basiscursus genetica

1

BASICS OF GENETICS

1.11.11.11.11.1 PRELIMINARIESPRELIMINARIESPRELIMINARIESPRELIMINARIESPRELIMINARIES

It is a matter of general observation that the cubs of a lion resemble the lion and pups of adog resemble the dog not only in shape but also in habits, strength and functional abilities. Thisphenomenon was described by a phrase ‘like begets like’ or ‘chips of the same block’. Evidently,there is something common between the parents and their offspring which was responsible for sucha tidy resemblance. Many speculations were made but none turned out to be correct unless modeof heredity in sexually reproducing animals was unraveled. It was Johann Gregor Mendel who gavethe concept of ‘factors’ being transmitted from parents to offsprings. These factors later turned outto be the structural units called genes.

Interestingly, variation is also a part of heredity because no two individual are identical in everyrespect. Even homozygous twins may differ in size, weight and in appearance due to acquiredcharacters. Thus, similarity and variation go hand in hand as the law of nature and constitute theraw material for a geneticist and breeder to work upon and evolve new strains, breeds, varieties etc.

It is not very far when gene was a hypothetical unit lying on the chromosome and somehowgoverning the expression of traits in an individual and imparting immortality to an individual bykeeping alive its characters from generation after generation albeit in different proportions. Breedingexperiments were the only tool available to a geneticist to speculate the genetic constitution of anindividual and plan his experiments for better production. It is only half of the century when thechemical nature of the gene was explored and they can now be replicated in a test tube, their naturecan be altered and they can be commanded to perform a desired function. In spite of these epochmaking discoveries, the fundamentals of genetics as laid down by early scientists have not changedand even today constitute the basis of green revolution in agriculture and white revolution in milkproduction. The purpose of this chapter is not to elaborate the theoretical part of genetics but tofamiliarise the readers with the basics of genetics and various terminologies used in the sequel. Forin-depth knowledge of the principles of genetics, the readers are advised to refer the books givenat the end of this chapter.

The universe consists of nonliving objects (inanimates) and living organisms (animates). Theformer are static, lifeless, nonsensual and nonreproducing whereas the latter have life, sensation,dynamism and are endowed with the capacity to reproduce. The structural and functional unit ofall living organisms is the cell. The ‘cell theory’ was propounded by Schleiden and Schwann in 1838and Virchow in 1855 which stated that all organisms are composed of one or more cells and thesecells arise only from preexisting cells. An organism may be as small as unicellular or as big as anelephant consisting of billions of trillion cells.

1

2 STATISTICAL ANALYSIS OF QUANTITATIVE GENETICS

CellCellCellCellCell

It is a microscopic unit circumscribed by a cell membrane in animals or a true cell wall inplants. The inner material (protoplasm) consists of a dense mass in the centre called ‘nucleus’ anda jelly like thick matrix surrounding the nucleus called ‘cytoplasm’. The latter contains manysubmicroscopic organelles like mitochondria, golgi bodies, centrosome, lysosomes, endoplasmicreticulum etc., which perform specialized functions in the cell. The nucleus contains a sap — thenucleoplasm and thread like bodies termed as chromosomes. The chromosomes are the structuralunit of inheritance and carry many genes on them which are the functional unit of a trait or character.The location on the chromosome where a gene rests is termed as locus. Each species has a specificnumber of chromosomes ranging from 4 in drosophila fly to 254 in shrimps (Eupagurus ochotensis).

In human beings, the number is 46 (23 pairs). Chromosomes always occur in pairs. All cellsof the body barring germ cells are called somatic cells and are characterised by the presence of thenumber of chromosomes specific for the species. On the other hand, gametic cells (ova, producedby ovary in female and sperm, produced by testes in male) contain half the number of chromosomespresent in a somatic cell deriving one member from each pair of chromosomes. In numerical terms,if n represents the number of pairs of chromosomes, then a somatic cell will have 2n chromosomesand the condition is known as diploid number whereas a gametic cell has only n chromosomes andis termed as haploid number. Occasionally, more than two chromosomes are present in a pair anddepending on its number, i.e., 3 or 4, the condition is known as triploid or tetraploid or in a generalterm polyploid.

Broadly, the chromosomes have been classified as autosomes ( )n − 1 pairs) and sex chromosomes(one pair). The former determine the traits of an individual while the latter are responsible fordetermining the sex of the individual. Again the sex chromosomes are of two types, X type and Ytype. The presence of two X (XX) chromosomes gives birth to a female, the condition is designatedas homogametic. Thus, the female produces only one type (X type) of gametes. On the other handa XY composition symbolises a male, the condition is called heterogametic and the individualproduces two types of gametes viz., X type and Y type. Thus, all females are homogametic and allmales are heterogametic. A condition of YY chromosomes does not exist. Rarely abnormal individualssuch as XXY (Klinfelter’s syndrome) and XO (Turner’s syndrome) have been detected. They arealways sterile.

