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STUDY UNIT 2

MENDELIAN GENETICS

Klug, Cummings & Spencer Chapter 3

3.1 Mendelian / Transmission genetics

How do we determine the inheritance patterns of traits?

Gregor Johann Mendel 1866 - reported rules that explain the inheritance of traits from parents to progeny.

Good experimental design:Pisum sativum, easy to cultivate and cross-breed.Seven characters, each with two contrasting forms.Experimental methodology.Accurate quantitative records.Scientific analysis.

3.2 Monohybrid cross reveals how ONE trait is transmitted from generation to generation

• Choose true breeding parents (P).• Cross parents with phenotypic differences in one trait

e.g. tall plants x dwarf plants.(trait = plant size)

• Obtain monohybrid progeny (F1).• Self-fertilize F1 to obtain next generation (F2).• Evaluate progeny phenotypes and numbers.• Propose genetic model.

Mendel’s results when considering one trait:

Plant size:

P:

F1:

F2:

RECIPROCAL CROSS � same results

� particulate unit factors determine each trait

Mendel’s results for seven traits:

Fig 3.1

Mendel’s first three postulates:

1. Unit factors in pairs

2. Dominance / recessiveness

3. Segregation

� Explanation for results of monohybrid crosses.

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Mendel in modern genetic terminology:

Phenotype: Physical expression of a trait.

Genotype: Genetic makeup of an individual.

Genes: Genetic factors (units of inheritance) that determine specific traits.

Alleles: Alternative forms of a single gene.

Homozygote: Identical alleles for a specific gene.

Heterozygote: Different alleles for a specific gene.

Convention for naming genes:

Trait � dominant form uppercaserecessive form � first letter of word � lowercase

E.g.Size � tall (dominant) D-allele

dwarf (recessive) � dwarf � d-allele

Combining Mendel’s postulates and modern terminology:

The monohybrid cross

Fig 3.2

Punnett squares:

Fig 3.3

This type of allelic interaction = (Complete) DOMINANCE

Dominant phenotype: phenotype expressed when genotype contains 1 or 2 copies of dominant allele

Dominant allele: allele expressed phenotypically in heterozygote, or allele of which one copy is sufficient to determine dominant phenotype

Recessive phenotype: phenotype expressed in absence of dominant alleles, when genotype contains only recessive alleles

Recessive allele: allele expressed phenotypically only in absence of dominant allele

In this example:

Dominant phenotype =

Dominant allele =

Recessive phenotype =

Recessive allele =

The testcross: one characterF2 plant with tall phenotype � is genotype DD or Dd?

Testcross:

Fig 3.4

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Examples:(Prac book 2.2)

In foxes, silver coat colour is governed by a recessive allele (b) and red by the dominant allele (B). Give the genotypic and phenotypic ratios of the progeny of:

(a) carrier red x silver (b) pure red x silver

(Prac book 2.5)

In rabbits short hair is due to a dominant allele (A) and long hair to its recessive allele (a). A cross between a short-haired � and a long-haired � produces 1 long-haired and 7 short-haired rabbits.

a) Give the genotypes of the parents.b) What was the expected phenotypic ratio of the

progeny?c) How many rabbits were expected to have long hair? d) Explain why only one with long hair was born.

3.3 Mendel’s dihybrid cross generated a unique F2 ratio

Dihybrid / two-factor cross involves two pairs of contrasting traits.

Fig 3.5

Consider each trait separately:

Colour Shape

F1: all yellow all round

F2: 9/16 + 3/16 = 12/16 yellow 12/16 round

3/16 + 1/16 = 4/16 green 4/16 wrinkled

� 3 yellow : 1 green � 3 round : 1 wrinkled

� Two pairs of contrasting traits are

� Can predict frequencies of F2 phenotypes

Fig 3.6

Mendel’s fourth postulates:

4. Independent assortment

REMEMBER:Rule of segregation: 2 alleles of a locus segregate, so that 1 allele of that locus is present in a gamete. However: every gamete receives one allele from everylocus.

E.g. for AaBb � A-allele has an equal chance to combine with the B- or the b-allele, etc.

