Unit 2- ch. 7, 8, 24, 10-12 Flashcards

Question

Understand why sex chromosome aneuploidy is more tolerated in humans than autosomal aneuploidy?

Answer 1

autosomal aneuploidies resulting in live births are less common because there is no mechanism of dosage compensation for autosomes -->most embryos with autosomal aneuploidies are spontaneously aborted --> aneuploidy of sex chromosomes is better tolerated than aneuploidy of autosomes (EX: turner or Klinefelter result from aneuploidy of sex chromosomes)

Answer 2

1. Primary down syndrome- (3 full copies of chromosome 21) arises from spontaneous nondisjunction during egg formation: about 75% of the nondisjunction events that cause Down syndrome are maternal in origin, most arising in meiosis I 2. Familial down syndrome- 4% of people with Down syndrome are not trisomic for a complete chromosome 21. instead, they have 46 chromosomes, but an extra copy of chromosome 21 is attached to another chromosome through translocation parents are carriers of chromosomes that have undergone a Robertsonian translocation (chromosome 14 and 21)

Answer 3

nondisjunction in a mitotic division may generate patches of cells in which every cell has a chromosome abnormality and other patches in which every cell has a normal karyotype this type of nondisjunction leads to a region of tissues with different chromosome constitutions, a condition known as genetic mosaicism EX: 50% of those with Tuner syndrome are mosaics, possessing some 45 X cells and some normal 46 XX cells

Answer 4

polyploidy: any organism that has more than two sets of chromosomes Autopolyploidy: all chromosome sets are from a single species Allopolyploidy: chromosome sets are from two or more species

Answer 5

DNA from different organisms varies greatly in base composition within each species, there is some regularity in the ratios of the bases: the amount of adenine is always equal to the amount of thymine (A = T) and the amount of guanine is always equal to the amount of cytosine (G = C)

Answer 6

disease-causing in smooth (S) form and nonvirulent rough form (R) -Griffith observed that small amounts of living type IIIS bacteria injected into mice cause them to die -When injected with IIR bacteria, they lived -Griffith knew that boiling killed all bacteria and destroyed their virulence; when he injected large amounts of heat-killed IIS bacteria, mice lived and no IIIS bacteria recovered from their blood surprise: -when injected mice with small amount of living type IIR and large amount of heat-killed type IIIS bacteria (both nonvirulent) -expected the mice to live but they died -live type IIIS bacteria observed in blood

Answer 7

Type IIR bacteria had somehow been transformed, acquiring the genetic virulence of the dead type IIIS bacteria, and this transformation had produced a permanent genetic change in the bacteria He theorized that some substance in the dead bacteria might be responsible. He called this substance the transforming principle

Answer 8

succeeded in isolating and partially purifying the transforming substance found that enzymes capable of destroying DNA eliminated the biological activity of the transforming substance Conclusion: the transforming principle - and therefore genetic information - resides in DNA

Answer 9

grew one batch of E. coli containing 32 P and infected bacteria with T2 phage so all the progeny phages would have DNA labeled with 32 P grew one batch of E. coli containing 35 S and infected these bacteria with T2 phage so that all the progeny phages would have proteins labeled with 35 S then infected separate batches of unlabeled E. coli with 35 S and 32 P progeny phages after allowing time for the phages to infect the E. coli cells, they placed the cells in a blender and sheared off the now-empty phage protein coats from the cell walls

Answer 10

most of the radioactivity was detected in the phage protein coats, and little was detected in the cells. when new phages emerged from the cells, they contained almost no 35 S results indicate that the protein component of a phage does not enter the cell and is not transmitted to progeny phages

Answer 11

when removed the phage protein coats, the bacteria were radioactive new progeny emerged and many of those phages emitted radioactivity demonstrating that DNA from the infecting phages had been passed on to the progeny phages

Answer 12

DNA, not protein, is the genetic material of the phages

Answer 13

a few viruses, use RNA, not DNA, as their genetic material demonstrated by Fraenkel-Conrat and Singer who worked with tobacco mosaic virus (TMV) which infects and causes disease in tobacco plants

