Heterozygote

individual possessing two different alleles at a locus

Homozygote

individual possessing two of the same alleles at a locus

Paracentric inversions

inversions that do not include the centromere

Pericentric inversions

inversions that include the centromere

Robertsonian translocation

long arms of two acrocentric chromosomes become joined to a common centromere --> metacentric chromosome with two long arms and another with two very short arms

Deamination

loss of an amino group (NH2) from a base - causes transition mutation - Usually repaired by enzymes that remove uracil whenever it is found in DNA - Normally cytosine deaminated to uracil which pairs with adenine

Nullisomy

loss of both members of homologous pair of chromosomes (2n-2)

Depurination

loss of purine base from nucleotide

Monosomy

loss of single chromosome (2n-1)

Unequal crossing over

misaligned pairing leads to one DNA molecule with an insertion and other with a deletion

Incorporated error

mismatched base has been incorporated into a newly synthesized nucleotide chain - type of spontaneous mutation

Neutral mutation

missense mutation that alters the AA sequence of the protein but doesn't change the function

Transposition

movement of TE from one location to another within the genome

Aneuploidy

number of chromosomes altered: one or more individual chromosomes are added or deleted (2n+1)

Expanding nucleotide repeats

number of copies of a set of nucleotides increase in number

Allele

one of two or more alternative forms of a gene

Polyploidy

one or more complete sets of chromosomes are added (3n, 4n, 5n)

Dihybrid

parents differ in two traits of interest

Tautomeric shifts

positions of protons in the DNA bases change - type of spontaneous mutation

Gain-of-function mutations

produce entirely new trait or causes a trait to appear in an inappropriate tissue or inappropriate time of development

Hybrid

progeny from crossing two individuals (or varieties) that are different in one or more traits

Transition base sub

purine replaced by purine or pyrimidine replaced by pyrimidine

Transversion base sub

purine replaced by pyrimidine or pyrimidine replaced by purine

Deamination in mammals

Cytosine bases are naturally methylated and deamination of 5mC --> thymine

A diploid species has a total of 52 chromosomes. How many chromosomes would be found in a monosomy, trisomy, and autotriploid? a. 51; 78; 53 b. 26; 53; 104 c. 78; 51; 53 d. 51; 53; 78 e. 51; 53; 104

D

Effects of deletions

- If deletion includes the centromere, chromosome will not segregate in meiosis or mitosis and will be lost - Many are lethal in homozygous state because all copies of any essential genes located in deleted region are missing - Heterozygous condition may produce imbalances in the amounts of gene products - Normally recessive mutations on the homologous chromosome lacking the deletion may be expressed when the wild-type allele has been deleted

Mutagenic effects of transposition in grapes

- In black grapes, the VvmybA1 gene regulates the synthesis of anthocyanin pigments - In white grapes, retrotransposon has inserted near the VvmybA1 gene, which disrupts the synthesis of anthocyanins - In red grapes, second mutation has removed most of the retrotransposon , but a piece is left behind. Anthocyanin production is partly restored

Metaphase II

- Individual chromosomes line up on metaphase plate, with sister chromatids facing opposite poles

Metaphase I

- Initiated when homologous pairs of chromosomes align along the metaphase phase - Microtubule from one pole attaches to one chromosome of a homologous pair, and a microtubule from the other pole attaches to the other member of the pair

Anaphase II

- Kinetochores of sister chromatids separate - Chromatids pulled to opposite poles

Anaphase I

- Marked by the separation of homologous chromosomes - Two chromosomes of a homologous pair are pulled toward opposite poles - Sister chromatids arrive at the spindle poles and cytoplasm divides

Process of mismatch repair in prokaryotes

- Mismatch-enzymes cut out a section of the newly synthesized strand and fill the gap with new nucleotides by using the original DNA strand as a template - E. coli differentiates using methyl groups on special sequence of the old strand - In DNA replication, a mismatches base was added to the new strand - Methylation at GATC sequence (of adenine) allows old and newly synthesized nucleotide strands to be differentiated: a lag in methylation means that, immediately after replication, the old strand will be methylated but the new will not be - The mismatch-repair complex brings mismatches bases close to the methylated GATC sequence and new strand is identified - Exonuclease removes nucleotides on the new strand between GATC sequence and mismatch DNA poly replaces the nucleotides, correcting the mismatch, and DNA ligase seals the nick

