Topic 3 Genetics

Question 1

Define “gene”.

3.1

Answer 1

A gene is a heritable factor that consists of a length of DNA
and infuences a specific characteristic.

Understanding: A gene is a heritable factor that consists of a length of DNA
and infuences a specific characteristic.

Question 2

Compare the number of genes in humans and other species.

3.1

Answer 2

Escherichia coli - 3,200
Drosophila melanogaster (Fruit fy) - 14,000
Homo sapiens (Humans) - 23,000
Daphnia pulex (Water fea) - 31,000
Oryza sativa (Rice) - 41,000

Application: Comparison of the number of genes in humans with other species.

Question 3

Outline where genes are found.

3.1

Answer 3

Each gene occupies a specific position on the type of chromosome where it is located. This position is called the locus of the gene.

Understanding: A gene occupies a specifc position on one type of chromosome.

Question 4

Define “allele”.

3.1

Answer 4

Alleles are alternative forms of the same gene, they occupy the same position on one type of chromosome
* They have the same locus
* Only one allele can occupy the locus of the gene on a chromosome.

Understanding: The various specifc forms of a gene are alleles.

Question 5

Outline the differences between alleles.

3.1

Answer 5

The different alleles of a gene have slight variations in the base sequence.
* Usually only one or a very small number of bases are different

Understanding: Alleles difer from each other by one or a few bases only.

Question 6

Outline how genetic variation can occur in people.

3.1

Answer 6

• “SNP” stands for single nucleotide polymorphism.
• SNP is when a single base pair is replaced with another
• Several snips can be present in a gene, but even then the alleles of the gene differ by only a few bases.

Understanding: Alleles difer from each other by one or a few bases only.

Question 7

Define “mutation”.

3.1

Answer 7

A gene mutation is a change in the nucleotide sequence of a section of DNA coding for a specific trait
* The most signifcant type of mutation is a base substitution.
* New alleles are formed by mutation

Understanding: New alleles are formed by mutation.

Question 8

Outline the types of mutations.

3.1

Answer 8

Gene mutations can be beneficial, detrimental or neutral
* Beneficial mutations change the gene sequence (missense mutations) to create new variations of a trait
* Detrimental mutations truncate the gene sequence (nonsense mutations) to abrogate the normal function of a trait
* Neutral mutations have no effect on the functioning of the specific feature (silent mutations)

Understanding: New alleles are formed by mutation.

Question 9

Outline the cause of the genetic disease of sickle cell anemia.

3.1

Answer 9

• Due to a mutation of the gene that codes for the alpha-globin polypeptide in hemoglobin.
• The symbol for this gene is Hb.
• Most humans have the allele HbA.
• If a base substitution mutation converts the sixth codon og the gene grom GAG to GTG, a new allele is gormed, called HbS.
• The mutation is only inherited by offspring if it occurs in a cell of the ovary or testis that develops into an egg or sperm.

Application: The causes of sickle cell anemia, including a base substitution mutation, a change to the base sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in hemoglobin.

Question 10

Outline the effect of the HbS allele being transcribed instead of the HbA allele.

3.1

Answer 10

• When the HbS allele is transcribed, the mRNA produced has GUG as its sixth codon instead of GAG
• The sixth amino acid in the polypeptide is valine instead of glutamic acid.
• change causes hemoglobin molecules to stick together in tissues with low oxygen concentrations.
• This distorts the red blood cells into a sickle shape.

Application: The causes of sickle cell anemia, including a base substitution mutation, a change to the base sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in hemoglobin.

Question 11

Outline the consequences of sickle cell anemia.

3.1

Answer 11

• The sickle cells may form clots within the capillaries, blocking blood supply to vital organs
• The sickle cells are also destroyed more rapidly than normal cells, leading to a low red blood cell count (anaemia)

Application: The causes of sickle cell anemia, including a base substitution mutation, a change to the base sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in hemoglobin.

Question 12

Define “genome”.

3.1

Answer 12

The genome is the whole genetic information of an organism.
* so a living organism’s genome is the entire base sequence of each of its DNA molecules.

Understanding: The genome is the whole of the genetic information of
an organism.

Question 13

Outline the HGP.

3.1

Answer 13

• The Human Genome Project began in 1990. Its aim was to fnd the base sequence of the entire human genome.
  **Findings: **
• Humans have ~23,000 genes, which was far less than expected.
• The majority of our genome is not transcribed/expressed.
• The genome that was sequenced was from one individual, there may be minor difference between individuals.
• Since the HGP, the genome of many other organisms has been sequenced

Understanding: The entire base sequence of human genes was sequenced in the Human Genome Project.