�

The dihybrid cross:

Fig 3.7

See p 48: A molecular explanation how Mendel’s peas became wrinkled

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The testcross: two characters

F2 plant with dominant (yellow, round) phenotype �what is its genotype?

Testcross:

Fig 3.8

Drosophila nomenclature for allele symbols:

One characteristic – two phenotypes

Wild type (common) Mutant (rare)

use description of mutant phenotype to choose letter(s) to describe this gene

capital letter if mutant allele dominantsmall letter if mutant allele recessive

same letter with + as superscript for wild type allele

E.g.• Eye colour: red or white

• potential allele symbols:

• Wild type (common) = red, mutant (scarce) = white→→→→

• Determine dominance (from crosses)→→→→→→→→

• White (mutant) allele = Red (wild type) allele =

Example:(Prac book 2.12)

In Drosophila, ebony body colour is determined by a recessive allele (e) and wild-type grey colour by the dominant allele (e+). Vestigial wings are determined by a recessive allele (vg) and wild type normal wings by the dominant allele (vg+). Wild-type, dihybrid flies are mated and produce 256 progeny. How many flies are expected in each phenotypic class?

3.4 The trihybrid cross: Mendel’s principles apply to inheritance of multiple traits

• Consider inheritance of 3 or more loci simultaneously.• 2 alleles / locus in population, with dominance.

Remember during gamete formation:•

•

Fig 3.9

Use a Punnett square to calculate progeny ratios:

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Alternative approach: Use the forked-line or branch diagram method to predict ratios:

Fig 3.10

• Consider each trait separately• Combine results• True only if genes assort independently

Determining gametes of multihybrids: Forked-line method E.g. Individual AaBBCcDdeeFf How many gametes?

F ABCDeFD e f

Cd e F

f ABCdefA B

D e F etc.c f

Fd e f

FD e f

Cd e F

fa B

D e Fc f

d e Ff

n heterozygous loci � 2n different gametes

General rules:

Table 3.1

Example:

Consider an individual with genotype JjMMNn.If self-fertilisation occurs, what is the expected genotypic ratio of the progeny?

Approach 1:

Determine gametes, use Punnet square, determine progeny genotypes and ratios.

Approach 2:

Consider each locus independently, combine genotypic ratios using forked-line approach.

3.5 Mendel’s work was rediscovered in the early 20th

century

1866: Publication of Mendel’s results.Darwin, natural selection, continuous variation.

3.6 Mendel’s postulates correlate with behaviour of chromosomes

1879: Chromosomes.Early 1900’s – hybridization experiments with plants.1902: Chromosome behaviour during meiosis �Mendelian principles of segregation and independent assortment.

→→→→ Chromosomal theory of heredity.

Unit factors, genes and homologous chromosomes

Fig 3.11

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Ensure that you can integrate and understand the following terms and concepts with respect to mitosis, meiosis, and Mendelian genetics:

Diploid number, haploid number, gamete, zygote, chromosome, chromatid, homologous pair, maternal parent, paternal parent, locus, gene , allele, segregation, independent assortment, genotype, phenotype.

Pedigrees reveal patterns of inheritance in humans

• Designed crosses not possible in humans

• Few offspring, long generation times

• Construct a family tree, indicate presence or absence of trait in question for each member

• Pedigree analysis

Pedigree conventions:

Fig 3.13

Pedigree for an autosomal recessive trait, e.g. albinism

Pedigree analysis

Define allelic interaction, genotypes, phenotypes:

Fig 3.14a

Pedigree for an autosomal dominant trait, e.g. brachydactyly or Huntington’s disease

Define allelic interaction, genotypes, phenotypes:

Fig 3.14b Table 3.4

Example:(Klug Ch 3 Q 26 p 67)

Consider the following pedigree.

Predict the mode of inheritance and the most probable genotypes of each individual. Assume that the alleles A and a control the expression of the trait.

3.8 Laws of probability help to explain genetic events

Genetic ratios are most properly expressed as probabilities, that predict the outcome of fertilization events.Probabilities range from 0 (event certain not to occur) to 1 (event certain to occur).

The symbol P( ) is used to indicate probability.P (xyz) = probability to obtain xyz.

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Analysis can be based on calculating frequencies.