Process of base-excision repair

- Modified base is first excised and then entire nucleotide is replaced - DNA glycosylases are enzymes that recognize and remove a specific type of modified base by cleaving the bond that links that base to the 1'-C atom of deoxyribose sugar - Produces an apurinic site or apyrimidinic site (AP site) - AP endonuclease cleaves the phosphodiester bond on the 5' side of the AP site and removes the deoxyribose sugar - DNA poly adds new nucleotides to the exposed 3'-OH group - Gap in the sugar-phosphate backbone is sealed by DNA ligase, restoring the original sequence

Effects of inversion

- Pronounced phenotypic effects - May destroy the gene

Mutagenic effects of transposition in maize

- TE of Ac-Ds system causes variegated corn kernels - Yellow kernel: Gene c - no pigment and results in yellow or colorless; Gene C interrupted by Ds element results in yellow - Purple kernel: Gene C - purple pigment - Variegated kernel: In cells where Ds jumps out early, the recovered C gene produces pigment in more cells during development as cell divides

How does depurination work?

- When covalent bond connecting purine to the 1'-C atom of a deoxyribose sugar breaks --> apurinic site (nucleotide that lacks its purine base) - Apurinic base cannot act as template for complementary bases - Incorrect nucleotide incorporated into newly synthesized DNA strand (most often adenine)

Examples of direct repair

- photoreactivation of UV-induced pyrimidine dimers: photolyase uses energy captured from light to break covalent bonds that link pyrimidines in a dimer - corrects O6-methylguanine --> guanine

How does meiosis create genetic variation?

1. Chromosome crossing over at prophase I (genetic recombinations) 2. Random distribution of chromosomes in anaphase I

4 Types of chromosomal rearrangements

1. Duplications 2. Deletions 3. Inversions 4. Translocation

What can silent mutations do?

1. Phenotypic effects 2. Affects splicing 3. Influences folding of mRNA

3 features of transposition

1. Staggered breaks are made by transposase in the target DNA (two "sticky" ends) 2. TE is joined to single strand "sticky" ends of the target DNA 3. DNA replicated at the single strand gaps

What 3 things cause CG --> TA mutation?

1. alkylating agents 2. deamination by nitrous acid 3. hydroxylamine adding hydroxyl groups to cytosine

A mutation that changes a GC base pair to AT is a a. transition. b. transversion. c. missense mutation. d. synonymous mutation.

A

Aneuploids of which chromosomes are less likely to result in a live birth? a. larger autosomes b. small autosomes c. sex chromosomes d. None of the above.

A

Crossing over usually occurs during: a. prophase of meiosis I. b. metaphase of meiosis I. c. prophase of meiosis II. d. metaphase of mitosis. e. Both a and c.

A

In Mendel's peas, purple flower color is dominant to white. From which of the following descriptions can you not infer the genotype completely? a. Purple b. White c. Pure-breeding purple d. pink

A

Meiosis I is called the ___________ division, and meiosis II is called the __________ division. a. reduction; equational b. interphase; prophase c. equational; reduction d. mitotic; interphase e. slow; rapid

A

Most body (nonreproductive) cells of humans and other multicellular eukaryotes have two sets of each chromosome. Such cells are ______ and the matching pairs of chromosomes are called _________. a. diploid; homologous chromosomes b. haploid; homologous chromosomes c. diploid; sister chromatids d. tetraploid; sister chromatids

A

Which of the following is NOT a reason that mutations are important? a. Mutations provide genetic stability. b. Mutations can cause diseases and disorders. c. Mutations are the raw material of evolution. d. Mutations help researchers understand biological processes.

A

Which of the following is a form of aneuploidy in which two members of the same homologous pair are absent? a. Nullisomy b. Monosomy c. Disomy d. Trisomy e. Tetrasomy

A

Which structural component of a eukaryotic chromosome is the attachment point for spindle microtubules during M-phase? a. the centromere b. the origin of replication c. the telomere d. Both a and b.

