All the lectures

Question 1

State Mendel’s first law

Answer 1

First law (the law of segregation): Two alleles at a locus segregate into separate gametes; half carrying one allele and the other half the other (genes are particles).

Question 2

What is the evidence from Mendel’s first law?

Answer 2

Mendel crossed pea plants from inbred lines that differed in one discrete trait (e.g. pea shape, height, flower colour) = monohybrid cross
P1: Yellow x Green
F1: All Yellow
He then self-fertilised the first filial generation (F1) generation and got a consistently predictable ratio
F2: 25% Green, 75% Yellow

These monohybrid crosses led Mendel to propose that:
• Each trait depends on two particles → the offspring can inherit three combinations of the pairs (homozygous for each particle or heterozygous for the particles)
• Information can be concealed but it still present - the trait (e.g. yellow) that is visible in the F1 generation is controlled by the dominant allele. The recessive trait (e.g. green) is masked
o This ruled out the theory that traits occurred due to mixing
• Segregation – during the formation of gametes, the paired particles segregate randomly so that each gamete receives each trait with equal likelihood

Question 3

What does the pedigree for dominant inheritance look like?

Answer 3

• Any unaffected individual has genotype aa
• Does NOT skip generations
• When Aa x aa → half the children are affected
• E.g. Brachydactyl, Huntington’s disease

Question 4

What does the pedigree for recessive inheritance look like?

Answer 4

• Skips generations because carriers/unaffected parents can have affected child
• Affected child must have heterozygous parents or affected parents
• Affected individuals must be homozygous recessive

Question 5

Why is Mendelian law too simplistic to predict genetic risk?

Answer 5

Human families are small and so in some families none or all of the children can be affected

each child has an independent chance of inheriting a particular allele

Question 6

State Mendel’s second law

Answer 6

Second Law (the law of independent assortment)- During the formation of gametes, the segregation of alleles at one locus is independent of that of the segregation of alleles at any other locus
I.e. genes for different characteristics such as colour or shape are inherited independently of each other

Question 7

What is the evidence of independent assortment?

Answer 7

The evidence for independent assortment resulted from a dihybrid cross (a cross in which plants differ in two characteristics). In which the observed F2 ratio was 9:3:3:1

The ratios of the F2 generation can be predicted using the product law – when two independent events occur simultaneously, the combined probability of the two outcomes is equal to the product of their individual probabilities.
This is why the same ratios in F1 and F2 are produced, regardless of whether the P1 is AB x ab or Ab x aB.

We can use a chi-squared test (tutorials) to determine whether two alleles segregate independently.
• In chi-squared the larger the deviation from the mean, the greater the effect of the value

Question 8

What is a recombinant?

Answer 8

Recombinants = new phenotypes in offspring that are not present in the parental generation (e.g. green wrinkled x yellow smooth → green smooth). This occurs because the alleles at a locus are inherited independently to the alleles at another locus.
Recombination  = the process that leads to new combinations of alleles on chromosomes due to independent assortment

Question 9

How many genotypically different gametes are produced from indendent assortment, when there are 5 different loci?
How many F2 genotypes?

Answer 9

2^5 = 32
3^5 = 243

Question 10

Why is recombination significant in agriculture?

Answer 10

Generating genetic variation
• Green revolution – advantageous characteristics from different lines are put together, selecting recombinants that have both.
o in India tall but productive pure lines of rice were easily damaged by weather, causing famine. These were crossed with a short but less productive variant to produce small and productive recombinants. The dwarfing gene was identified, giberllin, the plant growth hormone. This can be engineered into other crops. The same principle can be applied to animals – turkeys that have high meat production are crossed with high fecundity line
• E.g. tomatoes are crossed for desired characteristics – ripening, shape, colour, resistance to various fungi
• E.g. hybrid corn – heterozygotes grow best (heterosis), this increased the wheat yields in 1930s. The next generation produces recombinants that are less fit (LU in glossary of textbook) due to recombinants containing recessive alleles which are harmful, but when heterozygote, these alleles are masked by dominants
• Potato famine in Ireland because all potatoes are genetically identical and were susceptible to fungal disease

Question 11

Why can recombination be harmful to humans?

Answer 11

Disease, antibiotic resistance
• H5N1 bird flu RNA gives swine flue (the 1918 flu epidemic differed in one base from H4N1). Bird and human influenza can simultaneously infect pig, where they can recombine and potentially form a new virulent version of the virus.
o Happens on farms where these animals are kept together
• HIV strains have recombined to become resistant to many antiviral drugs
o Individuals have more than one strain, which combine to form new variants – this is hard to control

Question 12

How did recombination produce a white skinned population that are resistant to malaria?

