LSM 1102 MENDELIAN GENETICS

Multiples AllelesLethal Genes

NondisjunctionLinkage

Multiple Alleles

When more than 2 alleles of a given gene exists in a species they are referred to amultiple alleles

Multiple Alleles - Example DrosophilaWhite-eye multiple allelic series in DrosophilaThe genotypes are arranged in order of dominance

Gardner, Simmons & Snustad, page 38, Table 2.5Genotype Eye colour Rel. conc. of total pigmentw/w white 0.0044wa/wa white apricot 0.0197we/we white eosin 0.0324wch/wch white cherry 0.0410wco/wco white coral 0.0798wts/wts dull red (wild-type) 0.6854_____________________________________________________If n = number of alleles of a genenumber of different genotypes = n (n+1)

2

14 of the most commonly detected blood-group systemsData of Race and Sanger, and others. Strickberger, pg 154.

Blood-group Date Probable No.system Discovered of alleles

* ABO 1900 5MNSs 1927 20+P 1927 4

* Rh 1940 30+Lutheran 1945 3Kell 1946 10Lewis 1946 4Duffy 1950 3Kidd 1951 3Diego 1955 2Yt 1956 2I 1956 4Xg 1962 2Dombbrock 1965 2

Multiple Alleles – Example Humans

Phenotype GenotypeA group IA IA, IA Io

B group IB IB, IB Io

AB group IA IB

O group Io Io

Antigens: Occurs on the surface of Red Blood CellsAntibodies: Naturally Occurring in Blood Plasma

(antibody-A and antibody-B)

Multiple Alleles - Example HumansABO blood group multiple allelic series in manOne gene locus, Three alleles : IA, IB Io

co-dominant recessiveCo-dominanceBoth alleles of a gene pair are fully expressed in aHeterozygote.

IA IA homozygote (dominant): A-antigen (galactosamine)IB IB homozygote (dominant): B-antigen (galactose)IAIB heterozygote (co-dominant): A-antigen & B-antigenIo Io homozygote (recessive): no antigens

Antigen-Antibody ReactionTable. Blood transfusion compatibility in ABO blood groups______________________________________________________________Blood Antigens Anti-bodies Red cell types TransfusionsGroup present present agglutinated accepted fromA A Anti-B B, AB A or O

galactosamine

B B Anti-A A, AB B or Ogalactose

AB AB none none A, B, AB & Ogalactosamine (universal recipient)

+ galactose

O none Anti-A + A, B, AB OAnti-B (universal donor)

Biochemical Basis of ABO Grouping

The Bombay Phenotype

ANTIGEN-ANTIBODYREACTION

Fig 7. Agglutination reactions

controlled by A-B-O blood type locus in

humans

Gardner, Simmons & Snustad, pg 39

Rh Factor Alleles in HumansWhat are the different genetic hypotheses?

• Original hypothesis: one gene locus with 2 alleles (D and d).DD homozygotes and Dd heterozygotes are Rh+.dd homozygotes are Rh-.

• Weiner’s hypothesis: one gene locus with a seriesof multiple alleles, 8 originally but more have beenadded.

• Fisher and Race’s hypothesis: Three gene loci (C, D and E) that are closely linked.

Rhesus (Rh) Factor in HumansRh factor discovered by immunizing rabbits with the bloodof the monkey, Macaca rhesus.Rh antibodies are used to test the blood of humans.

(D) Rh+ (Rhesus positive) - presence of the Rh antigen(d) Rh- (Rhesus negative) - lack of the Rh antigen

Malays and Chinese are almost 100% Rh+

Eurasians and Indians are almost 5% Rh-

Europeans (Caucasians) are almost 15% Rh-

Rh and ABO compatibilities are important in:1. Blood transfusions and organ transplants2. Maternal-fetal incompatibility (Rh factor)

mother, Rh- (dd homozygote); fetus, Rh+ (Dd heterozygote)

Figure of interaction between Rh- mother and fetus.If a man who is Rh+ and a woman who is Rh- conceive a child who is Rh+, the woman’s body manufactures antibodies that attack future Rh+offspring, who may die due to haemolyticdisease

Lewis, Pg 283

GENE INTERACTION or EPISTASISInteraction between non-alleles, i.e. interaction between 2 or more gene pairs

• Any gene or gene pair that masks the expression of another non-allelic gene is epistatic to that gene.