GGGGGeneeneeneeneene

Genes are the structural and functional units lying on the chromosomes. Like chromosomesthey exist in pairs, one on each member of the chromosome and determine the biological characterof an individual. The two sister genes on a particular locus may or may not be identical but arealways complementary to each other. They are technically known as alleles, the abbreviated formof the word allelomorph. Chemically, the genes are composed of deoxyribonucleic acid (DNA) anddirect the cell to perform a particular function by using chemical triplet code by way of messengerribonucleic acid (mRNA).

There are a large number of genes existing on each chromosome but their exact number is notknown. According to a rough estimate, by the use of chromosomal maps, the number of genes inhuman beings has been found to be around 30,000 which may be far from reality. Bo-yuan and hiscolleagues of Ohio State University estimated that the Human Genome Consortium is 65,000–75,000.

BASICS OF GENETICS 3

Cell DivisionCell DivisionCell DivisionCell DivisionCell Division

There are two chief processes of cell division (i) Mitosis (ii) Meiosis.

MitosisMitosisMitosisMitosisMitosis

In the process of mitosis which is normally present in all somatic cells, the sister chromosomesof each pair become distinct, thickened and arrange themselves in a single plane about the centreof the cell. Each chromosome is then duplicated, when for a short time the number of chromosomesis doubled with their genes. One of the replicated chromosome passes to each pole of the cell. Thecell wall constricts and the cytoplasm is divided into two equal halves giving rise to two daughtercells which are exactly similar to that of the mother cell and each daughter cell carries diploid numberof chromosomes. Thus, mitosis is the process of increasing the cell number without disturbingthe genetic constitution of the cell.

MMMMMeiosiseiosiseiosiseiosiseiosis

This is also known as the reduction division. The main feature of this process of cell divisionis that one partner of each pair of chromosomes passes into daughter cells with the result that eachdaughter cell has half the number of chromosomes (n) of the parent cell. This process is observedin gametogenic cells and the gametes (ova, sperm) so formed have only half the number ofchromosomes (haploid number).

ReproductionReproductionReproductionReproductionReproduction

As mentioned earlier, the living organisms possess the capacity to reproduce i.e., to create thenew ones of their own kind to maintain the continuity of the species in space and time. Reproductionis also a necessity to replace the individuals which grow old and die. In general, two types ofreproduction has been observed namely, (i) asexual (ii) sexual.

Asexual ReproductionAsexual ReproductionAsexual ReproductionAsexual ReproductionAsexual Reproduction

This type of reproduction is normally observed in unicellular organisms and certain plants. Herean organism is divided into two parts such that each part is regenerated into complete organismsimilar to that of the parent organism. In some plants, a small portion such as root, stem or leafmay establish itself as a new plant. This process does not contribute to genetic variability and hardlyholds any importance in genetic analysis.

Sexual ReproductionSexual ReproductionSexual ReproductionSexual ReproductionSexual Reproduction

This occurs in animals, human beings and plants as well as in organisms where distinct dimor-phism exists i.e., where male and female individuals can be distinctly identified. It can be said thatthe vignettes of reproduction would have not been so prismatic, had there been no sexual reproduction.In bisexual animals, reproduction is possible only with conjugal cooperation and coordination of thefemale and male. The male gonad, the testis produces male gametes called sperms.

The peculiarities of the male gametes are :

(i) they are produced in large numbers (in billions).

(ii) they are unicellular in nature.


(iii) they are motile.

(iv) each gamete contains half of the number of chromosomes.

(v) they are heterogametic i.e., half of the sperms produced contain X chromosomes and theother half Y chromosomes.

On the other hand, the female gamete,

Ovum is produced by the ovary and it contains the following properties.

(i) only one or few ova are produced at a time.

(ii) each ovum has half the number of chromosomes (n) and they are homogametic i.e., allthe ova produced contain X-chromosomes.

(iii) ovum is unicellular.

(iv) ovum is nonmotile.

After pairing, the male and female gametes unite restoring the normal 2n number of chromosomes.If the ovum with X chromosome happens to unite with the sperm containing X chromosome, theoffspring produced is a female. But if the ovum unites with a sperm having Y chromosome, theresulting young one is a male. Thus, the bridge between the offspring and the parents are thechromosomes with their genes which determine the traits or characters of an individual. Hence, thephenomenon of transmission of traits from parents to offsprings is known as heredity and thescience dealing with the resemblance and differences among the related organisms has been termedas Genetics (Beatson 1905).

1.21.21.21.21.2 FUNDAMENTALS OF GENETICSFUNDAMENTALS OF GENETICSFUNDAMENTALS OF GENETICSFUNDAMENTALS OF GENETICSFUNDAMENTALS OF GENETICS

It is a matter of common experience that the individuals of different species differ widely bothin structure and function. Not only that, even the individuals within a species are highly variable. Theoffsprings inherit characters from their parents but still neither the parents and their offsprings, northe real brothers and sisters are identical in all respects. The variation between individuals is attributedcompositely to genetic composition, environmental factors and their interaction. A number of theorieswere propounded from time to time to explain the possible mechanism of inheritance based onexperience and speculations, but none stood the test of time. Later, the theory of quantitative orbiometrical genetics came into existence on the basis of experiments conducted by Gregor J. Mendelhonoured as the ‘Father of Genetics’ on garden pea in 1866. Mendel proposed two laws of inheritancewhich were not recognised until the same were rediscovered and reaffirmed by Derries in Holland,Correns in Germany and Tschermark in Austria in 1900. Mendel died in 1884 and never lived to seethat he had opened the water gates to a new science called genetics.

