Final Exam

Question 1

Outline the structure of nucleosomes

Answer 1

DNA is wound around an octamer of histones (146 bases and 1.65 turns of the helix per octamer)
The octamer and DNA combination is secured to a H1 histone, forming a nucleosome

Question 2

nucleosomes

Answer 2

They protect DNA from damage
They allow long lengths of DNA to be packaged (supercoiled) for mobility during mitosis / meiosis
When supercoiled, DNA is not accessible for transcription
Cells will have some segments of DNA permanently supercoiled (heterochromatin) and these segments will differ between different cell types

Question 3

Intron

Answer 3

A non-coding sequence of DNA within a gene (intervening sequence) that is cut out by enzymes when RNA is made into mature mRNA

Question 4

Exon

Answer 4

The part of the gene which codes for a protein (expressing sequence)

Eukaryotic DNA contains introns but prokaryotic DNA does not

Question 5

Helicase

Answer 5

Unwinds the DNA and separates the two polynucleotide strands by breaking the hydrogen bonds between complementary base pairs
The two separated polynucleotide strands act as templates for the synthesis of new polynucleotide strands

Question 6

DNA Polymerase

Answer 6

Synthesises new strands from the two parental template strands
Free deoxynucleoside triphosphates (nucleotides with three phosphate groups) are aligned opposite their complementary base partner and are covalently bonded together by DNA polymerase to form a complementary nucleotide chain
The energy for this reaction comes from the cleavage of the two extra phosphate groups

Question 7

DNA replication occurs in a 5’ - 3’ direction

Answer 7

DNA replication is semi-conservative, meaning that a new strand is synthesised from an original template strand
DNA replication occurs in a 5’ - 3’ direction, in that new nucleotides are added to the C3 hydroxyl group such that the strand grows from the 3’ end
This means that the DNA polymerase enzyme responsible for adding new nucleotides moves along the original template strand in a 3’ - 5’ direction

Question 8

Helicase

Answer 8

unwinds and separates the double stranded DNA by breaking the hydrogen bonds between base pairs
This occurs at specific regions (replication origins), creating a replication fork of two polynucleotide strands in antiparallel directions

Question 9

RNA primase

Answer 9

synthesises a short RNA primer on each template strand to provide an attachment and initiation point for DNA polymerase III

Question 10

DNA polymerase III

Answer 10

adds deoxynucleoside triphosphates (dNTPs) to the 3’ end of the polynucleotide chain, synthesising in a 5’ - 3’ direction
The dNTPs pair up opposite their complementary base partner (adenine pairs with thymine ; guanine pairs with cytosine)
As the dNTPs join with the DNA chain, two phosphates are broken off, releasing the energy needed to form a phosphodiester bond
Synthesis is continuous on the strand moving towards the replication fork (leading strand)
Synthesis is discontinuous on the strand moving away from the replication fork (lagging strand) leading to the formation of Okazaki fragments

Question 11

DNA polymerase I

Answer 11

removes the RNA primers and replaces them with DNA

Question 12

DNA ligase

Answer 12

joins the Okazaki fragments together to create a continuous strand

Question 13

DNA transcription in terms of the formation of an RNA strand complementary to the DNA strand by RNA polymerase

Answer 13

sequence of DNA that is transcribed into RNA is called a gene,Transcription occurs in the nucleus (where the DNA is) and, once made, the mRNA moves to the cytoplasm (where translation can occur)
Messenger RNA (mRNA):  A transcript copy of a gene used to encode a polypeptide
Transfer RNA (tRNA):  A clover leaf shaped sequence that carries an amino acid
Ribosomal RNA (rRNA):  A primary component of ribosomes

Question 14

transcription direction

Answer 14

works in a 5 to 3

reads in a 3 to 5

Question 15

translation

Answer 15

process of protein synthesis in which the genetic information encoded in mRNA is translated into a sequence of amino acids in a polypeptide chain

Question 16

sense and antisense strands of DNA

Answer 16

antisense strand is transcribed into RNA
Its sequence will be complementary to the RNA sequence and will be the “DNA version” of the tRNA anticodon sequence
The sense strand is not transcribed into RNA
Its sequence will be the “DNA version” of the RNA sequence (identical except for T instead of U)