• Epistatic interactions between non-alleles for 2 gene pairs result in the modification of Mendel’s classical

F2 ratio of a dihybrid cross - 9 : 3 : 3 : 1

• Mendel’s classical ratio is obtained when -– Two gene pairs assort independently

– Two gene pairs exhibit complete dominance

– For autosomal genes

GENE INTERACTION or EPISTASIS

Two gene pairs (no interaction)F2 ratio : (9:3:3:1)

Two gene pairs (with interaction)F2 ratio : deviation from (9:3:3:1)

Three examplesa) 9 : 3 : 4 (recessive epistasis)b) 12 : 3 : 1 (dominant epistasis)c) 9 : 7 (complementary loci)

RECESSIVE EPISTASISF2 ratio: 9 : 3 : 4 Example : mice coat colourCoat pattern : B = black, b = brown, Coat colour: C = colour formation, c = no colour (albino)

RECESSIVE EPISTASIS cc epistatic to B and b

P (Cross) (black) BB CC × bb cc (albino)

F1 (Interbreed) Bb Cc (black)

RECESSIVE EPISTASIS (cont.)F2 ratio : 9 black (B- C-) : 3 brown (bb C-) : 4 albino (-- cc)

4/16albino

9/16BlackBb

3/16brown

bb

bB

BB

cccCCcCC

DOMINANT EPISTASISF2 ratio 12 : 3 : 1 Example : Summer squash fruit colourA = white fruit, a = coloured fruitB = yellow fruit, b = green fruitDOMINANT EPISTASIS:A epistatic to the B and b genes

P Cross (white) AA bb × aa BB (yellow)

F1 Selfing Aa Bb (white)

DOMINANT EPISTASIS (cont.)F2 ratio:12 white (A- --) : 3 yellow (aa B-) : 1 green (aa bb)

3/16yellow

12/16white

Bb

1/16green

bb

bB

BBaaaAAaAA

EPISTASIS (Complementary loci)F2 ratio: 9 : 7

Example : Sweet pea, Lathyrus odoratusExperiment by Bateson and Punnet: Cross between 2 varieties of sweet pea with white flowers

C = colour production c = no colourP = purple colour p = white colour

white variety white varietyP (Cross) CC pp × cc PP

F1 (Selfing) Cc Pp (purple)

EPISTASIS (Complementary loci)

F2 ratio: 9 purple (P- C-) : 7 white (-- cc, pp --, pp cc)

9/16purpleCc

7/16whitecc

cC

CC

pppPPp PP

What are Lethal Genes?Lethal genes are so serious that the organism is unable to live. Semi-lethal genes are harmful to the organism but do not cause their death.1. Lethal genes in humans

a) Recessive lethal (autosomal)The gene is lethal when homozygousExample: Brachyphalangy

B - normal fingersb - brachyphalangic, fingers have only 2 joints

(short fingers)BB - homozygote, normal fingersBb - heterozygote, brachyphalangybb - recessive lethal, lethal at birth

Lewis, pg 72

Lethal Genes in HumansDominant lethal (autosomal)

Both the dominant homozygote and the heterozygote are lethal.

For example: Huntington chorea in humansH - dominant lethal causes the illness Huntington chorea. HH and Hh individuals do not survive. However, the disease does not manifest until the victim is in their late 30’s when deterioration of their nervous system starts.h - normal gene, recessive

Lethal Genes in HumansRecessive lethal (sex-linked)

Only males are affected. They inherit therecessive lethal sex-linked gene from theirmothers, who are heterozygous carriers.

Example: Pseudotrophic muscular dystrophy humansThis disease is characterized by wasting away of muscles, which begins in the early teens and usually results in death in a few years.

Lethal Genes in HumansExample: Pseudotrophic muscular dystrophy in humansL - normal gene, l - lethal gene

X-chromosme

female (carrier) male (normal) Y-chromosomeL × Ll

lethall carrier

l male L female

normalL normal

L maleL femaleL

L

l

Nondisjunction (chromosomal mutation)

C.B. Bridges 1916 (His classic experiment)

A cross between white-eye female Drosophila and red-eye male produced exceptional offspring. i.e. red-eye male offspring and white-eye female offspringFrequency - 1 in 2000 offspringThis is too high to be due to gene mutation.

The observation by Bridges is due to nondisjunction of the pair of X chromosomes in the female Drosophila.Nondisjunction also occurs in other organisms

The failure of disjunction or separation of homologous chromosomes during mitosis or meiosis, resulting in an extra chromosome in some daughter cells and a deficiency of a chromosome in other daughter cells.

female Drosophila male DrosophilaII II

III III

IV IV

I XX I XY

What is Nondisjunction?