Mendel’s first principle of inheritance is called ‘the law of segregation’. It states that “theallelic genes in zygote do not blend or contaminate each other but segregate and pass into differentgametes”. The segregation occurs at the meotic division.

Mendel’s second principle is known as the law of independent assortment. It states that,“during meiosis one of the chromosome in the pair is contributed independently in the gamete withoutbeing influenced by other chromosomes or cytoplasmic factors”. To explain this law, it may berecalled that the chromosomes and the genes always exist in pairs (all allelomorph) and at the timeof reduction division during gamete formation, any one partner of the pair goes to one gamete andthe other to the other gamete. Thus, the gametes formed may have similar number of chromosomes


and the genes but may not be identical. To elaborate it further, suppose a species has two pairs ofchromosomes with one allelic gene on each pair of chromosome as follows :

Thus, if there are two pairs of allelic genes, four types of gametes with different geneticcomposition are formed, provided both the pairs are heterozygous. If the number of allelic pairs isn, the possible number of gametes is 2n. This is due to uninhibited combination of any one allele foreach pair of gene and is called independent assortment.

A third principle recognised by Mendel is called ‘law of dominance’. The law emphasisesthat the two alleles of a particular gene may not be identical and may impart different character toa particular trait e.g., one allele of a particular gene, say R, may impart red colour to a flower whileits allele, r, may be responsible for white colour to the flower. When either is present in a homozygouscondition, the colour of the flower is either red (RR) or white (rr). On the other hand, if they arepresent in a heterozygous condition (Rr), the expression of the white colour is masked by thedominant red colour and the flower emerges as red. In this case, red colour allele (R) is said to bedominant and white colour gene (r) is said to be recessive.

From the above, it is evident that an organism inherits genes from its parents which areresponsible for the expression of characters. However, these genes are not expressed in vacuum butneed proper environment to manifest their maximal capacity. If the required environment is notavailable, the expression of the character may be inhibited fully or partially. For example, a plant mayhave the gene to grow tall but if the soil is deficient in nitrogen, minerals and water, the plant’sgrowth may remain stunted belying the presence of gene for tallness. Thus, an individual can bedeemed as genotype depending on the genetic material inherited from its parents and what it appearsto be is phenotype. Ideally, the genotype can be defined as the inherited genetic constitution or thesum total of heredity that an individual receives from its parents. Thus the total genic composition(genome) carries the blue print of structure and function for an individual. On the other hand, aphenotype is the appearance or performance of an organism as a result of the interaction of agenotype with that of a given environment. In nut shell, a genotype determines, what an individualshould be while phenotype reveals what it is. Further, a genotype is constant from birth till deathwhereas phenotype changes with time, place, age, food, environment etc.

Qualitative and Quantitative CharactersQualitative and Quantitative CharactersQualitative and Quantitative CharactersQualitative and Quantitative CharactersQualitative and Quantitative Characters

With the gain in knowledge about genes, their interaction and expression, it may be emphasizedthat all characters are not controlled by a single gene but a trait may be controlled by one, two, threeor more genes. Accordingly, the traits may be classified into (i) qualitative characters (ii) quantitativecharacters.

Qualitative characters are those which are characterised by the presence or absence of a trait.They are governed by one, two or at the most three pairs of genes. Examples of such charactersare, presence or absence of flowers of a particular colour, shape of comb in male fowl, plumagecolour in Andalusian fowl, shape of seeds etc.

Quantitative characters, on the other hand, are characterised by the presence of continuoustype of variation (range) and are controlled by multiple number of genes, each gene contributing its


share to the totality of the character. Mendelian laws can not directly be applied to such characters.Examples of quantitative characters are milk yield of a cow, yield of a crop, egg production, sizeof fruits, etc.

Qualitative GeneticsQualitative GeneticsQualitative GeneticsQualitative GeneticsQualitative Genetics

It deals with the qualitative characters. The individuals are classified into a number of distinctclasses and as such the qualitative variation can be studied by considering the frequencies intovarious classes. To test whether the observed frequencies are in agreement with the hypotheticalfrequencies based on Mendelian laws, one usually applies chi-square test.

Quantitative GeneticsQuantitative GeneticsQuantitative GeneticsQuantitative GeneticsQuantitative Genetics

It deals with the quantitative characters of a population. Inferences are drawn about thepopulation from samples collected from the same population. Each individual is represented by anumerical value usually considered as a continuous variable and all statistical tools used for measuringvariation of continuous variable(s) can be applied. Normally, the methods involved are mean, variance,analysis of variance, correlation and regression, path coefficients, etc.