Give an example of human recombination

Answer 12

Sickle cell
• A particular variant of haemoglobin, which gives protection against malaria when heterozygous
• Generally found in blacks in Africa
• However, this variant is often found in Portugal in white skinned people because the incidence of malaria made the HbS favourable and those with this variant survived and reproduced so that the HbS was kept at a high frequency, whereas skin colour genes were diluted out
• Now there are recombinants between sickle cell and white skin colour at high frequency

Question 13

Discuss applications of Mendel’s second law

Answer 13

Forensic genetics - independent assortment can generate huge numbers of different genotypes by recombination. Therefore no two individuals share the same set of alleles at freely recombining loci (the basis of a genetic fingerprint).
Green revolution – advantageous characteristics from different lines are put together, selecting recombinants that have both.

Question 14

What do these F2 ratios show?

3: 1
1: 2:1
1: 2

Answer 14

3: 1 mendelian inheritance
1: 2:1 incomplete dominance
1: 2 lethal alleles

Question 15

Define and give examples of incomplete dominance

Answer 15

Incomplete, partial, dominance = the expression of the heterozygote is an intermediate between the dominant homozygotes
• Mirabilis japonica (four o’clock plant) is normally red but there is a white variety, when you cross white and red, the F1 generation is pink
• The phenotypic ratio and the genotypic ratio of F2 is 1:2:1 because the neither allele is dominant
• Tay-Sachs disease – homozygous recessive individuals have a fatal lipid storage disorder, due to almost no activity of the enzyme, hexosaminidase. Heterozygotes, with only one copy of the mutant gene, only express 50% of the normal activity of this enzyme. However, this is enough to achieve normal biochemical function.

Question 16

Define and give examples multiple alleles

Answer 16

Multiple alleles = more than two alleles at a locus that control a specific phenotypic characteristic within a population
• Eye colour in Drosophila and humans (also incomplete dominance)
o In Drosophila, white-eye is caused by a recessive mutation, there are many alleles that can occupy this locus to produce a wide range of phenotypes within the population
• ABO blood group – can be A, B, O (three alleles) which encode for the expression of different glycoprotein antigens on red blood cell surface
o A has an added n-acetyl galactosamine
o B has an added galactose
o These A and B antigens are derived from the H-substance, a precursor molecule

Question 17

Define and give examples of lethal alleles

Answer 17

Lethal alleles = recessive homozygotes may be lethal
The time of death depends on when the product which the gene encodes for is necessary.

Yellow mice
•	The allele for yellow mice is dominant when expressed heterozygously with the wild-type and lethal when expressed homozygously  
o	Agouti (yy) x (Yy) Yellow → agouti (yy) 1 : 1 yellow (Yy)
o	Yellow (Yy) x (Yy) Yellow → 1 agouti (yy) : 2 Yellow (Yy) and one dies so is not seen because it is (YY) homozygous for a lethal allele

Manx cat – heterozygous, the cat has no tail, lethal allele if homozygous

Brachydactyly – lethal allele if homozygous

Dominant lethal alleles in a population are rarely observed because the affected individual must reproduce before the lethal allele is expressed. If the lethal allele is expressed before the individual can reproduce, it is no longer inherited and disappears from the population
Huntington’s disease is dominant autosomal lethal
• the presence of one copy of the allele is lethal for an individual
• the disease onset of the heterozygote is delayed into adulthood
• motor and neural degeneration until death
• often an effected individual will have offspring by the age of the disease onset, and each offspring has a 50% chance of having inherited the allele

Question 18

Define and give examples of pleiotropy

Answer 18

Pleiotropy – one gene has multiple phenotypic effects
e.g. all blue eyed cats are deaf, tabby cats are aggressive

Sickle-cell haemoglobin involves change in a single base/one amino acid. This single change has many detrimental effects in the body e.g. become anaemic, heart failure, paralysis, enlarged spleen, bossed skulls, brain damage, poor circulation - as well as malaria resistance in heterozygotes.

Marfan syndrome results from an autosomal recessive mutation encoding the connective tissue protein fibrillin. Fibrillin is important to the structural integrity of the lens of the eye, the lining of vessels such as the aorta and bones. Therefore Marfan syndrome results in lens dislocation, aortic aneurysm, and lengthened long bones in limbs.

Porphyria variegata is a human autosomal dominant disorder. Afflicted individuals cannot adequately metabolise the porphyrin component of haemoglobin, when this respiratory component is broken down when red blood cells are replaced. Excess porphyrin is evident in urine, which is deep red. The build up of porphyrin in the body is toxic and the phenotype includes abdominal pain, muscular weakness, fever, racing pulse, insomnia, headaches, vision problems, delirium, convulsions.
George III had this.

