Unit 3: Genetics

Question 1

Compare and contrast the structure of DNA vs RNA.

Answer 1

Both are polymers made of nucleotides in the shape of strands
However,
DNA is double stranded, RNA is single stranded
DNA contains the sugar deoxyribose (Deoxyribose Nucleic Acid = DNA), RNA contains the sugar ribose (Ribose Nucleic Acid = RNA)
DNA has bases of A, T, G, C, but RNA has the bases A, U, G, C
DNA forms a helix with it’s complementary bases while RNA does not

Question 2

Define nucleotides.

Answer 2

The basic building block of nucleic acids such as DNA and RNA
Composed of a pentose sugar, a nitrogenous base and a phosphate group
The pentose sugar + nitrogenous base is a nucleoside

Question 3

State the complementary base pairing rule.

Answer 3

Adenine (A) always pairs with Thymine (T)
Guanine (G) always pairs with cytosine (C)

Question 4

State Chargaff’s rule.

Answer 4

The ratio of purines to pyrimidines must be equal
More specifically,
% of A = % of T
% of G = % of C
(A and G are purines, T and C are pyrimidines)

Question 5

State the 7 main components of DNA replication.

Answer 5

Helicase
Single Strand Binding (SSB) proteins
Gyrase
RNA primase
DNA polymerase III
DNA polymerase I
DNA ligase

Question 6

Describe the function of helicase, SSB proteins and gyrase in DNA replication.

Answer 6

Helicase: unwinds strands by breaking the hydrogen bonds

SSB proteins: bind to DNA strands after helicase separation to prevent them from re-annealing (rewinding into a helix)
These fall off once replication is complete

Gyrase: reduces torsional strain created when helicase unwinds DNA

Question 7

Describe the function of DNA/RNA primase, DNA polymerase III, DNA polymerase I, DNA ligase. Include Okazaki fragments in your explanation.

Answer 7

RNA primase (aka DNA primase): binds to exposed DNA strand and generates a short RNA primer with a length of ~10-15 nucleotides

DNA polymerase III: binds to the RNA primer and extends it to create a complementary strand; this is required as DNA polymerase can extend a nucleotide chain but not start one
On the leading strand, DNA pol III moves towards the replication fork (spot where strands disconnect) and continuously synthesizes
On the lagging strand, DNA pol III moves away from the replication fork and must synthesize discontinuously
This occurs because DNA pol III can only synthesize in the 5’-3’ direction

DNA polymerase I: the lagging strand has multiple RNA primers placed in order to repeatedly initiate synthesis via DNA pol III
DNA pol I removes the RNA primers and replaces them with DNA nucleotides
These repeated pieces are called Okazaki fragments

DNA ligase: joins Okazaki fragments together by joining sugar-phosphate backbones together with a phosphodiester bond

Question 8

What did Meselson and Stahl prove about DNA replication, and how did they prove this?.

Answer 8

That DNA replication is semi-conservative; each new piece of DNA consists of one old strand and one new strand

Other, disproved methods of replication:
Dispersive: the new DNA alternates between new and old DNA
Conservative: the new strand is completely brand new

Process:
- E. coli was cultured in the presence of a heavy nitrogen isotope, N-15
- It was transferred to a medium with a lighter isotope, N-14
- After taking DNA samples several generations later, they subjected the samples to caesium chloride equilibrium density gradient centrifugation
- This formed lines on test tubes based on the weight of the sample - N-14 drawing a higher line, N-15 drawing a lower line
- Testing of the first generation resulted in a line down the middle of the N-14 and N-15 sections, disproving conservative (which would have one strand at N-14 and one at N-15)
- Testing of the second generation resulted in a line at the N-14 level and one between N-14 and N-15, disproving dispersive (which would only have one band at the level between N-14 and N-15)
- Thus, they concluded DNA replicated semi-conservatively

Question 9

Define template/antisense strand, coding/sense strand, and mRNA strand, in the context of DNA transcription.