TTTTTest Crossingsest Crossingsest Crossingsest Crossingsest Crossings

Crossing between individuals of prebreeding history or known genetic constitution for a particulartrait is known as the test crossing. This is a useful tool to explore the ingenuity of the genic natureof a character. Depending on the number of genes involved in governing a trait, different terminologieshave been used. A cross between two individuals controlled by a single pair of genes is called amonohybrid cross. In this case, if both male and female are homozygous dominant, all offsprings willbe homozygous dominant as depicted in Fig. (1.2.1).

Fig. 1.2.1 All homozygous tall.

Similarly, if both sexes are homozygous recessive, all offsprings will be true homozygousrecessive. The same phenomenon is exhibited in Fig. (1.2.2).

Fig. 1.2.2 All homozygous recessive.

On the other hand, if one mate is homozygous dominant and the other is homozygous recessive,all offsprings will genetically be heterozygous but will reveal dominant character because the presenceof dominant gene masks the expression of recessive gene as presented diagrammatically in Fig. (1.2.3).


Fig. 1.2.3 All heterozygous tall.

But the selfing among F1 generation to yield F2 generation results into a genotypic ratio of1 : 2 : 1 and phenotypic ratio 3 : 1. The proportion between different types of individuals in F2generation is known as monohybrid ratio. A cross between F1 generation is shown in Fig. (1.2.4).

Fig. 1.2.4 Homozygous tall Heterozygous tall Homozygous recessive

Another method to test the genotype for a monohybrid cross is to cross an F1 individual withits homozygous recessive parent (tt). This is known as back cross or test cross. As a special case,two crosses, A × B in which A has the female parent of P1 and male parent of P2 and vice-versaare known as reciprocal crosses. They are of great utility in understanding the genetic compositionof monohybrid characters. The crosses involving two pairs of genes or two characters differentiablyseparable in inheritance is called a dihybrid cross. Similarly, a cross involving three pairs of genesor three characters is known as trihybrid cross and those involving many genes are called polyhybridcrosses.

Genetic and Phenotypic VGenetic and Phenotypic VGenetic and Phenotypic VGenetic and Phenotypic VGenetic and Phenotypic Variationariationariationariationariation

Variation is the law of nature. All living beings whether plants, animals or humans differ inshape, size, colour, appearance and functions from each other. This difference is called variationand can be classified into genetic and phenotypic variation.

Genetic VGenetic VGenetic VGenetic VGenetic Variationariationariationariationariation

It is well documented that each character of an individual is governed by one or morenumber of genes. Further, it has been established that species has particular number of chromosomeswith thousands of genes on them. The genetic orientation of these genes varies from individualto individual and gets multiplied in each generation as per the laws of segregation and independentassortment advocated by Mendel.

Another factor which constitutes to genetic variation is the sudden alteration in genic structuredue to natural or man made causes such as X-rays, ultra violet rays, radiations from radioisopicsubstances such as radium, uranium, etc. These genic modifications are called mutations. Changescaused by mutations are permanent and heritable to next generation. Thus, the variation resulting dueto difference in genetic constitution between individuals is called genetic variation. As explainedlater, the genetic variation provides raw material for selection to improve the variety of a trait in plantsor animals.


Phenotypic VPhenotypic VPhenotypic VPhenotypic VPhenotypic Variationariationariationariationariation

As discussed, genes are the basic architects responsible for the characters of an organism butthey do not express in vacuum. They need proper environment to manifest themselves. The geneexpression becomes modified depending on the suitability of the environment to which an organismis exposed. Thus, the variation brought about by the environment effect(s) is called phenotypicvariation. This fact can be demonstrated by the illustration that two cuttings taken from achrysanthemum plant will have exactly the same genetic constitution. Let these be planted in twoseparate pots. If one is provided with adequate manure and water, while the other is deprived of theseessentials, it will be observed that the former grows rapidly whereas the later has poor or no growth.This substantiates the role of environment in gene expression.

It is emphasized that the genotypic variation is inherited but phenotypic variation is temporaryand is not transmitted to the next generation.

Thus, in the sequel it will be discussed that how the total variation observed between individualscan be attributed and partitioned into genetic variation, phenotypic variation and the variation due tointeraction between genetic and environmental effects. Further, how can it be harnessed to improvethe quality of a trait by using selective breeding as a tool.

1.31.31.31.31.3 PROBABILITY AND HEREDITYPROBABILITY AND HEREDITYPROBABILITY AND HEREDITYPROBABILITY AND HEREDITYPROBABILITY AND HEREDITY

Laws of probability are of great help to determine the chance of getting an offspring havingcertain desired traits. To delve deep on probability, an inquisitive mind should read a book onprobability theory namely, Feller, W. (1966), Haight, F.A. (1981).