Question 14

Distunguish between prokaryote and eukaryote chromosomes.

3.2

Answer 14

Eukaryotes:
* linear DNA molecule
* associated with histone proteins
* no plasmids
* 2 or more different chromosomes

Prokaryotes:
* circular DNA molecule
* naked DNA
* plasmids often present
* only 1 chromosome

Understanding: Prokaryotes have one chromosome consisting of a circular DNA molecule.

Question 15

Define “plasmids”.

3.2

Answer 15

• Plasmids are small extra DNA molecules that are commonly found in prokaryotes but are very unusual in eukaryotes.
• They are usually small, circular and naked, containing a few genes that may be useful to the cell but not those needed for its basic life processes.
• Copies of plasmids can be transferred from one cell to another, allowing spread through a population.

Understanding: Some prokaryotes also have plasmids but eukaryotes
do not.

Question 16

Outline Cairn’s technique for measuring the length of DNA molecules by autoradiography.

3.2

Answer 16

1. Cells are grown in a solution containing radioactive thymidine (to only label DNA and not RNA)
2. Cells walls are digested to isolate the DNA and placed over a membrane
3. The membrane is coated with a photgraphic film and left in the dark for 2 months
4. When the film was developed, lines of black bts showed position of the DNA molecules from E.Coli

Application: Cairns’s technique for measuring the length of DNA molecules by autoradiography.

Question 17

Outline the finding’s of Cairns’s images.

3.2

Answer 17

• showed that the chromosome in E. coli is a single circular DNA molecule with a length of 1,100 µm.

Application: Cairns’s technique for measuring the length of DNA molecules by autoradiography.

Question 18

Outline chromosomes in eukaryotes.

3.2

Answer 18

• Chromosomes in eukaryotes are composed of DNA and protein.
• The DNA is a single immensely long linear DNA molecule.
• It is associated with histone proteins.

Understanding: Eukaryote chromosomes are linear DNA molecules
associated with histone proteins.

Question 19

Outline how chromosomes in eukaryote species are different.

3.2

Answer 19

• Each chromosome has a constriction point called a centromere, which divides the chromosome into two sections (or ‘arms’)
• Eukaryotic species possess multiple chromosomes that may differ in both their size and the position of their centromere
• Each chromosome will carry specific genes and the position of a particular gene on a chromosome is called the locus

Understanding: In a eukaryote species there are different chromosomes
that carry diferent genes.

Question 20

Define “homologous chromosomes”.

3.2

Answer 20

If two chromosomes have the same sequence of genes they are homologous. Homologous chromosomes are not usually identical to each other because, for at least some of the genes on them, the alleles are different.

Understanding: Homologous chromosomes carry the same sequence of
genes but not necessarily the same alleles of those genes.

Question 21

Outline how homologous chromosomes allow interbreeding.

3.2

Answer 21

If two eukaryotes are members of the same species, we can expect each of the chromosomes in one of them to be homologous with at least one chromosome in the other.
* This allows members of a species to interbreed.

Understanding: Homologous chromosomes carry the same sequence of
genes but not necessarily the same alleles of those genes.

Question 22

Compare the genome sizes of T2 phage, Escherichia coli, Drosophila melanogaster, Homo sapiens and Paris japonica.

3.2

Answer 22

• T2 phage (virus) - 0.18
• E.coli - 5
• Drosophila melanogaster (fruit fly) - 140
• Homo sapiens - 3,000
• Paris japonica (a plant) - 150,000

Genome size measured in million base pairs

Application: Comparison of genome size in T2 phage, Escherichia coli, Drosophila melanogaster, Homo sapiens and Paris japonica.

Question 23

Distinguish between haploid and diploid cells.

3.2

Answer 23

Diploid: nuclei has pairs of homologous chromosomes
Haploid: only has one chromosome of each type
* Gametes such as sperm and egg are haploid
* two haploids fuse during fertilization to produce one diploid cell (the zygote)

Understanding: Haploid nuclei have one chromosome of each pair.

Question 24

Contrast the number of chromosomes in haploid vs diploid nuclei in humans.

3.2

Answer 24

• Haploid nuclei in humans contain 23 chromosomes.
• Diploid nuclei in humans contain 46 chromosomes

Understanding: Diploid nuclei have pairs of homologous chromosomes.

Question 25

Outline what happens to the zygote.