Answer 9

The mRNA strand is the strand that copies, or transcribes, the template/antisense strand
The other DNA strand, the one that is not copied but rather attached to the antisense strand is called the sense strand
The sense strand and the mRNA strand are sequentially equal, as they are both complementary to the antisense strand, but the mRNA strand has a U for every T in the sense strand

Question 10

State the three phases of DNA transcription.

Answer 10

Initiation, elongation, termination

Question 11

Describe the initiation phase of DNA transcription.

Answer 11

RNA polymerase binds to a promoter site, a site on DNA that indicates the start region of transcription

Question 12

Describe the elongation phase of DNA transcription.

Answer 12

RNA polymerase moves along the coding region in the 5’-3’ direction
As it does so, it uncoils the DNA and synthesizes an mRNA complementary to the antisense strand

Question 13

Describe the termination phase of DNA transcription.

Answer 13

The RNA polymerase unbinds from the DNA once it reaches and recognizes a terminator site
The sense and antisense DNA strands then re-anneal, or reconnect

Question 14

Describe the concept of post-transcriptional modifications, when they are used, and provide two examples.

Answer 14

Prokaryotes do not need post-transcriptional modifications after DNA transcription, but eukaryotes do
Eukaryotes need post-transcriptional modifications to protect the newly transcribed mRNA molecule from conditions and enzymes outside the nucleus
Two major examples include:
Poly-A tail: a chain of adenine nucleotides to protect the mRNA from enzymes in the cytosol
5’ cap: a sequence of seven guanines placed on the 5’ end that are recognizable by ribosomes

Question 15

Describe how small nuclear ribonucleoproteins (snRNP’s) remove introns, and include exons in your explanation.

Answer 15

Small nuclear ribonucleoproteins (snRNP’s) remove introns without cutting any nucleotides from the exons
Spliceosomes are when snRNPS attach to DNA
The snRNP’s recognize the intron areas and loop them such that the adjacent exons are touching
The spliceosomes are activated and cut the intron off of the DNA strand, where it loops back on itself and eventually degrades
The spliceosome also links the now-touching exons together

Question 16

State the codon that signals the start of protein translation.

Answer 16

AUG, corresponding amino acid methionine (Met)

Question 17

Define codon.

Answer 17

Also known as a triplet, it is a set of three consecutive bases in a DNA or RNA sequence
Together, they code for a singular amino acid

Question 18

Describe the initiation phase of DNA translation.

Answer 18

A small, ribosomal subunit binds to the 5’ end of the mRNA and moves along until it finds the start codon, AUG
The appropriate tRNA molecule, carrying the amino acid Met, binds to the AUG codon
The large ribosomal subunit, which consists of three sites, E, P and A, aligns itself such that the tRNA molecule is in its P (middle) site, and forms a complex with the small subunit

Question 19

Describe the translocation phase of DNA translation.

Answer 19

After elongation, the ribosome moves 1 codon forward
The tRNA that was in the P site has now moved to the E site, and is detached
The next required tRNA molecule comes in and attaches to the unoccupied A site
The bond between the tRNA and its AA is broken and the energy forms a bond between the two adjacent AAs
This process repeats until the termination phase

Question 20

Describe the termination phase of DNA translation.

Answer 20

When the ribosome reaches a stop codon, a release factor is signaled for translation to stop
The polypeptide is released from the ribosome and the ribosome disassembles back into its small and large subunit

Question 21

Define nucleosomes.

Answer 21

Eukaryotic DNA is packaged with histone proteins to created a compact structure called a nucleosome, which help supercoil DNA in order to achieve a more compacted structure that allows for more efficient storage

Question 22

Outline how the rate of DNA transcription can be increased or decreased.

Answer 22

Adding methyl to DNA, a process called methylation, decreases DNA transcription rate:
- Causes nucleosomes to pack tightly together
- Disallows transcription factors to bind to the DNA
- Prevents genes from being expressed

Adding acetyl groups to the DNA strands, a process called acetylation, increases DNA transcription rate:
- Causes looser packing of nucleosomes
- Allows transcription factors to more easily bind to DNA
- Allows genes to be more easily expressed

Question 23

Define reprogramming.

