Mendelian Inheritance Mesut Erzurumluoglu epmmeebristol ac uk Who

Mendelian Inheritance Mesut Erzurumluoglu epmmee@bristol. ac. uk

Who was Gregor Mendel? � Augustinian monk, Czech Republic. � Founder of ‘modern’ genetics � Studied segregation of traits in the garden pea (Pisum sativum) beginning in 1854 � Presented & published his theory of inheritance in 1865. ◦ “Versuche über Pflanzen-Hybriden” ◦ “Experiments in Plant Hybridization” � Mendel was “rediscovered” in 1902 ◦ Posthumously 2

Why did Mendel choose peas? Source: Mawer S. Gregor Mendel: Planting the Seeds of Genetics 3

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Mendel’s experiments • In his breeding experiments, Mendel did the following, and tracked heritable characteristics for three generations: • 1) Mated two true breeding parents to produce a hybrid. True-breeding parents are called the P generation. Hybrid offspring are called the F 1 generation. • 2) He then allowed the F 1 generation to self-pollinate. The offspring of this group are called the F 2 generation. 5

Parental Generation (True breeding plants) F 1 Generation (Hybrids) F 2 Generation (Ratio 3: 1) Purple Flowers White Flowers X All plants had purple flowers

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Terms • Locus: A location in the genome (e. g. ‘flower colour locus’, Chr 1 at position 5005). Plural: loci • Allele: Alternate forms of a gene (e. g. pink flower allele, white flower allele, P, p) • Genotype: Both alleles at the locus form a genotype (e. g. PP, Pp, pp) • Homozygous: Contains identical alleles for a character • e. g. AA, TT, CC, GG (or PP, pp) • Heterozygous: Contains two different alleles for a character • e. g. AT, GC, CA (or Pp) • Phenotype: An organism’s traits (e. g. eye colour, height, disease) 8

Mendel’s Laws • Similarities between parent and offspring are due to the transmission of discrete elements called ‘genes’. • There are multiple versions of the same gene (each alternate version of the gene is called an allele). • Each parent has two alleles. The two alleles separate during sex cell formation (i. e. ‘Meiosis’) Segregation (1 st Law) • Each organism inherits two (2) alleles; one allele from each parent. • Which allele an organism inherits from a parent is random (50% probability) Independent assortment (2 nd Law) 9

P Purple Flowers White Flowers X pp PP F 1 Generation Purple flowers Pp P P F 2 Generation p p PP Pp p. P pp Punnett Square

*Dominant and Recessive alleles u If a single copy of an allele results in the same phenotype as two copies irrespective of the second allele, the allele is said to be dominant over the second allele. – E. g. PP and Pp both produce purple flowers. Therefore P is dominant to p. u If an allele must occur in both copies of the gene to yield a phenotype, then it is recessive. – E. g. only pp produces white flowers. Therefore p is recessive to P. *This is Mendel’s third law (i. e. Law of Dominance) 11

Exercise 1 u You have two pea plants. One is true breeding for wrinkled seeds, the other is a round seed/wrinkled seed hybrid. Smooth seeds are dominant to wrinkled seeds. If we were to cross the two, what is the expected ratio of smooth to wrinkled seeds. u Use a Punnett Square to illustrate your answer. Smooth Ss X Wrinkled ss 12

Answer s S s s Ss Smooth ss ss Wrinkled Smooth : wrinkled 50 : 50 13

Problem u Let us assume for the moment that eye colour is a simple monogenic trait which follows simple Mendelian inheritance (the reality is slightly more complicated*) u Let us assume that the eye colour locus consists of two alleles, a brown allele and a blue allele. Brown is dominant to blue, therefore heterozygous individuals (Bb) exhibit brown eyes. u If this is the case, why do so many people have blue eyes? *see http: //www. ncbi. nlm. nih. gov/pubmed/20944644 14

Test Cross • Using a Punnett Square to reveal genotype: Carry out a ‘testcross’ • By breeding an organism of unknown genotype with an organism with a homozygous recessive trait, we can determine the genotype of the unknown individual. • The ratio of phenotypes in the offspring is used to determine unknown genotype. 15

Dominant Phenotype Recessive phenotype (Genotype pp) X (Genotype PP or Pp? ) Genotype PP? p P P Pp Pp Genotype Pp? p p p Pp Pp Pp p pp pp All offspring purple 1: 1 ratio of colours

Mendelian inheritance in Humans u We are unable to manipulate mating patterns of humans for experimentation. u Traits are studied by gathering information and placing it into a family tree. u The inter-relationships among parents and children across generations are called the family pedigree. 17

Example Pedigree Deceased male Affected female Unaffected male Affected male Unaffected female 18