NondisjunctionNONDISJUNCTION

PRIMARY AND SECONDARY

(MEIOSIS OR MITOSIS)

VARIATION IN CHROMOSOME NUMBER

(CHROMOSOMAL MUTATION)

MONOSOMY (2N-1) (one chromosome missing)ANEUPLOIDY

TRISOMY (2N+1) (one chromosome extra)

Primary nondisjunctionof the X chromosomes in Drosophila. Experiment by C. B. Bridges

Strickberger, page 210

Origin of exceptional types in Drosophila melanogaster through nondisjunction of the X chromosomes in white-eyed females.

Frequency

1/2000

Normal progeny

Except. progeny

Secondary Nondisjunction in Drosophila

Strickberger, page 211

Results of secondary nondisjunction in the cross between an XXY white-eyed female and a normal XYred-eyed male.

Frequency

1/600

NONDISJUNCTION RESULTS IN ANEUPLOIDY (MONOSOMY AND TRISOMY)

• Definition of nondisjunction

• How was this discovered in Drosophila?

• Primary Nondisjunction in Drosophila

• Secondary nondisjunction in Drosophila

• Nondisjunction in humans

Nondisjunction can happen to any pair ofchromosomes in any organism

Normal Human Karyotype

A micrograph of 46 human chromosomes (female). Brooker, Fig 1-2, pg 5

Karyotype of Down’s Syndrome, trisomy 21

Due to nondisjunction of chromsome 21 (Triplo-21)

Burns, pg 299

Graph showing incidence of Down’s syndrome against maternal age.Hartl, page 172

Incidence of Down’s Sydrome increases dramatically with age of the motherMother’s age Incidence<30 years 1 per 1,400

40 years 1 per 10045 years 1 per 16

NONDISJUNCTION IN HUMANSof the sex chromosomes in females

May occur at the 1st or 2nd Meiotic DivisionNondisjunction of sex chromosomes in females

XX X sperm Nullo-X Y sperm

XXY MaleKlinefeltersyndrome

OY dies

××

XX eggs Nullo-X eggs

Fertilization with normal sperm

NONDISJUNCTION IN HUMANSof the sex chromosomes in males

May occur at the 1st or 2nd Meiotic Division

Nondisjunction of sex chromosomes in males

1st meiotic division

2nd meiotic division

½ XY

½

½

¼ XX ¼ YY

sperms

sperms

Male with Klinefelter’s Syndrome (47, XXY)

Klug & Cummins, page 159

Klinefelter Syndrome (47, XXY)

Female with Turner’s Syndrome (45, X)

Klug mad Cummins, page 159

X

Turner Syndrome (45, X)

Non-MendelianModes of Inheritance

Maternal Inheritance• Due to mitochondrial genes.• Transmitted to sons and daughters by mother only• Reason: Sperm only contributes nuclear DNA but

oocyte contributes both nuclear and cytoplasmic DNA

• Features– No crossing over– No DNA repair– Many copies per mitochondrion and per cell– High exposure to oxygen free radicals– No histones, no introns

Uniparental disomy

One parent transmits 2 alleles of a gene, the other parent transmit none. Eg– Prader-Willi syndrome

• In 40% of cases, lack father’s DNA (sometimes chr 15)

• In the 60%, a piece of missing DNA in one gene allowed only 1 parent’s gene to be expressed.

– Angelman syndrome• Reverse of PW syndrome, mother’s DNA missing

– Both cause mental retardation but each has other distinctive clinical features

– Different gender-directed manifestations of the same genetic variant?

Exceptions to Mendel’s Law of Segregation

• Uniparental disomy• Law of Segregation holds true only for traits

that are – on different chromosomes (physically unlinked)– same chromosome but very far apart (genetically

unlinked).• Traits that are close to each other on the

same chromosome tend NOT to segregate independently, but usually inherited together (co-segregate).This is the basis genetic linkage.

Linkage groups

(1911) Thomas H. Morgan discovered that genes are linked together in groups are calledchromosomes.

Each linkage group refers to genes on the same chromosome. Linked genes are found on the same chromosome. In a species there are as many linkage groups as there are homologous pairs of chromosomes.

Linkage groupsDrosophila melanogaster: 4 pairs of chromosomes i.e. 4 linkage groups

1st linkage group: X-chromosome2nd linkage group: 2nd chromosome3rd linkage group: 3rd chromosome4th linkage group: 4th chromosome

The 4th chromosome is very short. About 12 genes have been located on it. About 150 genes have been located on each of the other 3 chromosomes.