3.2

Answer 25

• When the zygote divides by mitosis, more cells with diploid nuclei are produced.
• Many animals and plants consist entirely of diploid cells, apart from the cells that they are using to produce gametes for sexual reproduction.

Understanding: Diploid nuclei have pairs of homologous chromosomes.

Question 26

Outline the fundamental characteristic of a species.

3.2

Answer 26

One of the most fundamental characteristics of a species is the number of chromosomes.
* Organisms with a different number of chromosomes are unlikely to be able to interbreed so all the interbreeding members of a species need to have the same number of chromosomes.

Understading: The number of chromosomes is a characteristic feature
of members of a species.

Question 27

Compare the diploid chromosome numbers of Homo sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris equorum.

3.2

Answer 27

Parascaris equorum (horse) - 4
Oryza sativa (rice) - 24
Homo sapiens - 46
Pan troglodytes (chimpanzees) - 48
Cannis familiaris (dog) - 78

Application: Comparison of diploid chromosome numbers of Homo sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris equorum.

Question 28

Outline the 2 sex chromosomes in humans.

3.2

Answer 28

There are two chromosomes in humans that determine sex:
* the X chromosome is relatively large and has its centromere near the middle.
* the Y chromosome is much smaller and has its centromere near the end.

Understanding: Sex is determined by sex chromosomes and autosomes are chromosomes that do not determine sex.

Question 29

Outline how sex is determined.

3.2

Answer 29

• Females possess two copies of a large X chromosome (XX)
• Males possess one copy of an X chromosome and one copy of a much shorter Y chromosome (XY)

Understanding: Sex is determined by sex chromosomes and autosomes are chromosomes that do not determine sex.

Question 30

Distinguish between sex chromosomes and autosomes.

3.2

Answer 30

• Because the X and Y chromosomes determine sex they are called the sex chromosomes.
• All the other chromosomes are autosomes and do not affect whether a fetus develops as a male or female.

Understanding: Sex is determined by sex chromosomes and autosomes are chromosomes that do not determine sex.

Question 31

Outline how karyograms show the chromosomes of an organism.

3.2

Answer 31

The chromosomes of an organism are arranged into homologous pairs according to size in decreasing length (with sex chromosomes shown last)

Understanding: A karyogram shows the chromosomes of an organism
in homologous pairs of decreasing length.

Question 32

Define “karyotype”.

3.2

Answer 32

A karyotype is a property of an organism
* it is the number and type of chromosomes that the organism has in its nuclei.

Application: Use of karyotypes to deduce sex and diagnose Down
syndrome in humans.

Question 33

Outline how karyotypes can be used to deduce sex and down syndrome in humans.

3.2

Answer 33

1. To deduce whether an individual is male or female. If two XX chromosomes are present the individual is female whereas one X and one Y indicate a male.
2. To diagnose Down syndrome and other chromosome abnormalities. If there are three copies of chromosome 21 in the karyotype instead of two, the child has Down syndrome. This is sometimes called trisomy 21.

Application: Use of karyotypes to deduce sex and diagnose Down
syndrome in humans.

Question 34

Outline the process of meiosis.

3.3

Answer 34

The process of meiosis consists of two cellular divisions:
1. The first meiotic division separates pairs of homologous chromosomes to halve the chromosome number (diploid → haploid)
2. The second meiotic division separates sister chromatids (created by the replication of DNA during interphase)

Understanding: One diploid nucleus divides by meiosis to produce four
haploid nuclei.

Question 35

Outline the process of fertilization.

3.3

Answer 35

• In eukaryotic organisms, sexual reproduction involves the process of fertilization.
• Fertilization is the union of sex cells, or gametes, from two different parents.
• Fertilization doubles the number of chromosomes each time it occurs. (therefore halving of chromosome number occurs in meioisis creating gamtes).
• Fertilisation of two haploid gametes (egg + sperm) will result in the formation of a diploid zygote that can grow via mitosis

Understanding: The halving of the chromosome number allows a sexual
life cycle with fusion of gametes.

Question 36

Outline the replication of DNA before meiosis.

3.3

Answer 36

• DNA in the nucleus is replicated during the interphase before meiosis, so each chromosome consists of two sister chromatids.

Understanding: DNA is replicated before meiosis so that all chromosomes
consist of two sister chromatids.

Question 37

Outlin the formation of bivalents and crossing over.