Question 19

Define and give examples of gene interaction

Answer 19

Gene interaction – several genes affect one character
The allele at one locus affects the phenotype produced by an allele at another locus.
• sex-limitation (the expression of a particular characteristic limited to one sexes)
o in males activated due to the presence of testosterone e.g. the peacock’s tail and male secondary sexual characters in humans (e.g. beards and baldness)
o in females BRCA1 (BRCA=BReast CAncer) mutations are associated with an increase in breast cancer in women because of gene interaction with cell division genes and oestrogen.
• Mouse coat colours – there at least five interacting loci
Mouse Coat Colours; many different inbred lines. At least 5 interacting loci.

A locus: determines the distribution of colour in hair shaft

Wild type mouse – “agouti”. Each hair has a dark shaft with a yellow band. A is wild type and dominant. Recessive allele a removes the yellow band, so that aa mice lacking a genetic change at any other colour locus are black.

B locus: Codes for colour of hair

B codes for black, b for brown. BB and Bb mice are black; bb are brown. Consider cross:

P AA BB (agouti) x aa bb (brown)
F1 all Aa Bb (agouti)

Cross these together (F1 x F1):
Ratio	Genotype	Phenotype
9	A- B-	Agouti
3	A- bb	cinnamon: brown hair with yellow fleck
3	aa B-	Black
1	aa bb	Brown

A-bb is a new phenotype (Cinnamon) formed by gene interaction

C locus: determines presence or absence of pigment

CC or Cc animals have pigment, cc are albino. The albino allele is epistatic to other loci; prevents them from expressing their effects.

P BB CC (black) x bb cc (albino)
F1 all Bb Cc (black)

Cross these together (F1 x F1):
Ratio	Genotype	Phenotype
9	B-C-	Black
3	bb C-	Brown
3	B- cc	Albino
1	bb cc	Albino

→ I.e. a 9 : 3 : 4 ratio. The C locus masks the phenotype at the B locus. Example of epistasis.

D locus: Controls expression of pigment.

DD and Dd allow full expression of coat colours, dd is dilute (makes colours look pale by producing an uneven distribution of pigment).

P BB DD (black) x bb dd (dilute brown)
F1 all Bb Dd (black)

Cross these together (F1 x F1):
Ratio	Genotype	Phenotype
9	B-D-	Black
3	bb D-	Brown
3	B- dd	Dilute black
1	bb dd	Dilute brown

The S locus controls pigment distribution on the body.

SS or Ss are not spotted, but ss is piebald - large patches of colour - say black and dilute black - on the same individual.

Gene interaction in humans
• Secretor (Se) locus/phenotype. Secretor phenotype involves interaction between ABO locus and Se locus.
o SeSe or Sese produce ABO blood group substances in their salvia and other bodily fluids such as semen

Question 20

Define and give examples of complementation

Answer 20

Complementation – two strains of an organism with different homozygous recessive mutations that produce the same phenotype

Foxgloves – there are two recessive mutations, that occur at different loci, which can cause a foxglove to be white. If a plant is recessive homozygous at W1 or at W2, then the foxglove will be white.
Crossing two plants from each line, produces an all purple F1 generation because all the plants are heterzygouse at both loci.
The F2 phenotype ratio is 9 purple: 7 white because the same phenotypic effect occurs from a recessive homozygous at either locus.

P1 w1w1 W2W2 (white line 1) x W1W1 w2w2 (white line 2)

F1 All W1w1 W2w2 (purple) Self-fertilise

F2		9 W1- W2- (purple)		  	9
		3 W1- w2w2 (white)		
		3 w1w1 W2- (white)		 	7
		1 w1w1 w2w2 (white)			
At least one dominant allele is needed from each pair of genes to ensure that both of the biochemical reactions needed to produce purple pigment is functional. In the cross, 7/9 plants did not have at least e one dominant allele at each locus. 

Human deafness and complementation

Many different loci affecting hearing – some affect outer ear, some the malleus, incus and stapes of the middle ear, some the cilia in the inner ear, some nerve connections to the brain or the brain itself. A break in the chain for any of them will lead to deafness.

Hearing phenotype: Sound → A → B → C → D → E → F → Perception of sound.
Genotypes: First individual AA → BB → cc → DD → EE → FF – deaf
Second individual AA → BB → CC → DD → ee → FF – deaf
Offspring AA → BB → Cc → DD → Ee → FF – can hear

COMPLEMENTATION between C and E locus gives normal hearing as Cc and Ee heterozygotes have functioning steps in auditory chain. As a result, two deaf people often have hearing children as each parent is homozygous for a recessive abnormality at a different locus and child is double heterozygote.