Frequency

Suppose 10 000 male students at Tuks9 000 female students at Tuks

19 000 TOTAL

Frequency (male):Frequency (female):

Probability that the next student you meet will be a female=

NB: Probabilities of all options for a single event = 1∴∴∴∴ 0.526 + 0.474 = 1

If two events are mutually exclusive (when “or” is used), the P of one or the other of the two events occurring is the sum of their individual P.

The sum law (“either / or” law)

E.g. a dice has 6 surfaces, numbered 1 to 6, each surface has an equal chance of landing face up.

What is the probability of rolling either a 1 or a 4 in the next throw?

• If you use the word OR, the SUM law applies.• Summate the respective probabilities.

∴P (rolling a 1 or 4) = P(1) + P(4) =

Genetic example:

Consider the cross Aa x Aa. What is the probability that any of the progeny will show the dominant trait? (I.e. what is the P for the AA or Aa genotypes in the progeny of this cross?)

When two events are independent of one another (“and”is used), the P that they will occur together is the productof their individual P.

The product law (“and” law)

E.g. a coin is tossed.P(head) = ½ = P(tail).If same coin is tossed again, or another coin is tossed, the P remains ½ for any of the events, as they are independent and do not influence each other.

What is the probability of obtaining a head on a 20c coin and a tail on a 50c coin if you toss them together?

• If you use the word AND, the PRODUCT law applies.• Multiply the respective probabilities.

∴∴∴∴ P (head on 20c) and P (tail on 50c)=

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Genetic example:

Consider the cross � PpRr x � Pprr. What is probability of obtaining a pprr homozygote in the progeny?

Sum- and product laws are often combined:

Again consider two coins that are tossed simultaneously.What is P of obtaining heads on the one coin and tails on the other?

There are now two possible outcomes, i.e. heads on 20c and tails on 50c or tails on 20c and heads on 50c.

∴∴∴∴P [ H on 20c and T on 50c] or P [T on 20c and H on 50c]=

Product- & sum rules may be extended to more than 2 events:

E.g. Deck of cards has 13 types of cards (aces, twos, threes, …kings). P to draw a particular card = 1/13.

• What is P of drawing a face card (jack or queen or king) on your first draw?

• What is the P of drawing first an ace, and then a two, and then a six?

Conditional probability (Pc):If the probability of an outcome is dependent on a specific condition related to that outcome.

E.g. In the F2 of a monohybrid cross involving tall and dwarf plants, what is the probability that a tall plant is heterozygous(and not homozygous)?

Consider only tall plants, as all dwarf plant are homozygous recessive. Some tall plants are heterozygous.As outcome and condition are NOT independent, we CAN NOT use the product law.

Pc = P (outcome of interest) / P (specific condition)

Applications in genetic counselling:If one child in a family is affected with a recessive disorder, what is the probability that normal siblings are carriers of the disease allele (heterozygotic)?

Binomial theorem

To determine probability in the following situations:

• There are a number of occurrences of a specific event (e.g. children born in a family),

• there are two possible outcomes (boy or girl),• events must be mutually exclusive (only one or

the other can occur in individual),• events are independent (eg. birth of one child

does not influence next pregnancy),• no order is specified

(if order is specified: simply use product rule).

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E.g. In a family with 5 children, there are 6 possible combinations of males and females:

all 5 females4 females + 1 male

� 3 females + 2 males2 females + 3 males1 female + 4 malesall 5 males

The children may be born in any order.

Consider option �, then order of birth may be: M, F, F, M, F or F, F, F, M, M or M, M, F, F, F ……..or seven other possible orders.

How do we determine the probabilities of these different alternatives?

Binomial theorem: (a + b)n = 1

a, b = respective probabilities of 2 alternative outcomesn = number of trials

(a + b)33

(a + b)4

etc4

(a + b)22(a + b)11

Expanded binomialBinomialn

Exponents are determined by using pattern:(a + b)n = an, an-1 b, an-2 b2, an-2 b3, …….., bn

Numerical coefficients are determined by using Pascal’s triangle:

Table 3.2

1 6 15 20 15 6 1

Eg numbers of events n = 6(a + b)6

n = 6:

1 coefficient where 6 coefficient where

15 coefficient where �� ��

1 coefficient where

=

The probability of particular outcome can also be calculated using the following formula:

NB:P = n!!!! as bt s + t = n

s!!!!t!!!! a + b = 10! = 1x0 = 1

n = total number of eventss = number of times outcome a occurst = number of times outcome b occursa = probability of one outcomeb = probability of other outcome

n! / (s!t!) gives the numerical coefficient for any set of exponents.