3.3

Answer 37

1. Firstly homologous chromosomes pair up with each other. A pair of homologous chromosomes is bivalent and the pairing process is sometimes called synapsis.
2. Soon after synapsis, a process called crossing over takes place. A junction is created where one chromatid in each of the homologous chromosomes breaks and rejoins with the other chromatid.
3. Crossing over occurs at random positions anywhere along the chromosomes.
4. Because a crossover occurs at precisely the same position on the two chromatids involved, there is a mutual exchange of genes between the chromatids.
5. As the chromatids are homologous but not identical, some alleles of the exchanged genes are likely to be different. Chromatids with new combinations of alleles are therefore produced.

Understanding: The early stages of meiosis involve pairing of homologous
chromosomes and crossing over followed by condensation.

Question 38

Outline the process of random orientation.

3.3

Answer 38

• Each chromosome is attached to one pole only, not to both.
• The two homologous chromosomes in a bivalent are attached to different poles.
• The pole to which each chromosome is attached depends on which way the pair of chromosomes is facing. This is called the orientation.
• The orientation of bivalents is random, so each chromosome has an equal chance of attaching to each pole, and eventually of being pulled to it.

Understanding: Orientation of pairs of homologous chromosomes prior to
separation is random.

Question 39

Outline the seperation of homologous chromosomes in the first division of meiosis.

3.3

Answer 39

• Initially the two chromosomes in each bivalent are held together by chiasmata
• When homologous chromosomes are seperated and no longer held by chiamsata it is called disjunction.
• One chromosome from each bivalent moves to one of the poles and the other chromosome to the other pole.
• Seperation of pairs of homologous chromosomes to opposite poles of the cell halves the chromosome number of the cell.

Understanding: Separation of pairs of homologous chromosomes in the
first division of meiosis halves the chromosome number.

Question 40

Outline the process of Amniocentesis.

3.3

Answer 40

Amniocentesis involves passing a needle through the mother’s abdomen wall, using ultrasound to guide the needle.
* The needle is used to withdraw a sample of amniotic luid containing fetal cells from the amniotic sac.

Application: Methods used to obtain cells for karyotype analysis e.g. chorionic villus sampling and amniocentesis and the associated risks.

Question 41

Outline the process of chorionic villus sampling.

3.3

Answer 41

A sampling tool that enters through the vagina is used to obtain cells from the chorion, one of the membranes from which the placenta develops.

Application: Methods used to obtain cells for karyotype analysis e.g. chorionic villus sampling and amniocentesis and the associated risks.

Question 42

Outline the methods used to obtain cells for karyotype anaylysis.

3.3

Answer 42

• Amniocentesis
• chorionic villus sampling
• chorionic villus sampling can be done earlier in the pregnancy than amniocentesis, but whereas the risk of miscarriage with amniocentesis is 1%, with chorionic villus sampling it is 2%.

Application: Methods used to obtain cells for karyotype analysis e.g. chorionic villus sampling and amniocentesis and the associated risks.

Question 43

Outline the first division of meiosis.

3.3

Answer 43

Prophase I
* Cell has 2n chromosomes (double chromatid): n is haploid number of chromosomes.
* Homologous chromosomes pair (synapsis).
* Crossing over occurs.

Metaphase I
* Spindle microtubules move homologous pairs to equator of cell.
* Orientation of paternal and maternal chromosomes on either side of equator is random and independent of other homologous pairs.

Anaphase I
* Homologous pairs are separated. One chromosome of each pair moves to each pole.

Telophase I
* Chromosomes uncoil. During interphase that follows, no replication occurs.
* Reduction of chromosome number from diploid to haploid completed.
* Cytokinesis occurs.

Skill: Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.

Question 44

Outline how meiosis is a source of genetic variation.

3.3

Answer 44

1. Random orientation of bivalents (In metaphase I)
2. Crossing over (allows linked genes to be reshuffled, to produce new combinations)

Understanding: Crossing over and random orientation promotes genetic
variation.

Question 45

Outline how fusion of gametes from different parents promotes genetic variation.

3.3

Answer 45

• It allows alleles from two different individuals to be combined in one new individual.

Understanding: Fusion of gametes from diferent parents promotes
genetic variation.

Question 46

Define “non-disjunction”.

3.3

Answer 46

When homologous chromosomes fail to separate at anaphase.

Application: Non-disjunction can cause Down syndrome and other chromosome abnormalities.

Question 47

Outline the effect of non-disjunction.