Question 21

How can you test for complementation?

Answer 21

Complementation tests show whether the alleles that cause a mutation are alleles of the same genes, or whether the alleles are mutations are causes by alleles in separate genes, at separate loci.
A blue-eyed fruit fly turns up. Recessive to wild type. Make a homozygous line. In a different line a green-eyed fly appears; do the same.
• Cross the two lines, where each line contains a different recessive mutation to the wild type
• If the F1 generation have wild-type phenotype, then the two recessive mutations are NOT alleles of each other as all the F1 offspring are heterozygous at both loci and there is complementation
• If the F1 generation are all mutants, then the alleles occur at the same locus. And so the F1 offspring are homozygous for a mutation at the locus.

Cross the green homozygote with the blue homozygote. If they are ALLELES at the same locus, the F1 will be blue or green, depending which one is dominant (or could be blue-green if intermediate dominance).

But if two LOCI involved then F1 are heterozygous at both loci and have red eyes – the loci complement. If we have many mutants can sort them into “complementation groups”; within a group they do not complement, between, they do

Question 22

How can a phenotype be affected by sex?

Give an example for men and women

Answer 22

Sex-limitation – the expression of a certain phenotype is limited to one sex. The autosomal genes are responsible for the existence of different phenotypes, but the expression of these genes is dependent on the hormone constituent of the individual
• Therefore the heterozygous genotype will express one phenotype in makes and another in females
• E.g. pattern baldness
o BB – bald in both males and females, females express the baldness to a lesser extent and later in life
o Bb – males are bald, females are not
o bb – not bald
• E.g. BRCA1 (BRCA=BReast CAncer) mutations are associated with an increase in breast cancer in women because of gene interaction with cell division genes and oestrogen.

Question 23

How are phenotypes affected chemically?

Answer 23

Chemicals - Also a link between deafness and the environment interaction. Women with ‘Derbyshire neck’ caused by iodine shortage often have children who are deaf. Some children with measles also become deaf.

Question 24

What are temperate sensitive mutations?

Answer 24

Temperature-sensitive mutations – Temperature affects the kinetic energy of reacting substances, which in turn effects chemical activity and so temperature influences phenotypes.
E.g. Siamese cats and Himalayan rabbits exhibit dark fur in regions where their body temperatures are slightly cooler (ears, nose, paws). The enzyme normally responsible for pigments production is functional only at lower temperatures which are present in the body extremities, but loses its catalytic function at warmer temperatures found throughout the rest of the body.

Question 25

Why does age affect human phenotypes?

Give three examples

Answer 25

Age – the age at which a genetic mutant exerts a noticeable phenotype depends on events during the normal sequence of growth and development. The critical expression of genes varies throughout the life cycle of all organisms. Gene products may play more essential roles at specific life stages and it is likely that the internal physiological environment changes with age
• Tay-Sachs disease – newborns appear phenotypically normal for the first few months and then have developmental retardation, paralysis, blindness and die at around 3
• Lesch - Nyhan syndrome – Newborns are normal for 6-8 month. It is an X-linked recessive disease. The mutation is in the gene encoding hypozanthine-gaunine phosphoribosyl transferase (HGPRT). It is characterised by abnormal nucleic acid metabolism, leading to accumulation of uric acid in blood and tissues, mental retardation, palsy, self-mutilation of lips and fingers.
• Huntington’s Disease - progressive cell death in the frontal lobes of cerebral cortex. Most frequently diagnosed at 38 but occurs between 30 and 50 years.

Question 26

Define penetrance

Answer 26

Penetrance = the percentage of individuals who show at least some degree of expression of a mutant phenotype (i.e. if 85% of all individuals with a homozygous recessive have wild-type phenotypes, then the penetrance is 15%)

Question 27

Define expressivity

Answer 27

Expressivity – the range of expression of a mutant genotype (e.g. mutant flies with the eyeless gene can have normal eye, smaller eyes and the absence of both or one eye)

Question 28

What are the parallels between Mendel’s particles and chromosomes?

Answer 28

Parallels between chromosomes and Mendel’s particles: come in pairs (e.g. humans will have two copies of each chromosome), segregate at meiosis during the formation of egg and sperm and are re-joined at fertilisation, show independent assortment (different chromosomes segregate independently).

Question 29

define pleiotropy and give examples

Answer 29

= one mutations has multiple phenotypic effects

• tabby cats are aggressive
• blue-eyed cats are deaf
• sickle cell haemoglobin -