Answer 23

Methylation and acetylation mark the DNA to affect transcription by using markers called epigenetic tags
Since new organisms need unmarked DNA to develop unspecialized, reprogramming involves erasing most epigenetic tags to allow a fertilized egg to properly develop

Question 24

State the three noncoding regions of DNA.

Answer 24

Promoter
Operator
Terminator

Question 25

Discuss the role of the operator noncoding region in DNA.

Answer 25

Operators are regions that affect the transcription rate of nearby genes
There are two sub-regions, called enhancers and silencers
- Enhancers increase rate of transcription of corresponding gene when bound to by an activator
- Silencers inhibit transcription of corresponding gene when bound to by a repressor

Question 26

Describe PCR and provide one real-world application.

Answer 26

PCR stands for Polymerase Chain Reaction and allows scientists to get vast amounts of DNA by amplifying minute amounts repeatedly

Three steps:
- Denaturation
- DNA sample is heated to separate the two strands
- Annealing
- Sample is cooled and primers are annealed at the ends strands
- Elongation
- Sample is heated to the optimal temperature for a heat-tolerant polymerase (Taq) to function
- The DNA has now been artificially replicated

Taq polymerase is an enzyme from the bacterium Thermus aquaticus
It has a high optimal temperature, allowing it to function during the elongation step
It is able to extend the nucleotide chain from the primers, allowing the sequence to duplicate

Allows for:
- Identification of dead
- Identifying criminals at crime scenes using minute amounts of DNA
- Sequencing the DNA of extinct species and other life forms

Question 27

Define gene, allele, locus, genome, genotype and phenotype.

Answer 27

Gene: small section of DNA that determines a characteristic
Allele: a slightly different yet specific variation of a gene
Locus: an exact location of a gene
Genome: entire genetic information of an organism
Genotype: set of alleles you have for a gene
Phenotype: the trait you get as a result of the genotype

Question 28

Define autosomal chromosomes and sex chromosomes, and state the amount of chromosomes a human has as well as the sex chromosomes of a female and male human.

Answer 28

Sex chromosomes: chromosomes that determine the sex of an organism
XX –> female
XY –> male

Autosomal chromosomes: chromosomes that are not sex chromosomes

Humans have 23 pairs of chromosomes (46 total)
22 pairs are autosomal, last pair is sex chromosomes

Question 29

Define mutations, and state two causes of them.

Answer 29

Mutation: the changing of the structure of a gene, resulting in a new variant
Causes:
- Mistakes in DNA replication
- Exposure to high energy radiation
- Exposure to chemicals that might cause mutations (mutagens)

Question 30

Define a base-substitution mutation, and give a prominent example.

Answer 30

A mutation that changes one nitrogenous base in a sequence
Example: sickle cell anemia

Question 31

Compare the number of protein-coding genes of:
- E. coli (bacteria)
- Homo sapiens (humans)
- Oryza sativa japonica (rice)
- Triticum aestivum (wheat)
- Gallus gallus (chicken)

Answer 31

Organism: approx. number of protein-coding genes

Bacteria/E. coli: 4000
Humans/Homo sapiens: 20 000
Rice/Oryza sativa japonica: 35 000
Wheat/Tritucum aestivum: 100 000
Chicken/Gallus gallus: 20 000

Question 32

Describe the difference between a person with sickle cell anemia and an unaffected person.