3.3

Answer 47

• Both of the chromosomes move to one pole and neither to the other pole.
• The result will be a gamete that either has an extra chromosome or is defcient in a chromosome.
• If the gamete is involved in human fertilization, the result will be an individual with either 45 or 47 chromosomes.

Application: Non-disjunction can cause Down syndrome and other chromosome abnormalities.

Question 48

Outline the syndromes and chromosome abnormalities caused by non-disjunction.

3.3

Answer 48

• Trisomy 21, also known as Down syndrome: individual has three of chromosome number 21 instead of two.
• Trisomy 18 (Edwards Syndrome) and trisomy 13 (Patau’s Syndrome)
• Abnormal numbers of sex chromosomes (ex. Klinefelter’s syndrome is caused by having the sex chromosomes XXY. Turner’s syndrome is caused by having only one sex chromosome, an X.)

Application: Non-disjunction can cause Down syndrome and other chromosome abnormalities.

Question 49

Outline the trend between the age of parents infuences chances of non-disjunction

3.3

Answer 49

Studies show that the chances of non-disjunction increase as the age of the parents increase
* There is a particularly strong correlation between maternal age and the occurrence of non-disjunction events

Application: Studies showing age of parents infuences chances of non-disjunction

Question 50

Outline Mendel’s experiments

3.4

Answer 50

Gregor Mendel developed the principles of inheritance by performing experiments on pea plants.
* First, he crossed different varieties of purebred pea plants, then collected and grew the seeds to determine their characteristics
* Next, he crossed the offspring with each other (self-fertilization) and grew their seeds to similarly determine their characteristics
* These crosses were performed many times to establish reliable data trends (over 5,000 crosses were performed)

Understanding: Mendel discovered the principles of inheritance with experiments in which large numbers of pea plants were crossed.

Question 51

Outline the number of alleles in a gamete.

3.4

Answer 51

Gametes contain one chromosome of each type so are haploid.
* The nucleus of a gamete therefore only has one allele of each gene.

Understanding: Gametes are haploid so contain one allele of each gene.

Question 52

Explain how fusion of gametes results in two alleles that may be the same or different alleles.

3.4

Answer 52

The nucleus of the zygote contains two chromosomes of each type so is diploid. It contains also two alleles of each gene.

Understanding: Fusion of gametes results in diploid zygotes with two alleles of each gene that may be the same allele or diferent alleles.

Question 53

Outline how alleles of each gene are seperated during meiosis and its outcome.

3.4

Answer 53

• During meiosis I, homologous chromosomes are separated into different nuclei prior to cell division
• As homologous chromosomes carry the same genes, segregation of the chromosomes also separates the allele pairs
• Consequently, as gametes contain only one copy of each chromosome they therefore carry only one allele of each gene
• It breaks up existing combinations of alleles in a parent and allows new combinations to form in the offspring.

Understanding: The two alleles of each gene separate into diferent
haploid daughter nuclei during meiosis.

Question 54

Define “genotype”.

3.4

Answer 54

The gene composition (i.e. allele combination) for a specific trait

Understanding: Dominant alleles mask the efects of recessive alleles but
co-dominant alleles have joint efects.

Question 55

Outline dominant and recessive alleles.

3.4

Answer 55

Most traits follow a classical dominant / recessive pattern of inheritance, whereby one allele is expressed over the other
* The dominant allele will mask the recessive allele when in a heterozygous state
* Homozygous dominant and heterozygous forms will be phenotypically indistinguishable
* The recessive allele will only be expressed in the phenotype when in a homozygous state

When representing alleles, the convention is to capitalise the dominant allele and use a lower case letter for the recessive allele
* ex. BB or Bb or bb

Understanding: Dominant alleles mask the efects of recessive alleles but
co-dominant alleles have joint efects.

Question 56

Outline co-dominance.

3.4

Answer 56

Co-dominance occurs when pairs of alleles are both expressed equally in the phenotype of a heterozygous individual

• Heterozygotes therefore have an altered phenotype as the alleles are having a joint effect

When representing alleles, the convention is to use superscripts for the different co-dominant alleles (recessive still lower case)
* ex. C^W

Understanding: Dominant alleles mask the efects of recessive alleles but
co-dominant alleles have joint efects.

Question 57

Distinguish between homozygous and heterozygous.

3.4

Answer 57

• Homozygous: two of the same allele
• Heterozygous: two different alleles.

Understanding: Dominant alleles mask the efects of recessive alleles but
co-dominant alleles have joint efects.

Question 58

Outline the inhertiance of ABO blood groups.