Question 17

promoter region

Answer 17

Responsible for the initiation of transcription (in prokaryotes, a number of genes may be regulated by a single promoter - this is an operon)

Question 18

Coding Sequence

Answer 18

The sequence of DNA that is actually transcribed (may contain introns in eukaryotes)

Question 19

Terminator

Answer 19

Sequence that serves to terminate transcription (mechanism of termination differs between prokaryotes and eukaryotes)

Question 20

Transcription

Answer 20

RNA polymerase binds to the promoter and causes the unwinding and separation of the DNA strands
Nucleoside triphosphates (NTPs) bind to their complementary bases on the antisense strand (uracil pairs with adenine, cytosine pairs with guanine)
RNA polymerase covalently binds the NTPs together in a reaction that involves the release of two phosphates to gain the required energy
RNA polymerase synthesises an RNA strand in a 5’ - 3’ direction until it reaches the terminator
At the terminator, RNA polymerase and the newly formed RNA strand both detach from the antisense template, and the DNA rewinds
Many RNA polymerase enzymes can transcribe a DNA sequence sequentially, producing a large number of transcripts
Post-transcriptional modification is necessary in eukaryotes

Question 21

Pre-Initiation:

Answer 21

Specific tRNA-activating enzymes catalyse the attachment of amino acids to tRNA molecules, using ATP for energy

Question 22

Initiation

Answer 22

The small ribosomal subunit binds to the 5’ end of mRNA and moves along it until it reaches the start codon (AUG)
Next, the appropriate tRNA molecule binds to the codon via its anticodon (according to complementary base pairing)
Finally, the large ribosomal subunit aligns itself to the tRNA molecule at its P-site and forms a complex with the small ribosomal subunit

Question 23

Elongation

Answer 23

The small ribosomal subunit binds to the 5’ end of mRNA and moves along it until it reaches the start codon (AUG)
Next, the appropriate tRNA molecule binds to the codon via its anticodon (according to complementary base pairing)
Finally, the large ribosomal subunit aligns itself to the tRNA molecule at its P-site and forms a complex with the small ribosomal subunit

Question 24

Elongation

Answer 24

A second tRNA molecule pairs with the next codon in the ribosomal A-site
The amino acid in the P-site is covalently attached via a peptide bond to the amino acid in the A-site

Question 25

Translocation:

Answer 25

The ribosome moves along one codon position, the deacylated tRNA moves into the E-site and is released, while the tRNA bearing the dipeptide moves into the P-site Another tRNA molecules attaches to the next codon in the newly emptied A-site and the process is repeated The ribosome moves along the mRNA sequence in a 5' - 3' direction, synthesising a polypeptide chain Multiple ribosomes can translate a single mRNA sequence simultaneously (forming polysomes)

Question 26

Termination

Answer 26

Elongation and translocation continue until the ribosome reaches a stop codon These codons do not code for any amino acids and instead signal for translation to stop The polypeptide is released and the ribosome disassembles back into subunits The polypeptide may undergo post-translational modification prior to becoming a functional protein

Question 27

7.4.7 State that free ribosomes synthesise proteins for use primarily within the cell, and that bound ribosomes synthesise proteins primarily for secretion or for lysosomes

Answer 27

Ribosomes floating freely in the cytosol produce proteins for use within the cell Ribosomes attached to the rough ER are primarily involved in producing proteins to be exported from the cell or used in the lysosome These proteins contain a signal recognition peptide on their nascent polypeptide chains which direct the associated ribosome to the rough ER

Question 28

Prophase

Answer 28

DNA supercoils, chromosomes are each comprised of two genetically identical sister chromatids joined at a centromere,centrosomes move to opposite poles of the cell and spindle fibres begin to form between them (in animals, each centrosome contains 2 centrioles) The nuclear membrane is broken down and disappears

Question 29

Metaphase

Answer 29

Spindle fibres from the two centrosomes attach to the centromere of each chromosome Contraction of the microtubule spindle fibres cause the chromosomes to line up separately along the centre of the cell (equatorial plane)