Disease locus model u For monogenic traits, we can specify the underlying genetic model with three parameters: - Mode of inheritance (e. g. Autosomal, X-linked or mitochondrial) Disease allele frequency (e. g. using H-W equation) Penetrance, or probability of being affected given a certain disease locus genotype - Slide in appendix - 19

First address for Mendelian geneticists! 20

Mode of Inheritance (Mo. I) 1: Autosomal dominant u u u u Affected individuals are usually heterozygous (e. g. Pp) Disease will occur in all generations Rare homozygotes can be equally or more severely affected Disease affects both genders equally Recurrence risk for sibs and children is ½ Often have late onset (and reduced penetrance) Affected people often able to reproduce even if disease is fatal. u Example: Huntington’s disease dd dd Dd Dd dd Dd D = Dominant disease allele d = Recessive normal allele 21

Mo. I 2: Autosomal recessive (AR) u u u u Affected individuals are homozygous; Dd Dd heterozygotes are normal Both sexes equally affected Sometimes causes ‘heterozygote dd dd DD advantage’ D = Dominant normal allele u E. g. Sickle cell disease and malaria d = Recessive disease allele Can ‘skip’ generations On average, 25% of children of two carriers are affected Consanguinity is often high between parents of patients u Especially for very rare AR disorders Often onset is in childhood u Examples: Cystic fibrosis, AR intellectual disability 22

Sex-linked inheritance u Males = XY, Females = XX u For genes carried on X chromosome inheritance depends on: (i) Whether the mother or father was the mutation carrier (ii) Sex of the offspring (iii) Dominant or recessive 23

Mo. I 3: X-linked Dominant u d/- D/d u d/- D/- d/d u D = Disease allele d = Normal allele D/d d/- Affected females are heterozygous (mostly), affected males hemizygous Frequency of affected females is twice the frequency of affected males Disease is transmitted by mothers to children of both sex; by fathers, only to daughters 50% of children of affected mothers, and 100% of daughters of affected fathers, are affected E. g. Fragile X syndrome 24

Mo. I 4: X-linked Recessive. D/- u D/d u . D/- u D/d d/- u u D/- D = Disease allele d = Normal allele d/- D/D Usually only males are affected Disease is transmitted by carrier mothers to 50% of sons On average, 50% of daughters of carrier mothers are also carriers 100% of daughters of affected males are carriers E. g. Red-green colour blindness (Slide in appendix) . D/d D/- 25

Why no Mo. I 5? Y-linked or mitochondrial? • No disorder linked to Y chromosome so far • Think about how it would be (slide in appendix)… • Mitochondrial DNA (mt. DNA) not nuclear thus does not fit into Mendelian inheritance (slide in appendix) • Strictly maternally inherited • To all children – regardless of sex • Complex disorders are called ‘complex’ because they do not follow a distinct inheritance pattern (slide in appendix) – due to polygenicity and environmental factors 26

Hardy-Weinburg equation u Genotype frequencies and allele frequencies are stable across generations u If population obeys H-W equation, it is said to be in H-W equilibrium u H-W equilibrium assumptions in appendix Genotype frequencies are equal to the product of the relevant allele frequencies. If locus alleles are: A = with frequency p a = with frequency q where p + q = 1 u u Then the three genotypes: AA Aa aa have respective frequencies: p 2 2 pq q 2 (p 2 + 2 pq + q 2 = 1) 27

Exercise 2 u How many people carry Cystic Fibrosis causal allele? u The prevalence of cystic fibrosis in Caucasians is about 1 in 2500 u Assuming that all cases of cystic fibrosis are due to the action of one particular recessive mutation, what is the allele frequency of this mutation in the population? u HINT: Use the Hardy-Weinberg equation u What is the frequency of the heterozygote “carriers” in the population? 28

Answer D = Dominant normal allele d = Recessive disease allele u p = frequency of “normal” allele q = frequency of cystic fibrosis allele u • Freq. of diseased individuals = Freq(genotype = dd) = 1 in 2500 • So q 2 = 1/2500 • Therefore q = 1/50 and p = 49/50 (1 – 1/50) • Freq of heterozygote carriers = 2 pq = 2 x (49/50) x (1/50) ≈. 0392 ≈ 4% • 1 in 25 White Europeans is a carrier!!! 29

Law of independent assortment • Save for certain exceptions (will be discussed in a moment), alleles at different loci segregate independently during gamete formation. • To observe this process you can follow two characters at a time, to demonstrate their ‘Independent Assortment’. • These experiments use a di-hybrid cross. If independent, the combination of two traits gives a 9: 3: 3: 1 ratio! 30

S smooth seeds s wrinkled seeds Y yellow seeds y green seeds Parental Mating: SSYY x ssyy 31

