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Flashcards in The Human Genome and Karyotypes Deck (22):
1

What are the two mechanisms for increased genome complexity and size.

1. Duplication: certain sequences were duplicated and then via selection and divergence became unique and were kept for their function

2. Incorporation of DNA from other species (lateral transfer) like viruses.

2

What is the argument for why RNA preceded DNA

RNA is more complex and diverse in it's functions, for example it can fold on itself and catalyze reactions or act as tRNA.

DNA has more chemical stability and that is a evolutionary advantage.

RNA is a simpler molecule.

3

What is the ENCODE project?

Tried to sequence all the functional regions of the human genome.

The conclusion was that at least 80 percent of the genome is likely functional, this would include regions that are "noncoding regions" and they are just as important as protein coding regions/ determinants of health and disease.

4

What are five common genome variations?

1. Single-nucleotide polymorphism: a variation of a single nucleotide in a specific position of the genome

2. indels: insertion and deletion

3. Block substitutions - GCATT > GTTCT

4. Inversion

5. Copy number variations: regions of the genome full of repeats. The number of repeats will vary between individuals and are prone to instability.

5

What are three kinds of repetitive sequences?

1. Tandem repeats: repeats of genes or blocks of genes
-we can tell how recent this repeat occurred by viewing the sequence homology between the repeat and the original. If there is >90% sequence identity, then that is a recent repeat.

2. Short repeats: each within a region of the genome which are present in many copies
a. satellite sequences - repeats of sequences that are around a few hundred base pairs long, mostly at centromere and telomeres
b. microsatellite - repeats of a few nucleotides

3. Retrotransposons: repeats which are a result of reverse transcription.
-mRNA that is transcribed is then reverse transcribed by a retrovirus to make cDNA. THis cDNA can then be incorporated into the human genome. This accounts for 25% of increased complex of the human genome.

6

Describe the mechanism behind red-green color blindness and the implication of it.

The implication is that repeats (tandem repeat) creates hot spots for recombination during meiosis.

When homologues match up they may match incorrectly and that will result in incorrect recombination and subsequent aneuploidies.

Red green color receptors are coded on the X chromosome. During recombination, the red and green receptor genes have high sequence homology so they match up and recombination moves the green receptor gene to the one with red and green. One gamete will then only have the red receptor gene. If this was a male, they'd 100% be colorblind.

7

Describe the 3 types of retrotransposons.

1. LINE is RNA that get reverse transcribed into cDNA and integrated into the DNA. This DNA encodes for reverse transcriptase.

2. SINE are small bits of RNA, when incorporated they don't really do anything other than contain a restriction cut site for AluI

3. Pseudogenes: copies of cellular mRNA will not be transcribed because they lack promoter sequences. They will have no function unless they disrupt a gene.

8

What are the 3 karyotyping techniques?

1. G-banding, incubate cells with colchicine which arrests cells in metaphase. You stain with Giemsa dye. Chromosomes are identified by three criteria:
a. size (1 is largest, 22 is smallest)
b. centromere position
c. banding pattern
Function: can detect large changes in chromosomes regardless of location, low resolution, so can't detect small changes.

2. FISH: chromatin or chromosomes are fixed onto a slide, probes are designed specific for each chromosome or region of the chromosome and binds to that complementary DNA. (Denature and hybridize)
-interphase FISH is faster because do it directly on sample rather than incubate with colchicine for metaphase FISH, however lower resolution because DNA is not condensed.

Function: it indicates if the Region complementary to the general probe is present but however says nothing about the presence of a single nucleotide deletion or mutations in other parts of the genome. A specific probe can reveal small deletions. Often times you use two probes, a control for detecting the chromosome and a more specific probe for the suspected mutation.
-Disadvantage; you have to know the sequence or chromosome that you are looking for, cannot detect single nucleotide changes.

3. Comparative Genome Hybridization: you have a microarray with all the sequences of the genome. Then you incubate control (red) with patient (green). If it is yellow then they are equally expressed and that is normal. If there is more red, then there is a deletion. If there is more green then there is a insertion.

Function: can detect small changes across the entire genome, detects on increases and decreases and cannot detect rearrangments.