Humans: 23 pairs of chromosomes corresponding to 23 linkage groups.

Maize: 10 linkage groups corresponding to 10 chromosome pairs

Linkage map of Drosophila melanogaster

Tamarin

Fig 6.13

TERMINOLOGIESLINKAGEThe association of gene loci on the same chromosome

CROSSING-OVERA process in which chromosomes exchange material through the breakage and reunion of their DNA molecules. Crossing-over between homologous chromosomes leads to recombination of linked genes

RECOMBINATIONThe non-parental arrangement of alleles in a progeny that can result from either independent assortment or crossing-over

IMPORTANT FEATURES ON THE CONCEPT OF CROSSING-OVER

(Gardner et al., pg 163)

• The location of a gene on a chromosome is called a locus (plural - loci).

• The loci of the genes on a chromosome are arranged in a linear sequence.

• The 2 alleles of a gene occupy corresponding positions in homologous chromosomes.

E.g. Aa heterozygote : A (homolog 1)a (homolog 2)

• Crossing over involves the breakage of each of two homologous chromosomes (actually chromatids) and exchange of parts.

IMPORTANT FEATURES ON THE CONCEPT OF CROSSING-OVER

(Gardner et al., pg 163)

• Crossing-over occurs at the synapsis of the homologous chromosomes during the pachytene stage of prophase I of meiosis, after each chromosome has doubled.

• Each chromosome doubles to give two chromatids (tetrad stage).

• Only two chromatids are involved in a single crossover.• Crossing over between sister chromatids is seldom

detectable.• Chromosomes with recombinant combinations of linked

genes are formed by crossing-over in the region between the two loci.

IMPORTANT FEATURES ON THE CONCEPT OF CROSSING-OVER

(Gardner et al., pg 163)

• The strength of linkage between two genes depends on the distance between them on the chromosome.

• The % crossing-over (recombination frequency) between two loci is roughly proportional to the distance (map distance) between them.

• 1% cross-over (recombination) = 1 map unit (cM)(centiMorgan)

1 cM ~ 1 million bp

Cytological basis of Crossing-over

• T. H. Morgan proposed crossing-over to explain formation of recombinant combination of linked genes.

Crossing-over Recombination

• Cytological studies on meiosis show cross-shaped structures called chiasmata (singular chiasma) in which 2 of the 4 chromatids appear to change partners.

• At any one point the process of exchange, crossing-over involves 2 of the 4 chromatids in a meiotic tetrad.

Multiple chiasmata in testis of fowl

Meiotic prophase I chromosomes of the chicken showing multiple chiasmata in the large bivalents Chickens have 76 pairs of chromosomes.Snustad fig 7.11

Five stages of Prophase I of Meiosis

Snustad & Simmons, Fig 7.12

Tetrad

Crossing-over between two Chromatids of Tetrad gives rise to Recombination

Snustad & Simmons fig 7.7

How to detect linkage between two gene loci? Gene Mapping (Two-point Testcross)

DIHYBRID CROSS - A locus and B locus• Analysis of two gene loci, each segregating two alleles is called a two-

point cross.

NO LINKAGE (2 gene loci on different chromosomes)• Mendel’s second law of independent assortment applies• Two-point testcross progeny shows 1 : 1 : 1 : 1 segregation ratio

Parental types = 50%, Recombinant types = 50%

LINKAGE PRESENT (2 gene loci on the same chromosome)• Mendel’s second law of independent assortment does not apply• Two-point testcross progeny shows deviation from the 1 : 1 : 1 : 1

segregation ratio

Parental types > 50%, Recombinant types < 50%

No linkage between two gene loci(Two-point Testcross)

Mendel’s second law of independent assortment only applies to genes that are

1. completely dominant, no epistasis2. found on non-homologous chromosomes. i.e. not linked

Example: Dihybrid cross, no linkage,

Testcross: F1 crossed with homozygous recessive tester stockP A b a B

×A b a B

F1 (testcross) A b a b tester

×a B a b stock

Two-point Testcross Results

Dihybrid cross, no linkage1 : 1 : 1 : 1

50% parental types : 50% recombinant types

recombinant¼a baa bba b

recombinant¼A BAa BbA B

Parental¼a Baa Bba B

Parental¼A bAa bbA b

TypeExp. ratioPhenotypea b

How to identify parental and recombinant types among the testcross offspring?