Answer 14

found that after separating DNA and protein of TMV, could remix the DNA and protein of different strains of TMV and obtain intact, infectious viral particles these results showed that RNA carries the genetic information in TMV

Answer 15

Nucleotide: the primary structure of DNA consists of a string of nucleotides joined together by phosphodiester linkages 3 parts: 1. sugar 2. phosphate group 3. nitrogen-containing base

Answer 16

-2 polynucleotide strands twisted together (double helix) -strands antiparallel (run in opposite direction) -sugar-phosphates on outside (phosphodiester bonds) -bases on inside of helix -bases are complementary and joined by hydrogen bonds --> 2 bonds between A and T --> 3 bonds between G and C

Answer 17

single-stranded bases = G, C, A, U (you do not have to draw the structure of the bases) sugar = ribose

Answer 18

eukaryotic DNA in the cell is closely associated with proteins. This complex of DNA and protein is called chromatin. The two basic types of chromatin are euchromatin and heterochromatin

Answer 19

undergoes the normal process of condensation and decondensation in the cell cycle. constitutes the majority of the chromosomal material and is where most transcription takes place

Answer 20

remains in a highly condensed state throughout the cell cycle, even during interphase

Answer 21

Euchromatin: various condensing throughout cell cycle—where most transcription takes place Heterochromatin: remains very condensed throughout cell cycle—what are some areas of a chromosome that contain heterochromatin? -all chromosomes have permanent heterochromatin

Answer 22

the most abundant proteins in chromatin are the histones, which are small positively charged proteins of five major types: H1, H2A, H2B, H3, and H4

Answer 23

all histones have a high percentage of arginine and lysine, positively charged amino acids that give histones a net positive charge these positive charged attract the negative charges on the phosphates of DNA; this attraction holds the DNA in contact with the histones

Answer 24

-the simplest level of chromatin structure, the nucleosome; the basic structural unit of DNA packaging in eukaryotes -Nucleosome: octamer of 8 histones + DNA wrapped around histones 1. octamer = 2 H2A, 2 H2B, 2 H3, 2 H4 2. DNA wraps around octamer core ~2 times 3. H1 clamps DNA in place

Answer 25

new mitochondria and chloroplasts arise by the division of existing organelle; these divisions take place throughout the cell cycle and are independent of mitosis and meiosis

Answer 26

propose that mitochondria and chloroplast were once free-living bacteria that became internal inhabitants of early eukaryotic cells

Answer 27

when a heteroplasmic (two distinct varieties of DNA within the cytoplasm) cell divides, the organelles segregate randomly into the two progeny cells in a process called replicative segregation and chance determines the proportion of mutant organelles in each cell when replicative segregation takes place in somatic cells it may create phenotypic variation within a single organism: different cells of the organism may possess different proportion of mutant and wild-type sequences most of the replicated cells are heteroplasmic

Answer 28

overview: to determine which of the three models of replication applied to E.coli cells, Meselson and Stahl needed a way to distinguish old and new DNA What they did: used two isotopes of nitrogen 14N and 15N grew culture of E. coli with 15N and then switched the bacteria that contained 14N 1. bacteria synthesized before change was all 15N 2. any DNA synthesized after switch was 14N to distinguish between 15N and 14N used equilibrium density gradient centrifugation Results: found that DNA from bacteria grown in 15N produced a single band at the position expected of DNA containing only 15N DNA from bacteria transferred to 14N and produced a band intermediate between the expected DNA containing only 15N/14N Results were inconsistent with conservative replication model (predict one heavy and one light band) a single band intermediate is predicted by semiconservative/dispersive model)

Answer 29

to distinguish between two models, Meselson and Stahl grew bacteria containing only 14N for second generation two bands equal intensity appeared (one intermediate and one at the position expected of DNA containing only 14N) Meselson and Stahl’s results were exactly as expected for _semiconservative_ replication and were incompatible with those predicted for both _conservative and dispersive_ replication

