Genetics

Question 1

Nucleosome

Answer 1

8 histones wrapped in DNA

Question 2

Chromatin

Answer 2

The entire DNA/protein complex (including small amount of rna)

Question 3

Heterochromatin

Answer 3

Chromatin that is tightly condensed

Question 4

Euchromatin

Answer 4

Chromatin that is uncoiled and able to be transcribed.

Only coiled during nuclear division.

Question 5

DNA methylation

Answer 5

Most common example of epigenetic regulation through chemical change.
Involves the addition of an extra methyl group to particular cytosine nucleotides.
Causes DNA to be wound more tightly so those sections cannot be transcribed.

Question 6

of chromosomes in human somatic cell

Answer 6

46 double stranded DNA molecules

23 homologous pairs of chromosomes

Question 7

Homologues

Answer 7

Two chromosomes that code for the same traits

Question 8

Diploid

Answer 8

Cell that contains homologous pairs of chromosomes

Question 9

Haploid

Answer 9

Cell that doesn’t contain homologues

Question 10

Histones

Answer 10

Globular proteins that sections of DNA not in use are wrapped tightly around.
Have basic functional groups that give these proteins a net positive charge at the normal pH of the cells.
Net positive charge attracts the negatively charged DNA and assists in the wrapping process.

Question 11

Cell Cycle - G0

Answer 11

non-growing phase
most cells spend majority of life in this phase
most protein production take place during this phase when the cell is not exerting energy in self-replicating

Question 12

Transcription

Answer 12

process by which RNA is manufactured from a DNA template.
3 main stages: initiation, elongation and termination.
Main level of activation or deactivation of genes.

Question 13

Transcription: Initiation

Answer 13

a group of DNA binding proteins called transcription factors identifies a promoter (a sequence of DNA nucleotides that designates a beginning point for transcription.
major enzyme of transcription is RNA polymerase. it unzips the DNA double helix.

Question 14

Transcription: Elongation

Answer 14

RNA polymerase transcribes one strand of the DNA nucleotide sequence into a complementary RNA nucleotide sequence. Reads DNA 3’ - 5’, builds RNA 5’ - 3’.
Transcribed strand is template or (-) antisense strand, other strand is coding or (+) sense strand.
No proof reading mechanism (errors in RNA are not called mutations and not transmitted to progeny).

Question 15

Transcription: Termination

Answer 15

end of transcription.

Question 16

Post-transcriptional processing

Answer 16

Takes place in the nucleus.
5’ cap serves as an attachment site in protein synthesis during translation and as a protection against degradation by enzymes that cleave nucleotides.
3’ end is similarly protected from exonucleases by the addition of a long series of adenine nucleotides (poly A tail).
Introns are removed through splicing which involves snRNPs (and lariat).

Question 17

Alternative splicing

Answer 17

allows the cell to incorporate different coding sequences (omitting certain exons, incorporating certain introns, utilizing variable splicing sites) into the mature mRNA.
an important contributor to the diversity of protein products.

Question 18

Start codon

Answer 18

5’ - AUG - 3’

Question 19

Stop codons

Answer 19

5’ - UAA, UAG, or UGA - 3’

Question 20

Translation

Answer 20

takes place using a ribosome (free-floating in cytosol or attached to outer surface of ER).
3 stages: initiation, elongation and termination.
within each ribosome there are 3 sites where tRNA can bind. A site - Anticodon matches up with the codon. P site - a Peptide bond between amino acids is formed. E site - the tRNA which now lacks an amino acid can Exit the ribosome.

Question 21

Cell Cycle - S phase

Answer 21

DNA replication takes place during the S phase of the interphase portion of the cell cycle.

Question 22

5 main steps of DNA replication

Answer 22

1. Helicase unzips double helix
2. RNA polymerase (primase) builds a primer
3. DNA polymerase assembles the leading and lagging strands
4. RNAse H removes the primers
5. DNA ligase joins the okazaki fragments together.

Question 23

Telomeres

Answer 23

repeated 6 nucleotide units that protect the ends of chromosomes.

Question 24

Mitosis

Answer 24

nuclear division without genetic change.

stages: PMAT

Question 25

Prophase

Answer 25

characterized by condensation of chromatin into chromosomes.
centrioles move to opposite poles of the cell.
First the nucleolus and then the defined nucleus disappear as the nuclear envelope breaks down.
Spindle apparatus begins to form, consisting of asters and spindle microtubules.

Question 26

Metaphase

Answer 26

chromosomes align along the equator of the cell.

Question 27

Anaphase

Answer 27

Sister chromatids split at their attaching centromeres and segregate to opposite sides of the cell. This split is termed disjunction.
Cytokinesis, the actual separation of the cellular cytoplasm due to constriction of microfilaments around the center of the cell, may or may not commence towards the end of this phase.

Question 28

Telophase

Answer 28

nuclear membrane reforms followed by reformation of the nucleolus.
chromosomes decondense
cytokinesis continues.

Question 29

silent mutation

Answer 29

type of neutral mutation in which the amino acid sequence is unchanged.

Question 30

missense mutation

Answer 30

base substitution changes a codon

Question 31

nonsense mutation

Answer 31

change to nucleotide sequence creates a stop codon where none previously existed.

Question 32

frameshift mutation

Answer 32

occurs when deletions or additions occur in multiples other than three. often result in completely non-functional proteins.

Question 33

Meiosis I

Answer 33

separate homologous chromosomes to produce two haploid cells (reductive division). Proceeds similarly to mitosis.

Question 34

Prophase I

Answer 34

homologous chromosomes line up and may exchange sequences of DNA nucleotides in a process called crossing over.
when genes on the same chromosome are located close together, they are more likely to cross over together and are said to be linked.

Question 35

Metaphase I

Answer 35

two homologues remain attached and move to the metaphase plate. 23 tetrads lined up in humans.

Question 36

Anaphase I

Answer 36

homologous chromosomes each separate from their partner independently assorting to create two haploid cells.

Question 37

Telophase I

Answer 37

nuclear membrane may or may not reform and cytokinesis may or may not occur. In humans, both do. When cytokinesis occurs, the new cells are haploid with 23 replicated chromosomes and are called secondary spermatocytes or secondary oocytes. In the case of female, one of the oocytes (first polar body) is much smaller and degenerates.

Question 38

Meiosis II

Answer 38

proceeds through PMAT II appearing much like mitosis. Final products are haploid gametes each with 23 chromosomes.
In the case of spermatocyte, four sperm cells are formed.
In the case of oocyte, a single ovum is formed after degeneration of the polar bodies (telophase II produces one gamete and second polar boy).

Question 39

Nondisjunction

Answer 39

when, during anaphase I or II the centromere of any chromosome does not split.

Question 40

Gamete production in females

Answer 40

Oogonium (diploid) undergoes mitosis to produce two primary oocytes (takes place before birth). Primary oocytes remain arrested in prophase I until puberty. In preparation for ovulation, a primary oocyte completes meiosis I producing a secondary oocyte (haploid). Secondary oocyte begins meiosis II but is arrested in metaphase II and only completes meiosis II when penetrated by a sperm during fertilization.

Question 41

Hardy-Weinberg Principle

Answer 41

conditions of a population to be in hard-weinberg equilibrium:
1. No selection for the fittest organism
2. Random mating
3. large population
4. immigration or emigration must not change the gene pool
5. mutational equilibrium
p2 + 2pq + q2 = 1 (p represents frequency of dominant alleles and q represents frequency of recessive alleles)
p + q = 1