BIOL1020 Lecture Notes - Lecture 13: Chronic Myelogenous Leukemia, Philadelphia Chromosome, Mutation

82 views4 pages

blueclam798

18 May 2018

School

Department

Biology

Course

BIOL1020

Professor

Dr Paul Ebert

For unlimited access to Class Notes, a Class+ subscription is required.

Lecture 13 - Phenotypes !

Heredity - the phenomenon by which oﬀspring resemble their parents !

Variation - why oﬀsprings are not identical !

Principles of heredity

•All individuals of a species have the same base set of genes (variation due to gene mutation) !

•Variants of the same gene – alleles – account for variant traits $

Genotype

AA - dominant homozygote !

aa - recessive homozygote !

Aa - heterozygote !

Mutations

•changes in genotypes arise from mutations !

•Mutations can be in the coding region or the non-

coding region of DNA. Non-coding regions can

change the regulatory system of the protein. !

•Mutations can occur in somatic cells or germ cells!

•Types of mutations!

•point mutations -

•Insertions or deletions

•Translocations, deletions or duplications

Translocation !

•2 non-homologous chromosome swap !

•Our cells can either try to repair the parts or trigger cell death !

Example: Philadelphia chromosome !

•A chromosomal translocation in chronic myelogenous leukemia !

•This chromosome mutation causes the leukaemia progression !

•The Philadelphia chromosome contains a a short fragment of chromosome 9 that is

translocated away from its control area of the gene and placed onto the tip of chromosome 22.

Under this new control region, the gene over produces !

•The mutation happens in a single white blood cell that will pass the mutated chromosome to its

descendants !

Causes of mutations

Spontaneous mutations !

•Induced mutations from external mutagens (eg: UV light and chemicals)!

•result from internal cellular processes !

In humans !

-somatic cells will not pass down mutations but

sex cells will !

-This is because sex cells to not arise from the

somatic cells !

This is not the case with plants that form their

sex cells later on in life from the somatic cells.

Mis-incorportation of nucleotides

during DNA replication !

Errors during meiosis causing the

chromosomal abnormalities (eg: !

Wrong pairing of homologous

find more resources at oneclass.com

Unlock document

This preview shows page 1 of the document.
Unlock all 4 pages and 3 million more documents.

Already have an account? Log in

Document Summary

Heredity - the phenomenon by which o spring resemble their parents. Principles of heredity: all individuals of a species have the same base set of genes (variation due to gene mutation, variants of the same gene alleles account for variant traits. Aa - dominant homozygote aa - recessive homozygote. Mutations: changes in genotypes arise from mutations, mutations can be in the coding region or the non- coding region of dna. Non-coding regions can change the regulatory system of the protein: mutations can occur in somatic cells or germ cells, types of mutations, point mutations , insertions or deletions, translocations, deletions or duplications. Somatic cells will not pass down mutations but sex cells will. This is because sex cells to not arise from the somatic cells. This is not the case with plants that form their sex cells later on in life from the somatic cells.

Related Questions

I'll be thankful if someone can help me to do this!

According to this:

Read the following New York Times article and then answer the questions below and then answer the questions New Clues to Sex Anomalies in How Y Chromosomes Are Copied

http://www.nytimes.com/2009/09/15/science/15chrom.html?scp=1&sq=sex%20anomalies&st=cse

This activity contains 5 questions.

When discarding bad genes, what does the Y chromosome recombine with?

	the other Y chromosome
	the compliment X chromosome
	itself
	nothing

What holds a pair of copied chromosomes together?

	chromatid
	centrosome
	centromere
	chromatin

A person with Turner's syndrome has which of the following?

	a single X chromosome
	two X chromosomes
	a single Y chromosome
	two Y chromosomes

Where is the male-determining gene on the Y chromosome?

	close to the end of the right arm
	close to the end of the left arm
	near the middle of the right arm
	near the middle of the left arm

Briefly explain how an individual can have some cells with no Y chromosome and other cells with a double-Y chromosome . (in your response, go beyond the article, search the web, use experiences from your life, broaden your horizons)

You must make 2 reflective, thoughtful, deep, constructive critiques of your class mates posts. as part of a threaded discussion following their initial post. The critiques must not about you, but about science!

Here's the first peson. Needs to reply for some of her answers as it says above.

1. When discarding bad genes, the Y chromosomes recombines with itself.

2. The centromere functions to hold a pair of copied chromosomes together.

3. A person with Turnerâs syndrome is born with a single X chromosome.

4. The male-determining gene on the Y chromosome is located close to the end of the left arm.

5. An individual can have some cells with no Y chromosomes and other cells with double Y chromosomes, due to the differing variants present in regards to the genotypes of the cells in a single individual. When the Y chromosome is compared to other chromosomes such as the X chromosome, mutations and deviation in the distance of the centromere are much more prevalent due to the Y chromosome lacking a backup system to promote repairs.¹ (Links to an external site.) Even if a mutation caused a condition such as the XYY syndrome where two Y chromosomes are present, that mutation and genotype would not be present in every single cell of the individual.² (Links to an external site.) The presence of different genotypes offers an tentative explanation to as why a single individual can have some cells with no Y chromosomes while also housing other cells with double Y chromosomes.