A

Can chromosome duplications cause negative effects to an organism? Why? a. Yes. Duplicated regions require DNA to be replicated, which delays the cell cycle and wastes energy. b. Yes. Duplicated regions increase gene dosage, which affects processes like development that require specific amounts of protein. c. No. Duplicated regions of chromosomes are quickly lost by looping out during meiosis. d. No. Duplicated regions of chromosomes cannot affect gene dosage because the cell can always compensate for extra copies of genes.

B

How are old and new strands distinguished from each other during mismatch repair in eukaryotes? a. The new strand is methylated. b. The old strand is methylated. c. The new strand is acetylated. d. The old strand is acetylated. e. Scientists do not yet know how strands are distinguished from each other in eukaryote

B

Wild-type Arabidopsis has 10 chromosomes (2n = 10). Trisomicplants are designated as "Tr" followed by the trisomic chromosome number—that is, Tr1 is trisomic for chromosome 1. Assuming that trisomy is fully viable and that all possible pairing configurations (including nonpairing) are possible at meiosis, what proportion of the progeny from the cross wt × Tr1 will have a wild-type chromosomal complement? a. 1/2 b. 1/3 c. 1/4 d. 1/9 d. 1/81

B

A(n) _______ error occurs when a mismatched base has been incorporated into a newly synthesized nucleotide chain. When this DNA is copied, this error leads to a(n) _____ error. a. replication;incorporated b. tautomeric;replication c. incorporated;replication d. incorporated;tautomeric e. tautomeric;incorporated

C

In Labrador retrievers, black coat color is dominant to brown. Suppose that a black Lab is mated with a brown one and the offspring are 4 black puppies and 1 brown. What can you conclude about the genotype of the black parent? a. The genotype must be BB. b. The genotype must be bb. c. The genotype must be Bb. d. The genotype could be either BB or Bb. e. The genotype cannot be determined from these data.

C

In Mendel's peas, yellow seeds are dominant to green. A pure-breeding yellow plant is crossed with a pure-breeding green plant. All of the offspring are yellow. If one of these yellow offspring is crossed with a green plant, what will be the expected proportion of plants with green seeds in the next generation? a. 0% b. 25% c. 50% d. 75% e. 100%

C

In the example below, how many chromosomes are in each cell in prophase and anaphase in mitosis? a. 8 and 8 b. 4 and 4 c. 4 and 8 d. 8 and 4 e. 2 and 8

C

What is responsible for primary Down syndrome? a. Disomy b. Inversion c. Nondisjunction d. Polyploidy e. Translocation

C

What would be the consequence of a diploid gamete (resulting from meiotic nondisjunction) being fertilized by a haploid gamete from the same species? a. Allodiploid b. Allotriploid c. Autotriploid d. Allotetraploid e. Autotetraploid

C

Which of the following enzyme activities is not a part of base-excision repair? a. DNA glycosylase b. DNA ligase c. Reverse transcriptase d. DNA polymerase e. All of the above are part of nucleotide-excision repair.

C

Which of the following types of gene mutation in a protein-coding gene usually have the least-severe (i.e., deleterious) phenotype? a. base deletions b. nonsense substitutions c. missense substitutions d. expansion of a trinucleotide repeat e. All of the above generate equally severe phenotypes.

C

Which type of chromosome mutation increases the amount of genetic material for all chromosomes? a. Translocation b. Aneuploidy c. Polyploidy d. Inversion

C

Which type of chromosome mutation results in a chromosome segment that is turned 180 degrees? a. Deletion b. Duplication c. Inversion d. Translocation e. Transversion

C

FMR-1 gene

Causes fragile-X syndrome due to hundreds or thousands of copies of CGG [example of nucleotide repeats]

How does nondisjunction in mitosis create aneuploidy?

Cell proliferation of monosomic and trisomic cells

What can occur before the G1/S checkpoint?