Answer 30

1. Theta Model—common in bacteria, results in 2 circular DNA molecules 2. Rolling Circle Model—some viruses and bacteria, results in multiple circular DNA molecules 3. Linear Model—eukaryotic chromosomes, larger linear chromosomes in eukaryotic cells contain too much DNA to be replicated from a single origin

Answer 31

the raw materials from which new DNA molecules are synthesized consist of deoxyribose sugar and a base (nucleoside) attached to three phosphate groups

Answer 32

1. dATP 2. dTTP 3. dGTP 4. dCTP

Answer 33

nucleotides are added to the 3' OH group of the growing nucleotide strand 3' OH group of the last nucleotide on the strand attacks the 5' phosphate group of the incoming dNTP

Answer 34

-nucleotides added to 3’ OH of growing chain, so chain grows in a 5’ to 3’ direction -template strand is read in a 3’ to 5’ direction because it is antiparallel to the new strand

Answer 35

because the two single-stranded DNA templates are antiparallel and strand elongation is always 5' → 3', if synthesis on one template proceeds from right to left, synthesis on the other strand proceeds left to right

Answer 36

short lengths of DNA produced by discontinuous replication of lagging strand Okazaki fragments on the lagging strand are linked together to create a continues new DNA molecule

Answer 37

DNA unwinds, template strand exposed in 3’ → 5’ direction and allows the new strand to by synthesized continuously in 5’ → 3’ new strand, which undergoes continuous replication is called the leading strand

Answer 38

the other template strand is exposed in the 5’ → 3’ direction the synthesis must proceed 5’ → 3’ that is in the direction OPPOSITE that of unwinding runs out of template --> DNA synthesis must start anew at replication fork until it runs into the previously replicated segment of DNA -->process repeated newly made strand that undergoes discontinuous replication is lagging strand

Answer 39

four stage: initiation, unwinding, elongation and termination

Answer 40

single origin of replication initiator proteins bind to origin and cause a short section of DNA to unwind unwinding allows helicase and other single-strand binding proteins to attach to the polynucleotide strand

Answer 41

DNA synthesis requires a single-stranded template, so double-stranded DNA must be unwound before DNA synthesis can take place Helicase: breaks H bonds between the bases cannot initiate the unwinding once bound to single-stranded DNA, helicase moves in 5' → 3' direction SSBs: after DNA unwound by helicase, single-strand-binding proteins attach tightly to exposed single-stranded DNA protect single-stranded nucleotide chains and prevent formation of secondary structures bind to any single-stranded DNA Gyrase: control supercoiling of NDA type II topoisomerase (create double-strand breaks) reduce strain that builds up ahead of replication fork

Answer 42

requires 3’ OH in order to add a nucleotide cannot initiate DNA synthesis on bare template DNA polymerase III: synthesizes DNA from primers by adding nucleotides to 3’ end 5’→ 3’ polymerase activity (DNA synthesis) 3’→ 5’ exonuclease activity (proofreading) --> correct errors High processivity: can add many nucleotides without releasing the template DNA polymerase I: removes and replaces primers 5’ → 3’ polymerase activity 3’ → 5’ exonuclease activity 5’ → 3’ exonuclease activity (remove primers and replace with DNA nucleotides by synthesizing in 5’ → 3’ direction)

Answer 43

creates primer (RNA) to initiate DNA synthesis --> Primase = RNA polymerase; does not require 3’ OH to start the synthesis of nucleotide strand -Catalyzes the formation of a phosphodiester bond without adding another nucleotide to the strand

Answer 44

Usually occurs when replication forks run into each other

Answer 45

-present in germ cells and some somatic cells -BUT most somatic cells have little or no telomerase activity and chromosomes in these cells progressively shorten with each cell division telomerase plays a role in cancer - cancer cells have capacity to divide indefinitely and telomerase is expressed in 90% of all cancer

Answer 46

the protruding end of telomere can be extended by telomerase (protein and RNA component) telomerase extends the 3' end of the chromosome without the use of a complementary DNA template

Unit 2- ch. 7, 8, 24, 10-12 Flashcards

(70 cards)