Source 1 http://www.nytimes.com/2009/09/15/science/15chrom.html

Source 2 https://www.healthline.com/health/xyy-syndrome

Here's the second person. Needs to do the same as it says above.

1. Itself

2. Centromere

3. A single X chromosome

4. Close to the end of the left arm

5. Briefly explain how an individual can have some cells with no Y chromosome and other cells with a double-Y chromosome.

In some cases, the Y chromosome reaches over to a neighboring counterpart and the two chromosomes fuse. This results in the loss of part of the chromosome. Jacobâs Syndrome is sometimes also called XYY syndrome because the person has two Y chromosomes. Although a person can survive with this genetic mutation, some physical traits may be observed such as taller than average height, weak muscle development or late speech development.

Source 1 https://www.healthline.com/health/xyy-syndrome

Source 2 http://www.isna.org/faq/y_chromosome

Read the summary and answer questions in bold below:

In this post, weâll be focusing on Tyler Jacks and his graduate student Francisco J. SÃ¡nchez-Riveraâs review, âApplications of the CRISPR-Cas9 system in cancer biology,â circa 2015 in Nature Reviews. So we understand the context of the authors, Jacks is the director of the Koch Institute for Integrative Cancer Research at MIT. His lab has a large mouse experimental component (MIT 2017). This sets the stage for the angle weâll be reading from. In all, itâs a nice roundup of his and colleagues' cancer applications of CRISPR-Cas9 with great graphics, but leaves a bit desired in crossing the bridge from mouse to human.

Briefly, the CRISPR-Cas9 system stems from the prokaryotic immune system, where the RNA-guided DNA endonuclease Cas9 uses a single guide RNA (sgRNA) containing a CRISPR RNA (crRNA) and trans-activating crRNA (tracrRNA) that binds Cas9 and directs it to the genomic sequence of interest, adjacent to a photospacer adjacent motif (PAM). Cas9 causes a precisely-directed double-stranded break (DSB) in the DNA, that gets repaired via non-homologous end joining (NHEJ) or homology-directed repair (HDR).

With NHEJ, this can disrupt the gene by creating Indels: frameshifts and stop codons. In cancer, individual oncogenes or tumor suppressors can be targeted, or drug response modulators, or synthetic lethality genes. With a multiplexed knockout approach, combinatorial vulnerabilities and epistatic relationships can be assessed.

For gene modification via HDR, target single or double-stranded DNA can be introduced, resulting in precise gene knock-ins, with potential effects of introducing point mutations to mimic cancers, fluorescent reporters/synthetic tags can be created, conditional (Cre-lox/Flp-FRT) constructs can used to turn off/on a mutation, non-coding elements like enhancers/insulators/promoters can turn on proto-oncogenes, and ultimately, HDR can be used for gene correction.

With Cas9 and 2 sgRNAs, there can be genomic rearrangements, leading to deletions, inversions (like EML4-ALK), and translocations (like BCR-ABL).

Dead Cas9 (dCas), which lacks its endonuclease activity via mutation to its HNH and RuvC domains, can be used for reversible, transcriptional repression or activation of coding and non-coding genes, or chromatin modification, via an effector.

Much of the practical applications of CRISPR-Cas9 is in ES cells and mouse models, both germinal and somatic (GEMMs and nGEMMs). With this technology, a plethora of GEMM/nGEMM models of cancer can be created, some with the ability to turn on/off the Cas9 system, and multiplexed sgRNAs can help gain further understanding of multi-hit cancers. Patient derived xenografts (PDX) mice can be assayed with CRISPR-Cas9 to determine different treatment efficacies. Entire libraries of sgRNA can be screened in cancer cell lines for enrichment/depletion with Cas9, and dCas9-activators/repressors can be used for complementarity, both of which providing insights into drug targets, synthetic lethality, and chemotherapy modulators.

The delivery mechanisms for the system are diverse, and include plasmids, lentivirus, adenovirus, nanoparticles, hydrodynamic gene transfer, and protein-based. The authors also cite integrating Cas9 into the GEMMs, so itâs one less thing needed in the delivery vehicle. This of course would not be applicable to human.

Ultimately, human cancer treatment is the goal. Yes, we know the boundaries between ES/mice experimental models and human embryos have already been broken, with an international team targeting the autosomal dominant mutation in MYBPC3, which causes cardiac myopathy (Ledford 2017). But the current compromise position of CRISPR-Cas9 technology toward therapeutics and treatment, beyond basic science understanding, is using primary cells, organoids, PDX, or personalized GEMMs to create genetic and drug screens.