Detecting Partial Linkage between 2 Gene Loci(Two-point Testcross)

DIHYBRID CROSS - A locus and B locus• Analysis of two gene loci, each segregating two alleles is

called a two-point cross

TESTCROSS • F1 crossed with homozygous recessive tester stock

Mendel’s second law of independent assortment• Does not apply to linked genes• There will be deviation from the 1 : 1 : 1 : 1 testcross ratio

among the progeny of the testcross.parental types > 50%, recombinant types < 50%

• The % of recombinants gives an estimate of the map distance between 2 linked genes. 1% r.f. = 1 map unit (cM)

Partial linkage between two gene loci (two-point testcross)

Mendel’s second law of independent assortment does not apply to linked genes

TESTCROSS: F1 crossed with homozygous recessivetester stock

P A b a B ×

A b a B

F1 (testcross) A b a b×

a B a b

Results of Testcross for Two Linked Genes

Deviation from the 1 : 1 : 1 : 1 testcross ratio> 50% parental types : < 50% recombinant types

recombinant<¼a baa bba b

recombinant<¼A BAa BbA B

parental>¼a Baa Bba B

parental>¼A bAa bbA b

TypeExp. ratiophenotypea b

How do you identify parental and recombinant types among the testcross offspring?

DETECTING PARTIAL LINKAGEExperiment using the two-point testcross in drosophila

Aim: Are the vg and b loci linked?vg+ = long wings (wild-type)vg = vestigial wingsb+ = grey body (wild-type)b = black body

Results showed that the map distance between the vg and b genes = 18 map units (centiMorgans, cM)

Testcross

Trans arrangement (repulsion): Two mutants are across from one another.

P A b a B×

A b a B

F1 A b

a BCis arrangement (coupling): One chromosome carries both mutants and the other carries both wild-type alleles

P A B a b×

A B a b

F1 A B

a b

ARRANGEMENT OF TWO LINKED GENES IN F1 HETEROZYGOTE

Some useful points to note

Recombination Fraction & Genetic Distance

• Recombination fraction (θ) = # of recombinant progeny ÷ total number of progeny.

• θ must lie between 0 and 0.5.• θ of 0 means the 2 genes are so close that

crossing over between these 2 genes is never observed.

• θ of 0.5 means the 2 genes are very far apart (genetically unlinked).

Recombination Fraction & Genetic Distance

• The maximum value of 0.5 comes from the fact that, on average, crossing over involves only one chromatid of each chromosome of the homologous pair.

• θ of 0.05 means that on average, crossing over occurs only once in 20 meioses.

• Genetic or map distance (χ) is measured in units of centimorgans (cM). A χ value of 1 cM≅ θ of 0.01.

Caveat to χ ≅ θ Rule

• Over large intervals, recombination fraction, and thus genetic distance, is underestimated.

• Why? Double (multiple) crossovers between loci.

A B C D E F

0.05 0.02 0.05 0.1 0.05θ :

θ=0.07, χ=7cM

θexp=0.27, χexp=27cM; θobs=0.2 χobs=20cM

Double Crossovers

A a

F f

A a

F f

A a

F f

Result: No apparent crossover between the A/a and F/f loci

Recombination frequency & distance

Genetic & Physical Distances

5 cM

GeneticMap

A B

Ordered overlapping clones

PhysicalMap 5 Mb

Application of Linkage Analysis in Humans

Mapping disease gene loci– Use polymorphic markers from each chromosome to

study segregation of the markers with respect to the disease in large families.

– Perform statistical analysis to obtain Lod scores (log of likelihood ratios). This will identify linked marker/s which cosegregate with the disease more often than would be expected by chance,• Lod score ≥ 3 strongly suggests linkage of the marker to the

disease gene.• Lod score ≤ -2 strongly suggests no linkage of the marker to

the disease gene.

Disease Gene Mapping

DC

dc

DC

dc

Dc

dC

D

B

d

b

D

B

d

b

D

b

d

B

D d

D dD d

A a

A a A a

25% 25% 25% 25% 25% 25% 25% 25% <50% >0% >0%<50%

Unlinked Markers Linked Marker

Polymorphic Markers

Repetitive DNA interspersed throughout the human genome (flanking DNA sequences are unique)– Microsatellites

• Dinucleotide repeats, e.g. (CA)n, trinucleotiderepeats, e.g. (CAG)n, tetranucleotide repeats, e.g. (GATA)n, etc.

– Minisatellites or variable number of tandem repeats (VNTRs)

• Tandem repeats of 15 to >35 nucleotide sequences.

Microsatellite Marker Analysis