Answer 32

The altered gene is called HBB, located on chromosome 11
This gene codes the beta subunits of hemoglobin, a protein involved in the transport of oxygen
The standard, unaffected Hb^A allele reads G A G at the 6th codon of the sense DNA strand
The Hb^S allele, the affected one, reads G T G
When the HBB locus is transcribed, the mRNA from Hb^A has the codon G A G, which codes for the amino acid (AA) glutamic acid
When mRNA transcribes Hb^S, it copies the codon G U G, which codes for the AA valine
Hemoglobin formed from glutamic acid is usually the same as that formed from valine, but under certain conditions, such as low oxygen, they behave differently
Hemoglobin formed from glutamic acid functions fine
Hemoglobin formed from valine gets polymerized into long fibres, significantly reducing its ability to carry oxygen, and changing the shape of the red blood cell to be that of a sickle
The shape of the cell clogs blood vessels, causing pain, failure of blood supply and strokes
Furthermore, sickled cells are broken down and eliminated by the body, straining the liver by doing so
Healthy blood cells must be made, straining bone structure as red blood cells are made from bone marrow

Question 33

Describe the differences between somebody with a heterozygous genotype Hb^A Hb^S, and when it is preferred to somebody with homozygous genotypes Hb^A Hb^A or Hb^S Hb^S

Answer 33

Hb^A Hb^S: cells will only sickle when infected by the parasite that triggers malaria
Malaria is killed along with sickled blood cells, offering protection against malaria
Preferred genotype in areas with malaria

Hb^A Hb^A: no sickling, but no protection from malaria
Hb^S Hb^S: cells will sickle under extreme conditions, and offers no protection from malaria

Question 34

Describe prokaryotic DNA (number of copies of genes, name, shape).

Answer 34

DNA only consists of one chromosome, which carries out the entire genome
- Only one copy of each gene, unlike in eukaryotes

Shape: closed loop, like a rubber band
Name: naked DNA, as it is not organized around histone proteins

Question 35

Define a plasmid and where they are found.

Answer 35

Plasmids are found in some prokaryotes as small loops of DNA
They may contain genes, which are often related to very specific functions, such as digesting food or resisting a very specific antibiotic

Not often found in eukaryotes, with some exceptions (e.g. Saccharomyces cerevisiae)

Question 36

Compare and contrast prokaryotic and eukaryotic DNA.

Answer 36

Prokaryotic DNA takes the form of a closed loop, whereas eukaryotic DNA is long and linear
Prokaryotic DNA is not found in a membrane-bound nucleus, whereas eukaryotic DNA is
Prokaryotes only contain one copy of each gene, whereas eukaryotes contain two

Question 37

Define chromatin, sister chromatids, centromere, homologous chromosomes/pair and homologue

Answer 37

Chromatin: the form DNA takes on during interphase, when DNA is being actively used
Sister chromatids: DNA takes on a X-shape structure during the first stages of mitosis and meiosis; the two lines that form the X are identical copies of the chromosome called sister chromatids
- Note that these are renamed to daughter chromosomes after separation (anaphase)
Centromere: the place where sister chromatids connect
Homologous chromosomes/pair: the two versions of the same chromosome (one inherited maternally, one inherited paternally)
Homologue: one chromosome in a homologous pair

Question 38

Compare the genome sizes of:
- T2 phage (virus that infects bacteria)
- E. coli (common bacterium)
- Drosophila melanogaster (fruit fly)
- Homo sapiens (human)
- Paris japonica (woodland plant)

Answer 38

Organism: genome size, in millions of base pairs
T2 phage: 180k
E. coli: 5 mil
Drosophila melanogaster: 140 mil
Homo sapiens: 3.2 bil
Paris japonica: 150 bil

Question 39

Why does genome size not always correlate with the number of genes?

Answer 39

Though there exists a general correlation, there are many exceptions
This is due to some parts of chromosomes not coding for polypeptides, especially in eukaryotes

Question 40

Define diploid, haploid, gamete, sperm, egg, and zygote.