One is often anxious to know the sex of a foetus, the colour of a flower, shape of corn, etc.From the angle of probability, it will not be wrong to assume that the probability of a male birth andthat of a female is half and half. We consider the case where a couple wants to know, what is thechance that out of two births, both the children will be boys. If first birth is a male, the sex of thechild at second birth is independent of the sex of the child at first birth. So the probability of both

the children being male, by multiplicative law of probability, is 12

12

14

× .= Again the probability of

a boy and a girl is 12

. Probability 12

is obtained by the consideration that the probability of a boy

at first birth and of a girl at second birth is 12

12

14

× = and that of a girl at first birth and of a boy

at second birth is also 12

12

14

× = . Hence, the probability of a boy and a girl is 14

14

12

+ = . Similarly,

the probabilities in case of three births to a couple having all the boys, two boy and one girl, oneboy and two girls or all the three girls can be calculated by the respective terms of the binomialexpansion of (p + q)3 where p is the probability of a male birth and q of a female birth, each equal

to 12

. We know, (p + q)3 = p3 + 3p2q + 3pq2 + q3. On substituting the value of p and q, the

probabilities of the four events are, 18

38

38

, , and 18

respectively.


Consider another problem. In a couple, the wife is albino and husband is normal. Let thedominant gene for normal skinned colour be denoted by A and that of albino by recessive a. Thehusband may be carrying the genes Aa with normal skin colour and also carrying an allele a foralbino while the albino wife will have the genes aa. On coupling, the probability of the offspring Aa,

normal skinned child is 12

i.e., the gene A from husband and a from wife and that of aa is also12

i.e., the gene a from husband as well as from the wife. Suppose the first child is albino. The coupleis interested to know the chance of the second child being albino. Since the skin colour of the firstchild has no bearing on the colour of the second child, the probability of the second child being albino

is also 12

.

Now consider the problem in a different manner. The couple is interested to know the chance

of a normal skinned male birth. We know that the probability of the birth of a boy is 12

and that

of a normal skinned child is 12

. So the probability of a normal skinned boy is 12

12

14

× = and so

is the probability for a normal skinned girl. It mean that the couple may expect one normal skinnedboy or girl out of four children.

1.41.41.41.41.4 ASCERTAINMENT OF HYBRID RATIOSASCERTAINMENT OF HYBRID RATIOSASCERTAINMENT OF HYBRID RATIOSASCERTAINMENT OF HYBRID RATIOSASCERTAINMENT OF HYBRID RATIOS

When a cross is made between a Yellow-Round (GGWW) and Green-Wrinkled (ggww) peavarieties, all the seeds in F1 are Yellow-Round (GgWw) as only GW type of gametes are producedby one parent and gw type by the other parent. F1 plants will produce four types of gametes (pollensas well as ova) and this will have sixteen types of zygotic combinations in F2 generation. The possibleoutcomes (zygotic combinations) are depicted in the chequer-board given below.

Parents Yellow-Round Green-Wrinkled(GG WW) (gg ww)

Gametes GW gwGeneration F1 will have gamete Gg Ww. However, in the F2 generation, individuals with

combinations of pea colour and shape will appear as shown in the following chequer board.


On counting, it is easy to note that the ratio of the four phenotypes namely, Yellow-Round,Yellow-Wrinkled, Green-Round and Green-Wrinkled is 9 : 3 : 3 : 1. The theoretical ratios of phenotypescan be obtained by considering the pairs of alleles separately and their ratio. Considering the pair ofalleles individually, the ratio of Yellow to Green is 3 : 1 and for Round to Wrinkled is also 3 : 1. Inthe cross under consideration, it is apparent that when one character is fixed, the other charactershows a clear cut ratio of 3 : 1. Therefore, it can be concluded that the characters under study areassorting (giving rise to different combinations) in an independent manner. In such a situation, theprinciple of inheritance is governed by Mendel’s second law known as the law of independentassortment. It states that—

“Different characters in the offsprings of hybrids are distributed independently of each other.”

The expected ratio for a hybrid cross can be obtained by vectors multiplication as follows:

To show the ratio, elements of the two rows of the right hand matrix are written at a stretchand the ratio for GW, Gw, gW and gw turns out to be, 9 : 3 : 3 : 1, the same as obtained throughthe chequer board. In the same way, the expected ratio for a trihybrid cross obtained by consideringthe three pairs of genes through vector multiplication is as follows:

Writing the elements of the two rows of the right hand matrix at a stretch, we get the ratio as,

27 : 9 : 9 : 3 : 9 : 3 : 3 : 1

↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓SGW SGw SgW Sgw sGW sGw sgW sgw

The above approach saves the labour of making the checker board.

TTTTTest of Goodness of Fest of Goodness of Fest of Goodness of Fest of Goodness of Fest of Goodness of Fititititit

A researcher is always anxious to confirm whether his frequencies in various classes followthe expected ratio or not. This test is known as a test of goodness of fit and is best performed bychi-square test. The test statistic is,

χ2 =( )O E

E

Ki i

ii

−

=∑

2

1

...(1.4.1)

Statistic χ2 is distributed with (K – 1) d.f. where K is the number of classes.