Cells can exit from the active cell cycle and pass into nondividing phase G0

Direct repair

Change nucleotides back to their original (correct) structures

Sx's of cri-du-chat syndrome

Child who is heterozygous for this deletion has a small head, widely spaced eyes, round face, and mentally retarded; Cry sounds like meowing of a cat

Acentric chromatid

Chromatid with no centromere

Dicentric chromatid

Chromatid with two centromeres

S phase

Chromosomes duplicate (two chromatids)

Cytokenesis

Cytoplasm divides; cell wall forms in plant cells

A man has a condition where all of his gametes undergo nondisjunction of the sex chromosomes in meiosis I, but meiosis II proceeds normally. He mates with a woman who produces all normal gametes. What is the probability that the fertilized egg will develop into a child with Klinefelter syndrome (XXY)? Assume that all gametes and zygotes are viable. a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 0

D

In a cross between pure-breeding tall plants with pure-breeding short plants, all of the F1 are tall. When these plants are allowed to fertilize themselves, the F2 plants occur in a ratio of 3 tall:1 short. Which of the following is not a valid conclusion from these results? a. The allele that produces the tall condition is dominant to the allele that produces the short condition. b. The difference between tall and short stature in these lines is controlled by a single gene pair. c. During production of gametes in F1 plants, the tall and short alleles segregatefrom each other equally into the gametes. d. The tall and short traits assort independently of each other in this cross.

D

In poodles, black fur is dominant to white fur. A black poodle is crossed with a white poodle. In a litter of four, all of the puppies are black. What is the best conclusion? a. The black poodle must be homozygous. b. The black poodle is definitely heterozygous. c. The black poodle is probably heterozygous. d. The genotype of the black poodle cannot be inferred with this information.

D

Transposons are generally characterized by: a. target site inverted sequence repeats. b. terminal direct sequence repeats. c. the presence of introns. d. None of the above.

D

Trisomy and monosomy can be a product of nondisjunction in: a. meiosis I. b. meiosis II. c. mitosis. d. All of the above.

D

Which DNA repair mechanisms use an undamaged complementary DNA strand as a template for replacing the excised nucleotides : a. base excision repair. b. mismatch repair. c. nucleotide-excision repair d. All of those

D

Which is NOT a type of chromosomal mutation? a. chromosome rearrangement b. aneuploidy c. polyploidy d. point mutation

D

Which of the following crosses would produce a 1:1 ratio of phenotypes in the next generation? a. AA x AA b. AA x aa c. Aa x Aa d. Aa x aa e. aa x aa

D

Which of the following events of meiosis creates genetic variation among the gametes? a. crossing over b. random separation of maternal and paternal chromosomes c. distribution of differing numbers of chromosomes to daughter cells d. Both a and b. e. All of the above.

D

Which of the following statements is true? a. The genotype is the physical appearance of a trait. b. Alleles, genes, and loci are different names for the same thing. c. The phenotype of a dominant allele is never seen in the F1 progeny of a monohybrid cross. d. A testcross can be used to determine whether an individual is homozygous or heterozygous. e. All of these statements are true.

D

___________ can cause genes to move from one non-homologous group to another. a. Inversions b. Deletions c. Polyploidy d. Translocations e. Unequal crossing over

D

What DNA poly do bacteria use?

DNA Polymerase I - has proofreading activity

What DNA poly do eukaryotes use?

DNA polymerase B - no proofreading activity

Transposable elements (TE)

DNA sequences that are capable of moving

What is prone to unequal crossing over?

Duplicated or repetitive sequences

A cell that has acquired damage to some of its DNA will be stalled at _______ of the cell cycle. a. G1 b. S c. G2 d. S/G2 checkpoint e. G2/M checkpoint

E

Upon transposing to a new site, transposable elements a. add methyl groups to bases of the surrounding DNA. b. delete about 100 base pairs of DNA on each side of them. c. duplicate their transposase gene. d. express a gene that confers sensitivity to some common antibiotics. e. create a duplication of a target sequence on each side of them.

E

Which of the following was not one of Mendel's conclusions based on his monohybrid crosses? a. Genes are carried on chromosomes. b. Alleles exist in pairs. c. Alleles segregate equally into gametes. d. Alleles behave as particles during inheritance. e. One allele can mask the expression of the other allele.

E

Meiosis II is also known as ________.