In theory, what would it take to bring CRISPR-Cas9 to the ultimate, targeted gene therapy via either NHEJ or HDR, in human cancer? Are there cancers that are better bets for CRISPR-Cas9 than others, delivery vehicles that are safer or better optimized? Is CRISPR-Cas9 really âon targetâ enough to introduce into a person, and how do we best assess?

1. In Humans, many types of cancers have been linked to?

A. Duplications
B. Inversions
C. translocations
D. deletions
E all of these

2. 5- bromoouracil is mutagenic because it is a(n)

a. base-modifying agent
b. base analog
c. intercalating agent
d. alkylating agent.

3.Ultraviolet radiation creates __________ in DNA

A. 5-methyl cytosine residues
B. breakages
C point mutations
D. mutational hot spots
E thymine dimers

4. The telomeres of chromosomes is composed of

A. heterochromatin
b. nucleosome
c. histone
d. achromatic

5. In eukaryotes, a replicon is

a. the distance between the origin of replication and thetermination of replication
b. a single chromosome
c. the entire genome
d. one DNA strand of a single chromosome

6. In a nonreciprocal interchomosomal translocation, a segment of achromosome is

a. exchanged with an identical segment on a differentchromosome
b. exchanged with an identical segment on a sister chromatid
c. transferred to a different chromosome without a reciprocalexchange of genetic material
d. transferred to a new location on the same chromosome without areciprocal exchange of genetic material

7. Somatic mutations are always.

a. spontaneous
b. carcinogenic
c. heritable
d. lethal
e. none of these

8. A woman with an XO genotype is

a. aneuploid
b. euploid
c. diploid
d. poluploid

9. Spontaneous mutations may occur as a result of

a. all of these
b. errors in DNA replication
c. transposition
d. none of these
e. normal cell biochemistry

10. In the semi conservative model of DNA replication, progenydouble helices consist of

a. one parental DNA strand and one new strand
b. one parental DNA strand and one new RNA strand
c. two new RNA strands
d. two parental DNA strands
e. two new DNA strands

11. A mutation in a gene may alter a cellâs

a. phenotype
b. genotype and phenotype
c. genotype

12. During replication, the direction of synthesis of new DNA fromthe leading strand is

a. both 5â to 3â and 3â to 5â
b. none of these
c. 3â to 5â
d. from left to right only
e. 5â to 3â only

13. A mutation that results in a switch from a purine-pyramidingbase pair to a pyramiding-purine base pair is a _________mutation.

a. neutral
b. transition
c. missense
d. transversion
e. nonsense

14. A trisomic human cell would contain how many chromosomes?

a. 45
b. 46,
c. 47.
d. 48

15. In a strand of DNA, a _____________ bond connects the phosphategroup of one nucleotide to the sugar of the adjacentnucleotide

a. hydrogen
b. ester
c. phosphodiester
d. phosphate
e. disulfide

16. Cri-du-chat syndrome is caused by which type of chromosomalmutation?

a. deletion
b. duplication
c. translocation
d. transversion
e. inversion

17. During replication, the role of DNA primase is to

a. add nucleotide to the growing DNA chain
b. synthesize a short chain for RNA
c. untwist the double helix at the origin of replication
d. relax the super coiling of the double helix
e. synthesize the primo some

18. A(n) ____________ inversion involves the Centro mere

a. chromocentric
b. paracentric
c. epicentric
d. percentric
e. concentric

19. Which of the following substances mutates DNA by intercalatingbetween adjacent base pairs?

a. methyl methane sulfonate
b. all of these
c. nitrous acid
d. 2-amino urine
e. acridine orange

20. A molecule of RNA is always

a. double stranded
b. helical
c. nonparallel
d. all of these
e. none of these

21. Mutagenesis of cells using radiation or a chemicalintercalating agent could be used to create __________ mutations ina region of DNA.

a. random
b. both random and site-specific
c. site-specific

22. In Griffithsâs tranformation experiments, the mice died whenthey were injected with

a. living, no virulent bacteria
b. living, virulent bacteria
c. heat-killed, no virulent bacteria
d. living, nonvirulent bacteria
e. both B and D

23. A monoploid human cells

a. shows only the excessive phenotype of all these genes
b. would be functionally impaired
c. is normal
d. would not survive

24. Gametes containing an abnormal number of chromosomes may occuras a result of

a. paracentric inversion
b. reciprocal translocation
c. disjunction of chromosomes
d. unequal crossing over

25. DNA may be altered as a result of _____________ events thatoccur normally in cells

a. depurination
b. deamination
c. depyrimidation
d. all of the above
e. none of the above