Question 30

Anaphase

Answer 30

Continued contraction of the spindle fibres cause the two sister chromatids to separate and move to the opposite poles of the cell Once the two chromatids in a single chromosome separate, each constitutes a chromosome in its own right

Question 31

Telophase

Answer 31

Once the two sets of identical chromosomes arrive at the poles, the spindle fibres dissolve and a new nuclear membrane reforms around each set of chromosomes The chromosomes decondense and are no longer visible under a light microscope The division of the cell into two daughter cells (cytokinesis) occurs concurrently with telophase

Question 32

stages in a cell cycle

Answer 32

The cell cycle is an ordered set of events that culminates in cell growth and division into two daughter cells It can roughly be divided into two main stages: Interphase The stage in the development of the cell between two successive M phases This phase of the cell cycle is a continuum of 3 distinct stages (G1, S, G2), whereby the cell grows and matures (G1), copies its DNA (S) and prepares for division (G2) Sometimes cells will leave the cell cycle and enter into a quiescent state (G0), whereby it becomes amitotic and no longer divides M phase The periods of nuclear division (mitosis) and cytoplasmic division (cytokinesis)

Question 33

2.5.2 State that tumours (cancers) are the result of uncontrolled cell division and that these can occur in any organ or tissue

Answer 33

The cell cycle is controlled by a complex chemical control system that responds to signals both inside and outside of the cell Tumor suppressor genes produce proteins which inhibit cell division, while proto-oncogenes produce proteins that promote growth and division Mutations to these genes result in uncontrolled cell division, resulting in the formation of a tumour Tumours can grow in size which causes damage local tissue; they may also spread to other parts of the body (malignant tumours) Diseases caused by the growth of tumours are collectively known as cancers

Question 34

2.5.3 State that interphase is an active period in the life of a cell when many metabolic reactions occur, including protein synthesis, DNA replication and an increase in the number of mitochondria and chloroplasts

Answer 34

Interphase is an active period in the life of a cell - many events need to occur before a cell can successfully undergo division: Protein synthesis: The cell needs to synthesise key proteins and enzymes to enable it to grow, copy its contents and then divide ATP production: The cell will need to generate sufficient quantities of ATP in order to successfully divide Increase number of organelles: The cell needs to ensure both daughter cells will have the necessary numbers of organelles needed to survive DNA replication: The genetic material must be faithfully duplicated before division (this occurs during the S phase) As none of these processes can occur during the M phase, interphase contains growth checkpoints to ensure division is viable G1: A checkpoint stage before DNA replication during which the cell grows, duplicates organelles, synthesises proteins and produces ATP S: The stage during which DNA is replicated G2: A checkpoint stage before division during which the copied DNA is checked for fidelity (mutations) and final metabolic reactions occur

Question 35

2.5.5 Explain how mitosis produces two genetically identical nuclei

Answer 35

During interphase (the S phase) the DNA was replicated to produce two copies of genetic material These two identical DNA molecules are identified as sister chromatids and are held together by a single centromere During the events of mitosis (as described in 2.5.4), the sister chromatids are separated and drawn to opposite poles of the cell When the cell divides (cytokinesis), the two resulting nuclei will each contain one of each chromatid pair and thus be genetically identical

Question 36

2.5.6 State that growth, embryonic development, tissue repair and asexual reproduction involve mitosis

Answer 36

Growth: Multicellular organisms increase their size by increasing their number of cells through mitosis Asexual reproduction: Certain eukaryotic organisms may reproduce asexually by mitosis (e.g. vegetative reproduction) Tissue Repair: Damaged tissue can recover by replacing dead or damaged cells Embryonic development: A fertilised egg (zygote) will undergo mitosis and differentiation in order to develop into an embryo