Answer 40

Diploid: containing two complete sets of chromosomes
Haploid: containing only one set of chromosomes

Gamete: a male or female reproductive cell capable of fusing with a gamete of the opposite sex to create a zygote
For males, gametes are sperms
For females, gametes are ova (ovum singular) aka eggs

Zygote: a fertilized egg cell that is the result of the union of two gametes of the opposite se, that is able to develop into an organism

Question 41

Compare the diploid chromosome numbers of:
Parascaris equorum (equine roundworm)
Oryza sativa (rice plant)
Homo sapiens (human)
Pan troglodytes (chimpanzee)
Canis familiaris (domestic dog)

Answer 41

Organism: diploid chromosome number

Parascaris equorum (equine roundworm): 4
Oryza sativa (rice plant): 24
Homo sapiens (human): 46
Pan troglodytes (chimpanzee): 48
Canis familiaris (domestic dog): 78

Question 42

Define a karyogram and karotype.

Answer 42

Karyotype:
The arrangement of homologous chromosomes in an organism

Karyogram:
A photograph of a karyotype, ordered by decreasing lengths, taken during metaphase of mitosis
Homologous chromosomes are aligned by length, location of centromere and by bands of colour differences produced with dyes

Question 43

Define nondisjunction, and the role of karyotyping in regards to it.

Answer 43

Nondisjunction is when sister chromatids fail to split during meiosis, resulting in a zygote having an extra or missing chromosome, conditions called trisomy and monosomy respectively
Karyotyping, the act of taking a karyogram, allows doctors to identify chromosomal abnormalities, such as trisomy and monosomy

Question 44

Describe how Down syndrome occurs, and why other trisomy disorders are not as well-known.

Answer 44

Down syndrome occurs when chromosome 21 of the human body has 3 copies; thus, it is also known as trisomy 21
Occurrences of nondisjunction in the majority of chromosomes are fatal and result in infertility, which explains why they are not as well known
Down syndrome allows its gamete to develop, and its harms are comparatively less impactful

Question 45

Describe a disadvantage of karyotyping (what it cannot detect), and provide an example of such.

Answer 45

Karyotyping cannot detect differences at the gene-level, only large-scale chromosomal differences such as trisomy and monosomy
For example, it cannot detect base substitution mutations such as sickle cell anemia

Question 46

Describe how Cairn measured the length of E. coli’s DNA.

Answer 46

He used autoradiography, a technique that uses X-ray film to visualize the 2D distribution of a radioactively labelled structure
The produced image is called a autoradiograph

The process:
Thymidine was produced with radioactive hydrogen isotope H-3
E. coli was grown in the presence of this
The cells were lysed (burst open), spilling out its DNA onto slides
The slides were covered with photographic emulsion and stored in the dark for two months
Over the course of this period, high-energy electrons emitted by the radioactive decay of the H-3 isotope within the DNA caused the appearance of dark spots on the photographic emulsion sheet
The pattern of dark spots showed the appearance of DNA, which he then measured to be 1 mm, about 1000 times longer than the typical E. coli cell

Question 47

Compare and contrast asexual and sexual reproduction (briefly).

Answer 47

Asexual reproduction: there is only one parent, and it produces a clone (genetically identical organism) through binary fission
Sexual reproduction: combines genetic information of two parents through meiosis; combines two opposite sex gametes to form a zygote

Question 48

List all phases of meiosis.

Answer 48

MEIOSIS I:
Prophase I
Metaphase I
Anaphase I
Telophase I

MEIOSIS II:
Prophase II
Metaphase II
Anaphase II
Telophase II

Question 49

Fill in the blanks: meiosis consists of ___ ________ nucleus splitting into ____ _______ nuclei.

Answer 49

Meiosis consists of one diploid nucleus splitting into four haploid nuclei.
Note that mitosis produces cells, but meiosis produces nuclei!

Question 50

Describe meiosis, and name the phases.