3.4

Answer 58

• The A and B alleles are co-dominant and each modify the structure of the antigen to produce different variants
• The O allele is recessive and does not modify the basic antigenic structure

A allele = IA ; B allele = IB ; O allele = i (recessive)

Application: Inheritance of ABO blood groups.

Question 59

Define “genetic disease”.

3.4

Answer 59

**A genetic disease is an illness that is caused by a gene. **Most genetic diseases are caused by a recessive allele og a gene.
* The disease therefore only develops in individuals that do not have the dominant allele of the gene, usually because they have two copies of the recessive allele.

Understanding: Many genetic diseases in humans are due to recessive
alleles of autosomal genes.

Question 60

Define “carrier”.

3.4

Answer 60

• If a person has one allele for the genetic disease and one dominant allele, they will not show symptoms of the disease, but they can pass on the recessive allele to their offspring. These individuals are called carriers.

Understanding: Many genetic diseases in humans are due to recessive
alleles of autosomal genes.

Question 61

Outline other causes of genetic diseases.

3.4

Answer 61

• A dominant allele (ex. Huntington’s disease)
• Co-dominant alleles (ex. sickle-cell anemia)
• sex linkage (ex. colour blindness and hemophillia)

Understanding: Some genetic diseases are sex-linked and some are due
to dominant or co-dominant alleles.

Question 62

Outline cystic fibrosis.

3.4

Answer 62

Cystic fibrosis is an autosomal recessive disorder caused by a mutation to the CFTR gene on chromosome 7
* Individuals with cystic fibrosis produce mucus which is unusually thick and sticky
* This mucus clogs the airways and secretory ducts of the digestive system, leading to respiratory failure and pancreatic cysts
* Heterozygous carriers who possess one normal allele will not develop disease symptoms

Application: Inheritance of cystic fibrosis and Huntington’s disease.

Question 63

Outline Hungtington’s disease.

3.4

Answer 63

• Huntington’s disease is due to a dominant allele of the HTT gene.
• This gene is located on chromosome 4 and the gene product is a protein named huntingtin.
• The dominant allele of HTT causes degenerative changes in the brain.

Application: Inheritance of cystic fibrosis and Huntington’s disease.

Question 64

Outline the inheritance pattern for sex-linked genes.

3.4

Answer 64

Sex-linked inheritance patterns differ from autosomal patterns due to the fact that the chromosomes aren’t paired in males (XY)
* This leads to the expression of sex-linked traits being predominantly associated with a particularly gender

As human females have two X chromosomes (and therefore two alleles), they can be either homozygous or heterozygous
* Hence, X-linked dominant traits are more common in females (as either allele may be dominant and cause disease)

Human males have only one X chromosome (and therefore only one allele) and are hemizygous for X-linked traits
* X-linked recessive traits are more common in males, as the condition cannot be masked by a second allele

Understanding: The pattern of inheritance is diferent with sex-linked genes due to their location on sex chromosomes.

Question 65

Outline the trends in the inheritance pattern for sex-linked genes.

3.4

Answer 65

• Only females can be carriers (a heterozygote for a recessive disease condition), males cannot be heterozygous carriers
• Males will always inherit an X-linked trait from their mother (they inherit a Y chromosome from their father)
• Females cannot inherit an X-linked recessive condition from an unaffected father (must receive his dominant allele)

Understanding: The pattern of inheritance is diferent with sex-linked genes due to their location on sex chromosomes.

Question 66

State the two examples of sex-linked conditions on the X chromosome.

3.4

Answer 66

red-green colour-blindness and hemophilia

Application: Red-green colour-blindness and hemophilia as examples of sex-linked inheritance.

Question 67

Outline red-green colour blindness.

3.4

Answer 67

Caused by a recessive allele of a gene for one of the photoreceptor proteins.

Application: Red-green colour-blindness and hemophilia as examples of sex-linked inheritance.

Question 68

Outline hemophillia.

3.4

Answer 68

Due to an inability to make Factor VIII, one of the proteins involved in the clotting of blood.
* The allele that causes hemophilia is recessive.

Application: Red-green colour-blindness and hemophilia as examples of sex-linked inheritance.

Question 69

Outline the symbols in pedigree charts.

3.4

Answer 69

• males are shown as squares
• females are shown as circles
• squares and circles are shaded or cross-hatched to indicate whether an individual is affected by the disease
• Roman numerals indicate generations

Skill: Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.

Question 70

Explain why genetic diseases are rare.