Example 1.4.1. The data of Mendel’s experiment obtained in a dihybrid cross using green peasbetween Yellow-Round seeds and Green-Wrinkled seeds is reproduced here. Out of 556 seeds, thefrequencies of seeds in four phenotypes were as follows:


Yellow-Round = 315, Green-Round = 108, Yellow-Wrinkled = 101, Green-Wrinkled = 32

We have to test the null hypothesis,

H0 : The frequencies in four phenotypes follow the ratio of 9 : 3 : 3 : 1.

H1 : H0 is not true.

Expected frequencies are obtained as,

E1 =9

16556 312 75 313× . ! ,= =

E2 =3

16556 104 25 104× . != =

E3 =3

16556 104× !=

E4 =1

16556 34 75 35× . != =

By the formula (10.4.1),

χ2=

( ) ( ) ( ) ( )315 313313

108 104104

101 104104

32 3535

2 2 2 2− + − + − + −

= 0.0128 + 0.1538 + 0.0865 + 0.2571

= 0.5102

Tabulated value of χ2 for 3 d.f. and α = 0.05 is 7.81. Since, the tabulated value of χ2 is more thanthe calculated value, the null hypothesis is accepted. Thus, the experimental data supports the expectedratio.

Only one example is cited here just to impress upon the investigators that chi-square is apopular and powerful test in genetic theory to confirm the law of segregation.

Probability is the base of most of the statistical methods. Sometimes one calculates probabilitydirectly or uses it indirectly. Now a number of statistical tools like path coefficient, coefficient ofinbreeding, heritability, sire index, repeatability, general combining ability (g.c.a.), specific combiningability (s.c.a.), genetic and phenotypic correlations, analysis of variance, analysis of diallel crosses,stability models and many other methods meant for measuring genetic variation and relationships aredealt with adequately in the following discussion.

1.51.51.51.51.5 CROSSING OVERCROSSING OVERCROSSING OVERCROSSING OVERCROSSING OVER

As mentioned earlier, the genes are located on chromosomes in pairs and segregate independentlyat meosis. The two genes located closely on the same chromosome go together expressingsimultaneously the characters of both genes in the offsprings. Such a situation is called linkage. Thefield trials conducted with characters of linked genes exhibited a few offsprings with recombinationof characters. The phenomenon was explained by crossing over of a section of homologouschromosomes between two linked genes in a few cells. This was demonstrated histochemically inthe dividing cells of gonads leading to formation of gametic cells. The chromosomes break at thepoint of crossing (chiasma) and join with the part of other chromatid thereby exchanging asection of homologous chromosomes and the genes located on it. The process results in recombinationof genes.


Diagramatic representation of crossing over of homologous chromosomes.

Recombination FractionRecombination FractionRecombination FractionRecombination FractionRecombination Fraction

Consider the cross of two doubly homozygous populations AB/AB × ab/ab under the assumptionthat they are linked. On crossing, F1 generation will be AB/ab. In case of complete linkage, theindividuals AB/ab will produce two types of gametes AB and ab. Back cross i.e., crossing F1 withthe recessive ab/ab, the progeny should have the type AB/ab or ab/ab only. But it has been observedthat in some cases, the progeny have a few individuals of the type Ab/ab or aB/ab in addition tothe type AB/ab or ab/ab. Though such an occurrence is seldom observed. Thus, the departure fromusual independent segregation giving rise to the new genetic combinations is termed as recombination.At the same time, the genotypes are named as recombinants. The phenomenon of exchanging thegenes between homologous chromosomes is known as crossing over. The proportion of recombinantsoccurring in the offspring due to crossing over is called recombination fraction and is denoted by

λ. The value of λ lies between 0 and 12

, i.e., 012

≤ ≤λ . λ = 0 indicate complete linkage between

loci. But when λ = 12

, one fails to detect whether there is no linkage or the loci are on different

chromosomes.

It is apparent that each pair of homologous chromosomes belongs to a linkage group. Obviously,there are as many linkage groups as there are chromosomes in a genome. The phenomenon oflinkage may be defined as,

“The tendency of two genes located on the same chromosome to be transmitted together inthe process of inheritance”.