Answer 50

Meiosis I:
Prophase I:
Nucleolus and nuclear membrane disintegrate
Chromosomes supercoil and are thus visible
Homologous chromosomes are duplicated, and each forms two sister chromatids
All four total sister chromatids collectively are called tetrads or bivalents
Non-sister chromatids inside the same tetrad may cross over at points as chiasmata and exchange equivalent segments of DNA
Centrioles, if present, migrate to opposite poles, and spindle fibres form

Metaphase I:
Homologous pairs move together along the metaphase plate or equator
Maternal and paternal homologues show random orientation in regards to facing the poles
The spindle fibres connect each centromere to a single pole
In metaphase I, there is no splitting or pulling apart chromosomes

Anaphase I:
Spindle fibres shorten and pull homologues away from their respective partner towards opposite poles

Telophase I:
The chromatids uncoil partially and a nuclear membrane reforms around each new nucleus

Meiosis II occurs next:
It is very similar to meiosis I, with a few exceptions:
There is no crossing over in prophase II, as the homologous chromosomes were separated
Metaphase II does not feature random orientation
Anaphase II involves spindle fibres separating - pulling apart - sister chromatids, similar to mitosis’s anaphase
Sister chromatids, like in mitosis, become daughter chromosomes after pulled apart

Note that the final products, four haploid daughter cells, are genetically distinct, unlike in mitosis, which are genetically identical

Question 51

Describe the process of crossing over, and when it occurs.

Answer 51

Crossing over occurs in prophase I of meiosis
When homologous chromosomes in a tetrad line up next to each other (an event known as synapsis), crossing over can take place
Homologous pairs of sister chromatids exchange genes at chiasmata (single: chiasma), which creates new combinations of alleles
E.g. two homologous chromosomes with genes AB and ab could crossover to become Ab and aB
Crossing over contributes to genetic variation, as it introduces new, usually unique alleles

Question 52

Describe the process of random orientation/assortment, and when it occurs.

Answer 52

When pairs of homologous chromosomes line up at the equator of a cell during metaphase I, both the maternal and paternal copy have an equal chance of facing each pole, allowing random assortment as the other stages of meiosis carry through
This contributes to genetic variation

Question 53

State the correlation between the age of the mother during pregnancy and the amount of chromosomal abnormalities.

Answer 53

Positive correlation; as mother’s age increases, the amount of chromosomal abnormalities tends to increase as well
There is also a similar correlation between the father’s age and nondisjunction, but less compelling, as research suggests that, out of the cases of Down syndrome through nondisjunction, 90% comes from the mother, and 10% from the father

Question 54

Define fetal karyotyping and state the two methods of doing so.

Answer 54

Fetal karyotyping: checking the karyotype before someone is born, or karyotyping a fetus
Methods:
Aminiocentesis
Chorionic villus sampling (CVS)

Question 55

Describe the process of amniocentesis, as well as its risks and the time period it is performed.

Answer 55

Performed between weeks 14 to 20 of a pregnancy
Ultrasound imagery is used to guide a syringe needle through the mother’s abdomen and uterine wall, without piercing the fetus
The needle extracts a minute amount of amniotic fluid
This fluid contains fetal cells, which are cultured and karyotyped
Risks:
- Leaking amniotic fluid
- Miscarriage
- Needle injury to fetus
- Rh sensitization
- Infection/infection transmission

Question 56

Describe the process of chorionic villus sampling (CVS), as well as its risks and the time period it is performed.

Answer 56

Earlier in pregnancies (during weeks 10-13) there is not enough amniotic fluid to perform amniocentesis, so CVS is used
Ultrasound imagery is used to guide a medical professional during sampling
A suctioning tool (catheter or syringe) is inserted through the vagina or abdomen to reach the fetal cells in the chorion
- The chorion is the membrane that surrounds the fetus and develops into part of the placenta
The fetal cells are sampled, cultured and karyotyped
Risks:
- Miscarriage
- Rh sensitization
- Infection
Poses a higher risk of failure than amniocentesis

Question 57

State Mendel’s three laws.

Answer 57

The Law of Segregation: the inheritance of each characteristic of an individual is a randomly selected combination of their parents’ alleles

The Law of Independent Assortment: the allele of one trait does not affect the inherited allele of any other trait
Exception: genes whose loci are close together, called linked genes

The Law of Dominance: in an organism with two different alleles (heterozygous), only one allele will determine the trait: the dominant one
The unexpressed is called recessive
Exceptions: codominance, incomplete dominance

Question 58

Define homozygous and heterozygous, as well as dominant, recessive and codominant alleles.