Question 37

4.2.1 State that meiosis is a reduction division of a diploid nucleus to form haploid nuclei

Answer 37

Meiosis is the process by which sex cells (gametes) are made in the reproductive organs: Most sexually reproducing animals are diploid - meaning they have two copies of every chromosome (one of maternal origin, one of paternal origin) In order to reproduce, these organisms need to make gametes that are haploid (have only one copy of each chromosome) Fertilisation of two haploid gametes (egg + sperm) will result in the formation of a diploid zygote that will grow into a new organism Meiosis consists of two cell divisions: The first division is a reduction division of the diploid nucleus to form haploid nuclei The second division separates sister chromatids (this division is necessary because meiosis is preceded by interphase, wherein DNA is replicated)

Question 38

4.2.2 Define homologous chromosomes

Answer 38

Homologous chromosomes are chromosomes that share: The same structural features (e.g. same size, same banding pattern, same centromere position) The same genes at the same loci positions (while genes are the same, alleles may be different)

Question 39

4.2.3 Outline the process of meiosis, including pairing of homologous chromosomes and crossing over, followed by two divisions, which results in four haploid cells

Answer 39

The process of meiosis involves two divisions, both of which follow the same basic stages as mitosis (prophase, metaphase, anaphase and telophase) Meiosis is preceded by interphase, which includes the replication of DNA (S phase) to create chromosomes with genetically identical sister chromatids

Question 40

Meiosis I

Answer 40

Homologous chromosomes must first pair up in order to be sorted into separate haploid daughter cells In prophase I, homologous chromosomes undergo a process called synapsis, whereby homologous chromosomes pair up to form a bivalent (or tetrad) The homologous chromosomes are held together at points called chiasma (singular: chiasmata) Crossing over of genetic material between non-sister chromatids can occur at these points, resulting in new gene combinations (recombination) The remainder of meiosis I involves separating the homologous chromosomes into separate daughter cells In metaphase I, the homologous pairs line up along the equator of the cell In anaphase I, the homologous chromosomes split apart and move to opposite poles In telophase I, the cell splits into two haploid daughter cells as cytokinesis happens concurrently

Question 41

Meiosis II

Answer 41

In meiosis II, the sister chromatids are divided into separate cells In prophase II, spindle fibres reform and reconnect to the chromosomes In metaphase II, the chromosomes line up along the equator of the cell In anaphase II, the sister chromatids split apart and move to opposite poles In telophase II, the cell splits in two as cytokinesis happens concurrently

Question 42

4.2.4 Explain that non-disjunction can lead to a change in chromosome number, illustrated by reference to Down syndrome (trisomy 21)

Answer 42

Non-disjunction refers to the chromosomes failing to separate correctly, resulting in gametes with one extra, or one missing, chromosome (aneuploidy) The failure of the chromosomes to separate may either occur via: Failure of homologues to separate during Anaphase I (resulting in four affected daughter cells) Failure of sister chromatids to separate during Anaphase II (resulting in two affected daughter cells)

Question 43

down-syndrome

Answer 43

ndividuals with Down syndrome have three copies of chromosome 21 (trisomy 21) One of the parental gametes had two copies of chromosome 21 as a result of non-disjunction The other parental gamete was normal and had a single copy of chromosome 21 When the two gametes fused during fertilisation, the resulting zygote had three copies of chromosome 21, leading to Down syndrome

Question 44

4.2.5 State that, in karyotyping, chromosomes are arranged in pairs according to their structure

Answer 44

A karyotype is a visual profile of all the chromosomes in a cell The chromosomes are arranged into homologous pairs and displayed according to their structural characteristics

Question 45

4.2.6 State that karyotyping is performed using cells collected by chorionic villus sampling or amniocentesis, for pre-natal diagnosis of chromosome abnormalities

Answer 45

Pre-natal karyotyping is often used to: Determine the gender of an unborn child (via identification of sex chromosomes) Test for chromosomal abnormalities (e.g. aneuploidies resulting from non-disjunction) Amniocentesis A needle is inserted through the abdominal wall, into the amniotic cavity in the uterus, and a sample of amniotic fluid containing foetal cells is taken It can be done at ~ 16th week of pregnancy, with a slight chance of miscarriage (~0.5%) Chorionic Villus Sampling A tube is inserted through the cervix and a tiny sample of the chorionic villi (contains foetal cells) from the placenta is taken It can be done at ~ 11th week of pregnancy, with a slight risk of inducing miscarriage (~1%)