Law of independent assortment u The biological basis of independent assortment lies in the fact that genes reside on different chromosomes and are therefore transmitted independently during meiosis u Mendel has been lucky in that the loci controlling most of the traits he studied were located on different chromosomes, or far apart on the same chromosome Chromosomal location of the genes that Mendel studied (see slide 4) Source: Fairbanks, DJ & Rytting B: American Journal of Botany, Vol. 88, No. 5 (May 2001) p. 746 32

What happens if genes lie on the same chromosome? u Genes which lie close to each other on the SAME chromosome tend to be transmitted together during meiosis u Therefore genes which lie close to each other on the same chromosome WILL NOT assort independently (i. e. are linked) u There will be an overrepresentation of PARENTAL type gametes 33

Meiosis and crossing over Crash. Course video : Meiosis http: //www. youtube. com/watch? v=q. CLm. R 9 -YY 7 o 34

Linkage Parental genotype A 1 A 2 Q 1 Q 2 A 1 Q 1 A 2 Q 2 Non-recombinant Parental genotypes A 1 Q 2 A 2 Q 1 Recombinant genotypes 35

Are flower and seed colours linked? PP YY pp x yy Pp Yy P Y ‘P’ and ‘Y’ must be on one chromosome and ‘p’ and ‘y’ on the other p y 36

If two loci are linked (i. e. on same chromosome)… P Y p y Parental type (LOTS) Y p y X p P y p y P y Recombinant type (FEW) p y p Y Recombinant type (FEW) p y

If two loci are on different chromosomes (i. e. unlinked)… P p y Parental type (1/4) y X p Y Y p p P y y p y P y p Y p y Recombinant type (1/4)

Learning Objectives � Be aware of Mendel and the significance of his early breeding experiments � Know Mendel’s principles of segregation and independent assortment and how they relate to current genetic knowledge � Be able to discuss concepts of ◦ Dominance ◦ Sex-linked inheritance � Be able to identify features of (monogenic) autosomal or X-linked, and dominant or recessive diseases 39

Thank You Any questions? Please look back at the slides again once you complete the short-course(s)

Appendices Test your knowledge: http: //biology. kenyon. edu/courses/biol 114/TUTORIAL/inherit 1 c. htm l

Penetrance u Penetrance is the conditional probability of having the phenotype P given the genotype G: Pr(P|G) u We have encountered examples of penetrance parameters before, they are the familiar dominant and recessive examples u i. e. For a Autosomal Dominant Mendelian disease where D denotes the disease allele: – Pr(P | G = DD) = 1 – Pr(P | G = Dd) = 1 – Pr(P | G = dd) = 0 u Penetrance may depend on various factors, such as the genotype at another locus, sex, age, imprinting 42

Hardy-Weinburg equilibrium assumptions u u u Random mating (i. e. no consanguinity or endogamy) No migration No inbreeding No selection related to genotype A large population (preferably infinite ) No new mutations Very strong assumption! 43

Red-Green Colour blindness u u Protanopia is a colour vision deficiency in which the red retinal photoreceptors are absent affecting red-green hue discrimination Deuteranopia is a colour vision deficiency in which the green retinal photoreceptors are absent affecting red-green hue discrimination Both due to a defective gene cluster on the X chromosome Much more common in males 44

Y-Linked Mo. I? 45

Mitochondrial Mo. I 46

Complex disorders/traits (e. g. Obesity, Height) u Polygenic u More than one gene involved u Sometimes tens or even hundreds of genes involved u Environmental factors can also affect outcome u E. g. smoking and lung cancer u Think of how a complex trait may look if we were to analyse a family pedigree… 47

Complications: X chromosome inactivation u Females have two copies of genes on their X chromosome (XX) whereas males only have one (XY) u To ensure the same gene dosage between males and females, one of these chromosomes is inactivated in females u Which chromosome is inactivated may differ from cell to cell and is random in most mammals The Calico Cat 48

Complications: Epistasis u One gene affects the expression of another gene u Phenotype depends on the interaction between genotypes at different loci. u In the case of 2 loci, the familiar 9: 3: 3: 1 ratio will not be observed u u E. g. Black and Yellow Labrador Alleles at the different loci may still assort independently 49

Complications: Epistasis 50

Complications: Epistasis Independent assortment, but ratio is not what we would predict (9: 4: 3) Two loci: B = Black b = brown E/e = gold locus More info: www. nature. com/scitable/topicpage/epistasis-gene-interaction-and-phenotype-effects-460 51

*Many other complications! u Complementation u Occurs when organisms with different homozygous recessive mutations produce the wild-type phenotype u E. g. a change in wing structure in flies u No simple examples in humans! u Co-dominance u Incomplete dominance *see www. biologycorner. com/bio 2/genetics/notes_incomplete_dominance. html 52