Answer 58

Heterozygous: the alleles an individual has are different (e.g. Aa)
Homozygous: the alleles an individual has are the same (e.g. AA, aa)
Dominant allele: the allele that determines the organism’s phenotype, denoted with a capital letter (e.g. A, b))
Recessive allele: the allele that is overshadowed by the dominant allele and is only expressed when an organism is homozygous with a recessive allele, denoted with a lowercase letter (e.g. a, b)
Codominant alleles: alleles that will not suppress nor be suppressed by one another, and will instead both be expressed, denoted with a capital letter that indicates the trait followed by a capital superscript that indicates the allele (e.g. C^W, C^R)

Question 59

Define incomplete dominance and codominance.

Answer 59

Incomplete dominance: when two alleles have their traits “blended” to produce an intermediate phenotype
E.g. if C^W = white petals, C^R = red petals, then the genotype C^W C^R = pink petals
Codominance: when two alleles are codominant, if an organism is heterozygous, both will be equally expressed
E.g. if C^W = white petals, C^R = red petals, then the genotype C^W C^R = red-and-white spotted petals

Question 60

Name the four main blood groups (not including positive/negative)

Answer 60

A, AB, B, O

Question 61

Name the three alleles that determine the blood group of an individual, as well as the relation between the alleles (dominant, recessive, etc.) and the required genotypes for each group

Answer 61

Alleles: I^A, I^B, and i
I^A and I^B are codominant with respect to each other, but dominant with respect to i
Genotype: group
I^A I^A or I^A i: group A
I^A I^B: group AB
I^B I^B or I^B i: group B
ii: group O

Question 62

State the blood types that each blood type can be donated (excluding positive/negative), and why this is the case.

Answer 62

Excluding positive and negative…
Type A can receive type A or O
Type B can receive type B or O
Type AB can receive type AB, A, B or O
Type O can only receive type O
This is because of antigens and antibodies:
Antigens are molecules in blood that serve as identifiers for the body
If the body detects a foreign antigen in the blood, it sends antibodies to attack it
Antigens have specific markers that depend on the blood type they are in, to ensure there isn’t any self-sabotage
For example, type A blood has antigens with A markers, therefore they have antibodies that attack any antigen except A
Type AB has both A and B markers on their antigen, and type O does not have any markers
The lack of markers on type O antigens gives it two notable characteristics:
- It will never be attacked by any antibody, thereby allowing type O blood to be the universal donor
- However, it will contain antibodies that attack both A and B markers, thereby disallowing type A, B or AB blood to be donated to it

In contrast, the possession of both markers for type AB blood allows it to receive any type of blood (lack of anti-A and anti-B antibodies) but its blood will always be attacked when donated to an individual that is not type AB blood (markers will get attacked by both anti-A and anti-B antibodies)

Question 63

State the patterns of genetic diseases.

Answer 63

Autosomal recessive
Autosomal dominant
X-linked dominant
X-linked recessive
Y-linked

Question 64

Provide an example and description of an autosomal recessive and an autosomal dominant disease.

Answer 64

Autosomal recessive: cystic fibrosis
Occurs due to a mutation in the CFTR gene on chromosome 7, which codes for a chloride channel in mucous membranes
The altered mucus builds up in the lungs, causing infections and damage to lung tissue, and the liver and pancreas, making it more difficult to digest food

Autosomal dominant: Huntington’s disease
Neurodegenerative disorder that starts to affect people between 30-50
Occurs due to a mutation in the HTT gene on chromosome 4
Symptoms include loss of muscle coordination, cognitive decline and psychiatric problems

Question 65

Provide two examples and descriptions of an X-linked dominant sex-linked genetic disease.