Question 46

4.2.7 Analyse a human karyotype to determine gender and whether non-disjunction has occurred

Answer 46

Every cell in the human body has 46 chromosomes (except anucleate red blood cells and haploid gametes) Males (X,Y) and females (X,X) can be differentiated on the basis of their sex chromosomes Non-disjunction during gamete formation can lead to individuals with an abnormal number of chromosomes (aneuploidy) These disorders can be classified according to the chromosome number affected and the number of chromosomes present Analysing Karyotypes

Question 47

Meiosis I

Answer 47

Prophase I: DNA supercoils and chromosomes condense, nuclear membrane dissolves, homologous pairs form bivalents, crossing over occurs Metaphase I: Spindle fibres from centrioles (at poles) attach to centromeres of bivalent, bivalents line up along the equator of the cell Anaphase I: Spindle fibres contract and split the bivalent, homologous chromosomes move to opposite poles of the cell Telophase I: Chromosomes decondense, nuclear membranes may reform, cell divides (cytokinesis) forming two haploid daughter cells

Question 48

Prophase I:

Answer 48

DNA supercoils and chromosomes condense, nuclear membrane dissolves, homologous pairs form bivalents, crossing over occurs

Question 49

Metaphase I

Answer 49

Spindle fibres from centrioles (at poles) attach to centromeres of bivalent, bivalents line up along the equator of the cell

Question 50

Anaphase I

Answer 50

Spindle fibres contract and split the bivalent, homologous chromosomes move to opposite poles of the cell

Question 51

Telophase I

Answer 51

Chromosomes decondense, nuclear membranes may reform, cell divides (cytokinesis) forming two haploid daughter cells

Question 52

Interkinesis:

Answer 52

An optional rest period between meiosis I and meiosis II, no DNA replication occurs in this stage

Question 53

Meiosis II

Answer 53

Prophase II: Chromosomes condense, nuclear membrane dissolves (if reformed), centrioles move to opposite poles (perpendicular to previous poles) Metaphase II: Spindle fibres from centrioles attach to centromeres of chromosomes, chromosomes line up along the equator of the cell Anaphase II: Spindle fibres contract and split the chromosome into sister chromatids, chromatids (now called chromosomes) move to opposite poles Telophase II: Chromosomes decondense, nuclear membrane reforms, cells divide (cytokinesis) resulting in four haploid daughter cells

Question 54

Prophase II:

Answer 54

Chromosomes condense, nuclear membrane dissolves (if reformed), centrioles move to opposite poles (perpendicular to previous poles)

Question 55

Metaphase II

Answer 55

Spindle fibres from centrioles attach to centromeres of chromosomes, chromosomes line up along the equator of the cell

Question 56

Anaphase II

Answer 56

Spindle fibres contract and split the chromosome into sister chromatids, chromatids (now called chromosomes) move to opposite poles

Question 57

Telophase II

Answer 57

Chromosomes decondense, nuclear membrane reforms, cells divide (cytokinesis) resulting in four haploid daughter cells

Question 58

10.1.2 Outline the formation of chiasmata in the process of crossing over

Answer 58

Crossing over involves the exchange of segments of DNA between homologous chromosomes during Prophase I of meiosis The process of crossing over occurs as follows: Homologous chromosomes become connected in a process called synapsis, forming a bivalent (or tetrad) Non-sister chromatids break and recombine with their homologous partner, effectively exchanging genetic material (crossing over) The non-sister chromatids remain connected in an X-shaped structure and the positions of attachment are called chiasmata Chiasma hold homologous chromosomes together as a bivalent until anaphase I As a result of crossing over, chromatids may consist of a combination of DNA derived from both homologues - these are called recombinants

Question 59

Explain how meiosis results in an effectively infinite genetic variety in gametes through crossing over in prophase I and random orientation in metaphase I