3.4

Answer 70

The reason for this is that most genetic diseases are caused by very rare recessive alleles which follow Mendelian patterns of inheritance.
* The chance of inheriting one allele for any specifc disease
is small but to develop the disease two alleles must be inherited and the chance of this is extremely small.

Understanding: Many genetic diseases have been identifed in humans
but most are very rare.

Question 71

Explain why genetic diseases are rare.

3.4

Answer 71

The reason for this is that most genetic diseases are caused by very rare recessive alleles which follow Mendelian patterns of inheritance.
* The chance of inheriting one allele for any specifc disease
is small but to develop the disease two alleles must be inherited and the chance of this is extremely small.

Understanding: Many genetic diseases have been identifed in humans
but most are very rare.

Question 72

Define “mutation” and outline the causes for mutations.

3.4

Answer 72

A mutation is a random change to the base sequence of a gene. Two types of factor can increase the mutation rate.
1. Radiation (Gamma rays, short-wave ultraviolet radiation and X-rays are all mutagenic.)
2. Chemical substances (ex. mustard gas)

Understanding: Radiation and mutagenic chemicals increase the mutation
rate and can cause genetic disease and cancer.

Question 73

Outline the consequences of radiation after the bombing of Hiroshima and Nagasaki.

3.4

Answer 73

Apart from cancer the other main effect of the radiation that was predicted was mutations, leading to stillbirths, malformation or death.

Application: Consequences of radiation after nuclear bombing of Hiroshima and Nagasaki and the nuclear accidents at Chernobyl.

Question 74

Outline the consequences of radiation after the nuclear accident at Chernobyl.

3.4

Answer 74

An explosion at the reactor core caused the release of radioactive material.
* thyroid cancer
* Bioaccumulation caused high levels of radioactive caesium in fish

Application: Consequences of radiation after nuclear bombing of Hiroshima and Nagasaki and the nuclear accidents at Chernobyl.

Question 75

Outline the process of gel electrophresis.

3.5

Answer 75

Gel electrophresis is a method of seperating mixtures of proteins or fragments of DNA, which are charged.
1. The mixture of DNA is placed on a thin sheet of gel, which acts like a molecular sieve.
2. An electric field is applied to the gel by attatching electrodes to both ends.
3. Depending on whether the particles are positively or negatively charged, they move towards one of the electrodes or the other.
4. The rate of movement depends on the size of the molecules (small molecules move faster than larger ones).

Understanding: Gel electrophoresis is used to separate proteins or fragments of DNA according to size.

Question 76

Outline the use of PCR to amplify small amounts of DNA.

3.5

Answer 76

Can rapidly copy small samples of DNA into larger amounts needed for analysis.
1. Denaturation: DNA sample is heated to separate it into two strands (~95 deg C)
2. Annealing: DNA primers attach to opposite ends of the target sequence (~50 deg C)
3. Elongation: A heat-tolerant DNA polymerase (Taq) copies the strands (~72 deg C)

Understanding: PCR can be used to amplify small amounts of DNA.

Question 77

Outline the process of DNA profiling.

3.5

Answer 77

1. A sample of DNA is obtained, either from a known individual or from another source such as a fossil or a crime scene.
2. Sequences in the DNA that vary considerably between individuals are selected and are copied by PCR.
3. The copied DNA is split into fragments using restriction endonucleases.
4. The fragments are separated using gel electrophoresis.
5. This produces a pattern of bands that is always the same with DNA taken from one individual. This is the individual’s DNA profle.
6. The profiles of different individuals can be compared to see which bands are the same and which are different.

Understanding: DNA profling involves comparison of DNA.

Question 78

Outline how DNA profiling is used in paternity and forensic investigations.

3.5

Answer 78

Forensic investigations
1. DNA profle of material from the crime scene is compared with the DNA profile of a sample of DNA taken grom the suspect or the victim.
2. If the pattern of bands matches exactly it is highly likely that the two samples of DNA are from the same person.
3. This can provide very strong evidence o who committed the crime.

Paternity investigations
1. DNA profiles of the mother, the child and the man are needed.
2. DNA profiles of each of the samples are prepared and the patterns of bands are compared.
3. If any bands in the child’s profle do not occur in the profle of the mother or man, another person must be the father.
4. Children inherit half their chromosomes from each parent and thus should possess a combination of parental fragments

Application: Use of DNA profling in paternity and forensic investigations.

Question 79

Define “gene transfer”.