Answer 65

Red-green colourblindness (aka protanopia)
Self-explanatory

Hemophilia
A protein needed for blood clotting, usually factor VIII, is not made
Small injuries may result in major blood loss

Question 66

Outline the conventions of a pedigree chart.

Answer 66

Females are circles
Males are squares
A shaded shape means the individual is affected by the disease under inspection
Parents and children are connected with a T shape
Sometimes, but not often, half-shaded shapes indicates a carrier (when dominant allele suppresses effect of recessive disease)
Roman numerals identify generations, and Arabic numerals identify individuals within a generation
Exact genotypes are not known, but phenotypes are (if they have the disease or not)

Question 67

Describe mutagens, and how they affect alleles.

Answer 67

A mutagen is any agent that causes/increases the frequency of mutations by triggering changes in the genetic material of an organism
Includes chemicals and high-level radiation
Could potentially form new, harmful alleles, and increase chances of cancer (in which case the mutagens are also called carcinogens)
Most mutations made by mutagens will not be passed to offspring, as they were made to non-sexual cells
They will only be passed on if present in the DNA of the cells that produce gametes

Question 68

Describe radiation and its effect on genes. Provide two prominent examples of radiation exposure.

Answer 68

High energy radiation, also known as ionizing radiation, can break bonds between atoms including DNA, which causes mutations
- E.g. UV light, X-rays, and alpha, beta and gamma radiation from the decay of radioactive materials

August 6, 1945, an atomic bomb devastated the city of Hiroshima, Japan
- Released a large amount of radiation, causing many deaths and cases of Acute Radiation Syndrome (ARS)

1986, the Chernobyl nuclear power station in Ukraine accidentally melted down, releasing a significant amount of radiation, much more than in Hiroshima
The radioactive material is still present to date
It resulted in a clear increase of thyroid cancer in children

Question 69

Discuss why the observed outcomes of a genetic cross may not conform to the predicted outcomes, outside of the context of linked genes.

Answer 69

Probability
The smaller the sample size, the more randomness affects results
Larger sample sizes tend to conform to predicted outcomes more

Question 70

Describe continuous and discontinuous variation.

Answer 70

Continuous variation: when there can be infinitely many possible phenotypes produced by genes
- Think continuous data, no concrete numbers
- Skin colour, height, etc.

Discontinuous variation: when there are a finite amount of possible phenotypes produced
- Think discrete data, concrete numbers
- Blood type, earlobe attachment

Question 71

Define monogenic and polygenic, and state the assumptions of the polygenic model.

Answer 71

Monogenic: when a trait is determined by only one gene loci
Tend to exhibit discontinuous variation

Polygenic: when a trait is determined by more than one gene loci
Tend to exhibit continuous variation

Assumptions:
Each contributing gene has a small, relatively equal and additive effect
The genes behave as if they are codominant
There is no linkage involved
The value of the trait depends solely upon genetics; environmental influences are ignored

Question 72

Define genetic linkage and recombinant or variant gametes.

Answer 72

Genetic linkage: the tendency of certain genes to be inherited together
Genetic loci that are on the same chromosome are physically close and tend to stay together during meiosis, thus making them genetically linked
Denoted with vertical pairs separated by a double line
When crossing over occurs, two linked genes may swap places (e.g. AB // ab –> Ab // aB)
These alternate alleles can create recombinant or variant gametes
These occur at a smaller rate compared to normal gametes, due to their dependency of crossing over

Question 73

Describe the process of DNA sequencing.

Answer 73

By using a technique called the Dideoxy Chain Termination Method, scientists take advantage of the fact that adding dideoxy nucleotides to a DNA strand stops DNA synthesis
Dye is added to four dideoxynucleotides (ddNTP’s), which are added to the DNA sequence such that, when the DNA synthesis stops, the final base sequence is recognized by the dye shown
Scientists do this repeatedly by using PCR (mass DNA cloning) and stop synthesis at every position to conclude the pattern of bases in a segment of DNA