Equational division

Insertions

addition of nucleotide vs. deletions - removal of nucleotide

Autopolyploidy

all chromosome sets are from a single species - i.e. autotriploid: AAA - i.e. autotetraploid: AAAA

Chromosome rearrangements

alters the structure of chromosomes (2n)

Forward mutation

alters wild-type phenotype

5-bromouracil (5BU)

analog of thymine - Has a bromine atom on the 5'-C atom instead of methyl group - Mispairs with guanine causing transition mutation - Incorporation of bromouracil followed by mispairing leads to a TA --> CG transition mutation

Mutagen

any environmental agent that significantly increases the rate of mutation above the spontaneous rate

Phenotype

appearance or manifestation of a character

Mitotic spindle

array of microtubules move chromosomes

2-aminopurine (2AP)

base analog of adenine - Mispairs with cytosine causing transition mutation

Missense mutation

base sub that results in a different AA in the protein

Backcross

between F1 genotype and either parental genotypes

Lethal mutations

cause premature death

Loss-of-function mutations

cause the complete or partial absence of normal protein function - frequently recessive

Telocentric

centromere at or very near the end of the chromosome

Submetacentric

centromere displaces toward one end

Acrocentric

centromere is near one end, producing a long arm and knob at the other end

Metacentric

centromere located approximately in the middle

Silent mutation

changes codon to synonymous codon that alters the DNA sequence without changing the AA sequence

Nonsense mutation

changes in sense codon to stop codon

Reverse mutation

changes mutant phenotype back into wild type

Base analogs

chemicals with structures similar to that of any of the four bases of DNA

Allopolyploidy

chromosome sets are from two or more species - Normally arises from hybridization between two species - i.e. Allohexploid: AABBCC

Karyotype

complete set of chromosomes posessed by an organism and presented as a picture of metaphase chromosomes lined up in descending order of size

Monohybrid

crossing parental varieties that differ in only one trait of interest

Induced mutations

results from changes by environmental chemicals or radiation

Reciprocal cross

reverse of an original cross

Genotype

set of alleles possessed by an individual organism

Sx's of Turner Syndrome

short and swelling, sterile

haploinsufficient gene

single copy of gene is insufficient

Monohybrid crosses

those between parents that different in a single characteristic

Dominance

traits that appeared unchanged in the F1 heterozygous offspring

Recessive

traits that disappeared in the F1 heterozygous offspring

Strand slippage

when one nucleotide strand forms a small loop - If the looped-out nucleotides are on the newly synthesized strand, insertion results - If the looped-out nucleotides are on the template strand, then the newly replicated strand has a deletion

Concept of dominance

when two different alleles are present in a genotype, the only trait encoded is the one observed in the phenotype ("dominant" allele)

What happens when crossing over happens during meiosis?

□ Anaphase I chromosomes separate - Two chromatids are no longer identical and the different alleles segregate in anaphase II

What happens when crossing over does not happen during meiosis?

□ Anaphase I chromosomes separate Two chromatids of each chromosome segregate in anaphase II and are identical

G1 phase

□ Cell grows + proteins necessary for cell division are synthesized □ Lasts several hours □ Critical point termed G1/S checkpoint holds the cell in G1 until the cell has all the enzymes necessary for replication of DNA --> division

Metaphase

□ Chromosomes arranged in single plane between two centrosomes □ Centrosomes at opposite ends radiate outwards and meet in the middle of the cell □ Lines up at metaphase plate □ Spindle-assembly checkpoint ensures that each chromosome is aligned on the metaphase plate and attached to spindle fibers from opp poles

Prophase

□ Chromosomes visible □ Each chromosome has 2 chromatids preceding the S phase attached at the centromere

Anaphase

□ Connection between chromatids separate and move to opposite spindle poles □ Chromosome movement due to disassembly of tubulin molecules at both kinetochore end (+) and spindle end (-) of fiber - Molecular motors disassemble tubulin molecules from spindle and generate forces to pull chromosome to spindle pole

Prometaphase

□ Disintegration of nuclear membrane □ Spindle microtubules enter nuclear region □ Ends of microtubules make contact with chromosomes - For each chromosome, a microtubule from one of the centrosomes anchors to the kinetochore of one of the sister chromatids

G2 phase

□ G2/M checkpoint at the end of G2 - Passed only if DNA is undamaged - Damaged DNA inhibits activation of some proteins necessary for mitosis to take place

Polyploidy in agriculture

○ Important in terms of evolution of a plant species ○ Important tool for breeders to develop new species by artificially crossing different species and then using colchicine to double the chromosome ○ Used to produce sterile or seedless triploid