Consider a single locus with K alleles A1, A2, ... AK. In this situation there can be possibly K

homozygotes of the type Ai Ai (i = 1, 2, ...., K) and K2FH

IK heterozygotes of the type Ai Aj (i = 1,

2, ..., K and i < j). So the total number of genotypes is,

K +K2FH

IK = K +

K(K K(K + 1)

2

K + 1

2− = =

FHG

IKJ

1

2

)...(1.5.1)

For example, there are three alleles A, B and O in the blood group system ABO. A and B are

dominant over O but are codominant to each other. So for three alleles, there will be 3 1

2+F

HIK = 6

genotypes. At the same time, the number of phenotypes can not be formulated as it depends on the


mode of dominance of alleles. In the present case, six genotypes and four phenotypes with threealleles A, B and O are,

Genotypes AA AO AB BB BO OO

Phenotypes A AB B O

If we consider n loci where the ith locus contains Ki alleles, the total number of genotypes inrespect of these n loci are,

K K K KK K1 2 n+F

HIK

+FH

IK

+FH

IK = +F

HIK = +

==∏∏1

21

21

21

212

111

... (in i i

i

n

i

n

...(1.5.2)

Now consider the case when organisms are polyploids say, 2m-ploids. Again suppose there areK alleles per locus and m homologous chromosomes, the number of genotypes is,

K + 2mm

−FH

IK

12

=K + 2

K 1m −−

FH

IK

1

and the number of gametes is,

K + mm

−FH

IK

1=

K +K 1

m −−

FH

IK

1

For instance, when m = 2, 2m = 4 i.e., tetraploid with two alleles A and a i.e., K = 2, thenumber of genotypes is,

2 4 14

+ −FH

IK =

54

51

5FHIK = F

HIK =

and the number of gametes is,

2 2 12

+ −FH

IK =

3

2

3

13

FHGIKJ =

FHGIKJ =

To elaborate further, for tetraploid all possible combinations of A and a are 24 i.e., 16. GenesA and a can occur in any manner on the four loci and the 16 combination result into 5 genotypesand 3 gametes as given below.

Combinations: AAAA AAAa AAaA AaAA aAAA

Genotypes: AAAA AAAa

AAaa AaAa AaaA aAaA aaAA aAAa

AAaa

Aaaa aAaa aaAa aaaA aaaa

Aaaa aaaa

Whereas the three gametes are AA, Aa, aa.

1.61.61.61.61.6 STATISTICAL TESTS FOR SEGREGATION AND LINKAGESTATISTICAL TESTS FOR SEGREGATION AND LINKAGESTATISTICAL TESTS FOR SEGREGATION AND LINKAGESTATISTICAL TESTS FOR SEGREGATION AND LINKAGESTATISTICAL TESTS FOR SEGREGATION AND LINKAGE

Mendel’s first law of segregation is stated in section 1.2. On the other hand, linkage may bedefined as, “The tendency of two genes on the same chromosome to remain united in the process


of inheritance”. The phenomenon of linkage is a sort of refutation of Mendel’s second law ofindependent assortment.

Statistical TStatistical TStatistical TStatistical TStatistical Test for Segregationest for Segregationest for Segregationest for Segregationest for Segregation

Consider heterozygous Aa type of individual. It can produce two gametes A and a. Differentsegregation ratios arise on the presumption that Aa produces the two gametes in equal number whichcan statistically be tested from the observed data. If we take its back cross with the recessive aa,it can result into two phenotypes Aa and aa in the ratio of 1:1. Suppose, an experiment or surveyhas a family of n individuals and it is observed that out of n, f are of the type AA and (n – f ) areof the type aa. Now to confirm whether the segretation ratio of 1:1 holds good, make use of chi-square test provided n is sufficiently large. We know for two classes having the observed frequenciesa and b for hypothetical ratio r : 1, the chi-square statistic is,

χ( )12 =

( )( )

a rb

r a b

−+

2

...(1.6.1)

In the situation under consideration, a = f, b = (n – f ) and r = 1. So the test statistic,

={ .( )}

( )f n f

f n f

− −+ −

11

2

...(1.6.2)

Suffix 1 of χ2 indicates its d.f.

If the calculated value of χ2 is greater than the tabulated value of χ2 for 1 d.f. and α = 0.05i.e., 3.841, One has to reject the segregation ratio 1:1. Rejection of segregation implies the presenceof linkage.

Consider now two loci with two factors A-a and B-b with dominance of A over a and of Bover b. Let the observed frequencies in an experiment with double back cross Aa Ba × ab ab beO1, O2, O3 and O4. The expected ratio of these frequencies in four phenotypic classes AB, Ab, aB,ab with no linkage between two factors will be 1 : 1 : 1 : 1. Let the total number of individuals under

study is n. So o nii

==∑ .

1

4

Under hypothetical ratio, the expected frequencies in each phenotype will

be n

4. The validity of the expected ratio can easily be tested by χ2-test. The test statistic in the

present case is,

χ2 =o

n

n

i

i

−FH IK=∑ 4

4

2

1

4

...(1.6.3)

Statistic χ2 has 3 d.f.

But if there is linkage between factors, the above situation does not prevail. We expect anexcess of Aa Bb and ab ab if in F1 they were in coupling (AB/ab) and an excess of Ab ab or aB abif they were in repulsion (Ab/aB). Thus the contrast

Aa Bb + ab ab – Abab – aBab ...(1.6.4)


is suitable for detection of linkage.

For single factor (A – a) detection of linkage, the appropriate contrast is,

Aa Bb + Ab ab – aB ab – abab ...(1.6.5)

Similarly, the contrast for single factor (B – b) detection of linkage is,

Aa Bb – aBab – Abab – abab ...(1.6.6)

Above concept can be summarised in the table below.