3.5

Answer 79

A gene determines a particular trait by encoding for a specific polypeptide in a given organism
* Because the genetic code is universal, an organism can potentially express a new trait if the appropriate gene is introduced into its genome
* The transfer of genes between species is called gene modification, and the new organism created is called a transgenic

Understanding: Genetic modifcation is carried out by gene transfer
between species.

Question 80

Outline the process of gene transfer to bacteria with plasmids using restriction endonucleases and DNA ligase.

3.5

Answer 80

1. **Restriction enzymes ‘cut’ the desired gene from the genome. **
2. E. coli bacteria contain small circles of DNA called plasmids.
3. The same restriction enzyme cuts into the plasmid.
4. Because it is the same restriction enzyme the same bases
   are left exposed, creating ‘sticky ends’
5. Ligase joins the sticky ends, using complementary base pairing, fixing the gene into the E. coli plasmid.
6. The recombinant plasmid is inserted into the host cell. It now expresses the new gene. An example of this is human insulin production.
7. Fermenters are used to produce large quantities of bacteria. The human insulin is then separated from the bacteria and purified.

Application: Gene transfer to bacteria with plasmids using restriction
endonucleases and DNA ligase.

Question 81

Outline potential benefits of GM crops.

3.5

Answer 81

1. Higher crop yield, more food, less pest damage
2. Less land needed for crop production
3. Less use of insecticides, expensive and harmful

Application: Assessment of the potential risks and benefts associated with genetic modifcation of crops.

Question 82

Outline potential risks of GM crops.

3.5

Answer 82

1. Insects that are not pests are killed (ex. toxin in Bt maize harms monarch butterflies)
2. Transfered gene may spread to population of other plants by cross pollination, makes them also toxic to insects
3. Insects pests may develop resistance to toxin, even harder to control afterwards

Application: Assessment of the potential risks and benefts associated with genetic modifcation of crops.

Question 83

Define “cloning”.

3.5

Answer 83

The production of genetically identical organisms is called cloning and a group of genetically identical organisms is called a clone.

Understanding: Clones are groups of genetically identical organisms,
derived from a single original parent cell.

Question 84

Define “cloning”.

3.5

Answer 84

The production of genetically identical organisms is called cloning and a group of genetically identical organisms is called a clone.

Understanding: Clones are groups of genetically identical organisms,
derived from a single original parent cell.

Question 85

Outline natural methods of cloning in plants and some animals.

3.5

Answer 85

Uses asexual reproduction.
* extra bulbs
* tubers
* runners

Understanding: Many plant species and some animal species have natural methods of cloning.

Question 86

Define “splitting or fragmentation”.

Answer 86

An embryo divides into two or more parts and each part to develop into a separate individual with all body parts.

Understanding: Animals can be cloned at the embryo stage by breaking up the embryo into more than one group of cells.

Question 87

Outline how identical twins are formed.

Answer 87

Formation of identical twins could be regarded as cloning by splitting.

Understanding: Animals can be cloned at the embryo stage by breaking up the embryo into more than one group of cells.

Question 88

Outline the process of somatic-cell nuclear transfer (SCNT)

3.5

Answer 88

1. Remove a differentiated diploid nucleus from the individual to be cloned.
2. Enucleate a donor egg cell. (remove nucleus)
3. Insert the diploid nucleus into the enucleated egg cell.
4. Implant into the endometrium of a surrogate mother and gestate.
5. The newborn will be genetically identical to the donor nucleus parent.

Understanding: Methods have been developed for cloning adult animals using diferentiated cells.

Question 89

Outline the process of therapeutic cloning.

3.5

Answer 89

1. Remove a differentiated diploid nucleus from the cell to be cloned.
2. Enucleate a donor egg cell.
3. Insert the diploid nucleus into the enucleated egg cell.
4. Stimulate it to divide and grow in vitro.
5. The resulting embryo is a rich source of stem cells which can be harvested or cultured.
6. The outer layer of cells is removed, so only the inner cell mass is used to culture the tissues needed.

Understanding: Methods have been developed for cloning adult animals using diferentiated cells.

Question 90

Define “therapeutic cloning.”

3.5

Answer 90

Creating an embryo as a source of stem cells, by transfer of a differentiated nucleus.

Understanding: Methods have been developed for cloning adult animals using diferentiated cells.

Question 91

Define “somatic-cell nuclear transfer.”

3.5

Answer 91

Creating a genetically identical organism through transfer of a differentiated diploid nucleus.

Understanding: Methods have been developed for cloning adult animals using diferentiated cells.