Answer 59

During anaphase I, homologous chromosomes separate, such that each resultant daughter cell (and subsequent gametes) contains a chromosome of either maternal or paternal origin The orientation of these homologues in metaphase I is random, such that there is an equal probability of the daughter cell having either the maternal or paternal chromosome As humans have a haploid number of 23 chromosomes, this means that there is 223 potential gamete combinations (over 8 million combinations) Crossing over in prophase I results in entirely new chromosome combinations, as recombination through gene exchange produces wholly original chromosomes containing both maternal and paternal DNA, resulting in near infinite genetic variability Other sources of genetic variation include random fertilisations, DNA mutations, chromosome mutations and non-disjunction

Question 60

Outline the processes involved in spermatogenesis within the testes, including mitosis, cell growth, the two divisions of meiosis and cell differentiation

Answer 60

Spermatogenesis describes the production of spermatozoa (sperm) in the seminiferous tubules of the testes The first stage of sperm production requires the division of germline epithelium by mitosis These cells (spermatogonia) then undergo a period of growth This is followed by two meiotic divisions that result in four haploid daughter cells These haploid cells then differentiate to form sperm cells The developing sperm cells are nourished throughout by the Sertoli cells

Question 61

List three roles of testosterone in males

Answer 61

Pre-natal development of male genitalia Development of secondary sex characteristics Maintenance of sex drive (libido)

Question 62

11.4.3 State the role of LH, testosterone and FSH in spermatogenesis

Answer 62

LH: Stimulates the interstitial cells (Leydig cells) to produce testosterone FSH: Stimulates the (first) meiotic division of spermatogonia Testosterone: Stimulates the (second) meiotic division of spermatogonia and the maturation of spermatozoa through differentiation

Question 63

LH

Answer 63

Stimulates the interstitial cells (Leydig cells) to produce testosterone

Question 64

FSH

Answer 64

Stimulates the (first) meiotic division of spermatogonia

Question 65

Testosterone

Answer 65

Stimulates the (second) meiotic division of spermatogonia and the maturation of spermatozoa through differentiation

Question 66

Epididymis

Answer 66

Testicular fluids are removed, concentrating the sperm | Sperm mature and develop the ability to swim

Question 67

Seminal Vesicle

Answer 67

``` Adds nutrients (including fructose) for respiration Secretes prostaglandins, causing contractions to the female system and helping sperm move towards the egg ```

Question 68

Prostate Gland

Answer 68

Secretes alkaline fluid which neutralises vaginal acids (changes pH from 4 to 6 which aids sperm motility)

Question 69

11.4.5 Outline the processes involved in oogenesis within the ovary, including mitosis, cell growth, the two divisions of meiosis, the unequal division of cytoplasm and the degeneration of polar body

Answer 69

Oogenesis describes the production of female gametes (ova) within the ovary The process begins during foetal development, when a large number of cells (oogonia) are formed by mitosis before undergoing a period of growth These cells begin meiosis but are arrested in prophase I until puberty At puberty, some follicles continue to develop each month is response to FSH secretion These follicles complete the first meiotic division to form two cells of unequal size The cell with less cytoplasm is a polar body (which degenerates), while the larger cell forms a secondary oocyte The secondary oocyte begins the second meiotic division but is arrested in prophase II (until fertilisation) It is released from the ovary (ruptured follicle develops into corpus luteum) and, if fertilisation occurs, will complete meiosis The second meiotic division will produce an ovum and a second polar body

Question 70

Compare the processes of spermatogenesis and oogenesis, including the number of gametes and the timing of formation and release of gametes

Answer 70

Similarities: Both processes result in the formation of haploid gametes Both processes involve mitosis, growth and meiosis ``` Differences: Spermatogenesis Life long production of gametes, Oogenesis: fixed amount only 400 mature timing of gamete release sperm any time, oogen monthly cycle timing of formation: sperm: any time oog: once a month (menstural cycle ```

Question 71

7.3.2 Distinguish between the sense and antisense strands of DNA

Answer 71

The antisense strand is transcribed into RNA Its sequence will be complementary to the RNA sequence and will be the "DNA version" of the tRNA anticodon sequence The sense strand is not transcribed into RNA Its sequence will be the "DNA version" of the RNA sequence (identical except for T instead of U)