Ex 1. What is the probability that 3 females and 2 males will be born in families with 5 children?

n = s = t = a = b =

Sequence of events NOT specified � use binomial expansion

Approach 1: Pascal’s triangle

Approach 2: Formula

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Ex 2. A husband and wife, both carriers of a recessive genetic disorder, are planning 6 children.

a. What is the probability that their first two children will be normal, and the next four affected?

b. What is the probability that only one of their children will be affected?

� Aa x Aa ��

GR: ¼ AA : ½ Aa : ¼ aa

PR: ¾ normal : ¼ affected

a = b =

a. Order of events specified - use product law.

b. Order of events not specified - use Pascal’s triangle or formula.

n = s = t =

3.9 Chi-square analysis evaluates the influence of chance on genetic data

• Deviations from expected ratios are often observed.

• Due to random fluctuations of chance events.

• Error much greater with small samples.

• How do you evaluate observed deviation?

E.g. In a monohybrid cross Aa (purple) x Aa (purple) you would expect progeny in the ratio 3 purple: 1 white.

Null hypothesis (H0): Assume data will fit a 3:1 ratio.

If 5 progeny individuals are analysed � by chance all may be purple.

If 1000 progeny individuals are analysed � more accurate data expected.

Expected number 750 purple : 250 whiteYou may observe 690 purple : 310 white

Is this deviation acceptable to still fit a 3:1 ratio?Will the H0 be rejected or not rejected?

The observed deviation from an expected ratio may be:• attributed to chance alone (sampling error) �

• not attributed to chance alone (probably other reasons) �

Statistical test for goodness of fit of H0

� chi-square (χχχχ2) analysis

χχχχ2 = statistical test to determine if

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1. Formulate hypothesis.

2. Calculate χχχχ2 value from results.χχχχ2 = ���� (observed - expected) 2

expected

3. Interpret χχχχ2 value.If deviation is small � small χχχχ2 value �

If deviation is large � bigger χχχχ2 value �

How do we decide to reject or not reject a hypothesis?

Compare calculated χχχχ2 value from the data with a critical χχχχ2 value from a standard chi-square table or graph.

Fig 3.12

Note:

Fig 3-12 in the 9th (new) edition of the textbook is correct.

Please correct the legend of Fig 3-12 in the 8th (old) edition of Klug p 55 so that it reads as follows:

(a) …….. (b) Table of χχχχ2 values for selected values of df and p. χχχχ2 values that lead to a p value of 0.05 or greater (darker blue ares) justify failure to reject the null hypothesis. Values leading to a p value of less than 0.05 (lighter blue areas justify rejecting the null hypothesis. For example ……

To determine the critical χχχχ2 table value you need two reference points:

1. Degrees of freedom (df)Df = Need to be taken into account as with more categories more deviation is expected as a result of chance.

2. ProbabilityStandard for biological samples usually p value of 0.05.

p < 0.05: Probability < 5% (very small) that

p > 0.05: Probability > 5% that

E.g. the critical χχχχ2 table value for df = 5 and p = 0.05 is 11.07

Fig 3.12 (b)

• If calculated χχχχ2 value < tabulated value �� ��

Deviation not significant , and probably due to chance.

• If calculated χχχχ2 value > tabulated value �� ��

Observed ratios differ significantly from the expected ratios, the deviation can not be explained by chancealone.

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Example:(Klug Ch 3 Q 23 p 67)

In one of Mendel’s dihybrid crosses, he observed 315 round yellow, 108 round green, 101 wrinkled yellow, and 32 wrinkled green F2 plants. Analyse these data using a χχχχ2 test to see if they fit a 9:3:3:1 ratio.

1. Formulate hypothesis.2. Calculate χχχχ2 value for data.3. Interpret χχχχ2 value.