26. Which enzyme prevents DNA from tangling up as the replicationfork migrates during replication?

a. DNA polymerase I
b. DNA helicase
c. topoisonmerase
d. DNA polymerase III
e. DNA gyrase

27. Which of the following enzymes play an important role in DNAreplication?

a. DNA helicase
b. DNA ligase
c. DNA primase
d. all of them
e. DNA polymerase III

28. A mutation gene sustains a mutation that restores the originalphenotype, This is a

a. frame shift mutation
b. transversion
c. reversion
d. transition
e. forward mutation

29. A dicentric chromosome has two

a. telomeres
b. mutations
c. arms
d. centromeres

30. Which type of chromosomal mutation cannot revert to thewild-type state

a. duplication
b. deletion
c. translocation
d. both A and C
e. all of them

31. Translation of histone mRNAs occurs during which phase of theeukaryotic cell cycle?

a. M
b. G1
c. G2
d. S
e. all of these

32. After removal of the RNA primers, Okazaki fragments are joinedthrough a reaction that involves which enzyme?

a. RNA polymerase
b. DNA helicase
c. DNA polymerase
d. DNA primase
e. DNA ligase

33. Which of the following is required for the synthesis of DNAduring replication?

a. DNA polymerase
b. magnesium ions
c. all of them
d. template DNA
e. dNTPs

34. Most of the genome of an active cell consists of

a. heterochromatin
b. histochromatin
c. poluchromatin
d. euchromatin

35. In eukaryotes, DNA replication begins,

a. a single origin of replication on each chromosome
b. a single origin of replication on each chromosome
c. multiple origins of replication on each chromosome
d. all of the above
e. none of the above

36. A human cell contains 2 sets of 24 chromosomes is

a. euploid
b. haploid
c. polyploid
d. aneuploid

37. Pseudodominance occurs as a result of which type ofmutation?

a. translocation
b. duppolication
c. all of these
d. point mutation
e. deletion

38. The enzymatic activity of a telomerase is best described asa

a. none of these
b. polymerase
c. ligase
d. reverse transcriptase
e. topoisomerase

39. A Barr body is an example of

a. constitutive heterochromatin
b. facultative heterochromatin
c. facultative euchromatin
d. constitutive euchromatin

40. An addition or deletion of one or two base pairs in a genecauses a ____________ mutation

a. Transition
b. frameshift
c. nonsense
d. transversion
e. neutral

41. A heterozygous chromosomal inversion can be detected

a. by cytogenetic observation of loops in chromosomes
b. through genetic studies indicating a decrease in the frequencyof recombination between genes
c. by observing a decrease in the number of viable progenyfollowing genetic crosses
d. all of the above

42. The mutagen 5-bromouracil is an analog of which base?

a. adenine
b. threonine
c. thymine
d. cystosine
e. uracil

43. A mutation that changes a codon from one that represents anamino acid to one that signals a chain termination is a______________ mutation.

a. neutral
b. missense
c. frame shift
d. nonsense

44. A chromosomal mutation that results in an increase in theamount of DNA is a(n)

a. duplication
b. inversion
c. deletion
d. both A and C
e. all of the above

45. Prior to replication, the DNA double helix must be untwisted bywhich type of enzyme?

a. topoisomerase
b. ligase
c. polymerase
d. helicase
e. primase

46. In his transformation experiments, Griffiths used cultures ofwhich bacterium?

a. Sacchromyces cerevisiae
b. Escherichia coli
c. Streptococcus pneumoniae.
d. Bacillus cereus
e. Eneterococcus faecalis

47. A mutation changes a codon from a UCG to a UAG. This is a____________ mutation?

a. point
b. nonsense
c. silent
d. both A and B
e. all of the above

48. In eukaryotes, DNA replication

a. is initiated at several sites on chromosomes
b. occurs bidirectional
c. is semidiscontinuous
d. all of the above
e. none of the above

49. Which of the following bacteria are used in the Amestest?

a. Pseudomonas aeruginosa
b. Escherichia coli
c. Vibrio cholera
d. Salmonella typhimurium
e. Rickettsia rickettsii

50. Which of the following changes represents a missensemutation?

a. UGA (stop) to UAG (stop)
b. AAA (lysine) to AGA (arginine)
c. UUU (phenylalanine) to UAA (stop)
d. AUG (methionine) to UGA (stop)
e. AAA (lysine) to AAG (lysine)

BIOL1020 Lecture Notes - Lecture 13: Chronic Myelogenous Leukemia, Philadelphia Chromosome, Mutation

Document Summary

Get access

Related Documents

BIOL1020 Lecture Notes - Lecture 13: Mutation, Dna Replication, Phenotype

BIOL1020 Study Guide - Homologous Chromosome, Gregor Mendel, Karyotype

BIOL1020 Study Guide - Homologous Chromosome, Gregor Mendel, Karyotype

Related Questions