9

Describe the three kinds of centromere positions

1. Metacentric: the centromere is right in the middle and the p and q arms are both same size.

2. Sub-metacentric: the centromere is closer to one end

3. Acrocentric: the centromere is much closer to one end,

10

Describe chromosome bands and numbering.

Theyre numbered on each arm extending out of the centromere.

example from top to bottom.
3
2
1
()
1
2
3
4
5
so within each band there is more numbering.
6q31.1 for example, band 3 of q arm of chromosome six, within it third band, within that .1

11

What is DiGeorge Syndrome?

Where is Prader-willi and Angelman syndrome mutation located?

Deletion on chromosome 22, q arm band 11.

del 22q11.

Failure of pharyngeal pouches to develop,

Deletion of chromosome 15 on q arm bands 11-13

12

What is euploidy, aneuploidy, and the exceptions to aneuploidy.

Translocations commonly occur between acrocentric autosomal chromosomes because their short p arms mostly contain genes for ribosomal RNA which has a large copy number (recombination vulnerable(

Euploidy: normal number of chromosomes: 46, XX

Aneuploidy, extra or missing chromosome

Aneuploidy is usually lethal except for three cases

1. X or Y chromosomes
-Klinefelter (XXY)
-Turner syndrome (XO)
-all other monosomies are lethal.

2. trisomy for some autosomes (13, 18, 21)
Down syndrome: 47, XY, +21

triploidy is lethal.

13

How is the identity of the chromosome and the number of chromosomes determined?

Its all in the centromeres. The number of centromeres is the number of chromosomes. The identity of the centromere is the chromosome

If you have centromere 22 and most of it is replaced with chromosome 1 arms, it is still chromosome 22.

14

How would you write a male with a reciprocal translocation between chromosome 1 and 3, at q arm 31 and q arm 24 respecitively.

46, XY, t(1;3)(q31;q24)

15

What is Robertsonian Translocation and how is it written?

The breakpoint is in the centromeres of chromosomes 14 and 21. The p arms are lost and then they fuse the two q arms.

A carrier is normal phenotype because both are represented. and losing p arms is not lethal.

45, XY, -14, -21, +rob (14q;21q)

The problem stems from meiosis. When homologous chromosomes align, they form a paring between three chromosomes (2 normal and robertsonian) The only okay segregation would be if the gamete contains only robertsonian chromosome and then it would be a normal carrier and the other gamete would have normal 14,21. The rest are bad. Trisomies with extra 14 or 21 is often lethal and monosomies as well.


16

Describe isochromosome 21 and whats the nomenclature.

45, XX, t(21;21)(q10,q10)

Q arms near the centromere attach onto one another and you lose the p arms. In this case either you get a trisomy or monosomy 21. (lethal)

So all births are down syndrome.

17

What is pericentric and how is it different from paracentric inversion.

Nomenclature for a pericentric inversion with break points in chromosome 6 and at p23 and q21

46, XY, inv(6) (p23;q21)

An inversion on the same chromosome.

Paracentric means the breakpoints are on the same arm and don't include the centromere.

18

How would you describe a duplication on chromosome 8 from q13 to the terminus of the q arm.

46, XX, dup8(q13 → qter)

19

How would you write a deletion on chromosome 19 from q13.1 to q13.3

46, XY, del19(q13.1;q13.3)

20

how many average genes per pair of chromosome?

Average length of a gene?

1. The number of genes in the genome = 22,000

22,000 / 23 = about 1000

around 1000 genes per pair of chromosome.


2. The entire genome is 3.2 x 10 ^ 9 base pairs long.

(3.2 x 10^9) / (22,000) = 150,000 base pairs per gene.

20

A patient carries a single paternal allele and two copies of the same maternal allele, what does this mean for meiosis?

A nondisjunction in maternal meiosis II.

Homozygosity for the centromere is a sign of non-disjunction in maternal meiosis II.

single paternal allele and two different maternal alleles would be nondisjunction at meiosis I. Heterozygosity.

21

If an individual is heterozygous for an inversion, why is there no progeny with recombinant genotypes for genes within the inversion?

Recombination within an inversion creates acentric and dicentric chromosomes which lead to an unbalanced gene contact. This is usually lethal.

Inversion wouldn't prevent synapsis or recombination.