SESSION 4 Flashcards

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1
Q

Why might there be a problem with producing the final okazaki fragment on the lagging strand?

A

At the ned of the strand, the last RNA primer has no space to attach to and initiate replication
Thus there is premature termination of DNA replication
There is no space for the DNA replication machinery to continue to function

Consequence on daughter DNA:
Shortened DNA molecule

Consequence on granddaughter DNA:
More shortened DNA molecules

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2
Q

What molecular mechanism do eukaryotic cells have to allow them to overcome this problem (shortening of DNA molecule)?

A

Presence of telomeres
Telomeres are expendable bits at the ends of the chromosomes that get shortened during each DNA replication
Telomeres themselves are physically protected by sheltering, RNA that is encoded by the telomeres themselves
Telomeres are replenished by telomerase reverse transcriptase

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3
Q

How might interfering with telomeres allow treatment of cancer?

A

Cancer cells divide continuously and so there is a heavy load on the telomeres
Telomeres try to make up for it but if their production or their replenishment is affected then the cancer cels DNA replication will be affected

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4
Q

What is xeroderma pigmentosa?

A

Xeroderma pigmentosum is a rare autosomal recessive genetic disorder of DNA repair in which the ability to repair damage caused by ultraviolet light is deficient
In extreme cases, all exposure to sunlight must be forbidden, no matter how small
UV damage to DNA is not repaired leading to skin cancer

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5
Q

Explain how genetic information in a cell is organised as chromosomes

A
Chromosomes are made up of chromatin
Chromatin is made up of:
- DNA
- non- histone proteins 
- RNA
- histone 

Of the histones H2A, H2B, H3 and H4 interact directly with the dna
H1 varies between species and H3, and H4 are highly conserved
The histones are responsible for the โ€˜beads on the stringโ€™ structure of chromatin forming nucleosomes

Euchromatin: lightly packed chromatin often under active transcription
Heterochromatin: tightly packed chromatin

DNA in the cell is organised into tightly folded chains around histone proteins to form chromosomes
In a human cell, there are 23 pairs of chromosomes.
These replicate during S phase of the cell cycle

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6
Q

Describe the chromosomal basis of sex determination

A

Normal male = 46, XY

Normal female = 46, XX

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7
Q

Describe numerical and structural chromosome abnormalities and their significance

A

Numerical- a number of chromosomes other than 46
polyploidy (e.g. Triploidy, tetraploidy)
aneuploidy (an abnormal number that is not a multiple of the haploid number)
- monosomy is a loss of one chromosome, e.g. One โ€˜chromosome pairโ€™ exists as a single chromosome
- trisomy is a gain of one chromosome, e.g. One chromosome pair exists as a triplet

Structural- physical changes to one or more of the chromosome

  • balanced, when the chine does not cause any missing or extra genetic information
  • unbalanced, when the changes cause missing or extra genetic information
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8
Q

Describe the chromosome mutations within one chromosome

A

Deletion- loss of genetic information

Duplication- some genetic material is double

Inversion- no loss of genetic material, but a rearrangement of genetic material

Ring chromosome- loss of telomeres or ends of both arms and formation of a ring

Isochromsome- creation of two non- identical chromosomes, one is a combination of the two short arms, the other is a combination of the two long arms

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9
Q

Describe chromosome mutations with two chromosomes

A

Inversion- no loss of genetic material, but as rearrangement of genetic material to a non- homologous chromosome

Reciprocal translocation- no loss of genetic material, but an exchange of genetic material between tow non- homologous chromosomes โ€˜

Robertsonian translocation- rearrangement of genetic material between two chromosome ; the q arm (long) of two acrocentric chromosome combine to form one โ€˜ super chromosomeโ€™ wth the loss of both p arms

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10
Q

Describe the concept of karyotyping

A

Karyotypes can be produced by โ€˜cut&pasteโ€™ of chromosomes pictures into a systemically organise set of metaphase chromosomes organised in pairs

Chromosome 1 is the largest chromosome and 22 the smallest.
Chromosomes are numbered and grouped according to their size and the position of their centromere
The 23rd Paris is the sex chromosomes

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11
Q

Recognise, comprehend and apply chromosome nomenclature

A

The karyotype formula starts with the total number of chromosomes in the cell, followed by a comma, then the X chromosomes, then the Y chromosomes

E.g. Normal female= 46,XX
Normal male= 46,XY

The plus (+) or minus (-) sign and hen a number indicate an extra/ missing nitre chromosome

A chromosome number then a p/ q and then a +/ - indicates an extra/ missing piece

E.g. 5p- means โ€˜missing a segment of the p arm on chromosome 5

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12
Q

Outline the reasons for referral of patients for karyotyping

A

Constitutional (congenital) abnormalities:
Prenatal screening:
- downโ€™s syndrome
- family history of chromosome abnormality
- abnormal ultrasound scan of foetus

Birth defects:

  • malformations
  • mental retardation

Abnormal sexual development:
- e.g. Klienfelterโ€™s syndrome

Infertility

Recurrent foetal loss

Acquired abnormalities
- leukaemia and related disorders

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13
Q

Explain how fluorescent in situ hybridisation (FISH) works and recognise its importance in the detection of chromosomal abnormalities

A

Fluorescent in situ hybridisation is a technique in which single- stranded nuclei acids (usually DNA, but RNA may also be used) are permitted to interact so that complexes, or hybrids, are formed by molecules with sufficiently similar complementary sequences

Through nuclei acid in hybridisation, the degree of sequence identity can be determined, and specific sequences can be detected and located on a given chromosome

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14
Q

Define non- disjunction

A

The failure of one or more pairs of homologous chromosomes or sister chromatids to separate normally during nuclear division, usually resulting in an abnormal distribution of chronometer in the daughter nuclei
Both copies go to one pole
Can occur in both meiosis I/ II and mitosis

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15
Q

Describe mitotic non- disjunction

A

Leads to aneuploidy (the presence of an abnormal number of chromosomes in a cell)
Leads to non- mosaic karyotype
It may occur in the first post- zygotic division or in later cell division

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16
Q

Describe mosaicism

A

Presence of two or more cell lines in an individual

Degree of mosaicism depends on when:

  • first post- zygotic division- non- mosaic
  • later mitotic division- mosaic
17
Q

Explain the importance of meiosis

A

Maintaining constant chromsome number from generation to generation

Generation of genetic diversity

  • random assortment of chromosomes
  • crossing over of genetic material
18
Q

Define multipolar spindles

A

More than two spindles

19
Q

Define amplified centrosomes

A

More than the right number of centrosomes