Table 1.6.1 Detection of linkage for backcrosses AaBb × ab ab

Phenotypes AaBb Abab aBab abab

AB Ab aB ab

Observed freq. O1 O2 O3 O4

Expected ratio 1 1 1 1

Contrasts for linkage

A vs. a +1 +1 –1 –1

B vs. b +1 –1 +1 –1

Linkage +1 –1 –1 +1

It can easily be verified that the three contrasts given in the table are orthogonal.

For testing of linkage in the three cases, there is the difference of the sum of frequencies oftwo phenotypes which are in the ratio 1:1. Therefore, the chi-square statistic each with 1 d.f. fortesting of linkage and segregation are :

For factor A – a,

χA−a2

={( ) ( )}O O O O1 2 3 4

2+ − +n

...(1.6.7)

For factor B – b,

χB−b2 =

{( ) ( )}O O O O1 3 2 42+ − +

n...(1.6.8)

For linkage,

χL2 =

{( ) ( )}O O O O1 4 2 32+ − +

n...(1.6.9)

Here, we test whether the families are homogeneous for two single factors A-a, B-b and forlinkage. The decision about the hypothesis is taken in the usual way by comparing the calculatedvalue of χ2 with table value of χ2 for 1 d.f. and α level of significance. From chi-square table,

χ12

05, . = 3.841.

The testing of segregation and linkage of two factors A – a and B – b for inbreedingexperiments in doubly heterozygous individuals or in F2 generation can be done by chi-square test.For an intercross Aa Ba × Aa Bb we have to test whether single factors A – a or B – b segregatein the ratio 3:1. Also the expected ratio of frequencies in the phenotypes A – B – , A – bb,


aaB –, aabb is 9 : 3 : 3 : 1. Without going into details, we give below the formulae for chi-squarestatistic each having 1 degree of freedom.

For the factor A – a,

χA−a2 =

{( ) ( )}O O O O1 2 3 423

3

+ − +n

(1.6.10)

For the factor B – b,

χB−b2 =

{( ) ( )}O O O O1 3 2 423

3

+ − +n

(1.6.11)

For linkage,

χL2 =

{( ) ( )}O O O O1 4 2 329 3

9

+ − +n

(1.6.12)

The decision about the expected ratio is taken in the usual manner.

Example 1.6.1. In rabbits, two recessive genes produce a solid body colour and long hairrespectively, in contrast to a spotted body colour and short hair which result from the dominantalleles. The results from a cross between the heterozygous spotted short haired rabbits and solidlong haired rabbits are as follows :

Class Frequency

Spotted/short 48

Spotted/long 5

Solid/short 7

Solid/long 40

If we want to perform a test for independent assortment in the above test cross to detect thelinkage between the genes, we can proceed in the following manner.

Denoting.

Spotted/solid body = Ss

Short hair/long hair = Ll

Under independent assortment, the frequencies in the classes SL, Sl, sL, sl should be in theratio, 1 : 1 : 1 : 1. Thus,

H0 : 1 : 1 : 1 : 1

Vs. H1 : H0 is not true,

can be tested by chi-square test by the formula (1.6.3). Here n = 100. Hence,

χ2=

( ) ( ) ( ) ( )48 2525

5 2525

7 2525

40 2525

2 2 2 2− + − + − + −

= 21.16 + 16.00 + 12.96 + 9.00

= 59.12.

Tabulated value of χ2 for 3 d.f. and a = 0.05 is 7.81, which is less than the calculated value


of χ2 = 59.12. So we reject H0. This leads us to conclude that there is no independent assortmentand therefore, the genes S and L are linked.

Example 1.6.2. In maize there is a dominant gene for coloured seed and another dominant genefor full seed. The recessive alleles of these genes produce colourless and shrunken seed.

A homozygous coloured and full seeded maize plant was crossed with a colourless andshrunken seeded one i.e., F1 plants selfed to raise the F2 progeny resulted into the followingphenotypic categories and their frequencies.

Coloured and Full (CF) : O1 = 920

Coloured and shrunken (Cf) : O2 = 310

Colourless and full (cF) : O3 = 285

Colourless and shrunken (cf) : O4 = 85

On the basis of the above experimental data, one would like to verify whether:

(i) The hypothetical ratio of 3:1 for phenotypic categories in F2 considering the segregationof only one character at a time stands true.

(ii)There exists a linkage between the genes for seed colour and shape.

For the character C – c, the hypothetical ratio of 3 : 1 can be tested by the chi-square testusing the formula (1.6.10). Thus,

χC c2

− ={ ( )}

×920 310 3 285 85

3 1600

2+ − +

Since n = 1600

= 120 1203 1600

××

= 3.00

Tabulated value of χ2 for 1 d.f. and α = 0.05 is 3.841, which is greater than χC c2

− = 3 00. .Hence we conclude that the genes pertaining to colour and colourlessness segregate in the ratio of3 : 1.

Similarly for the characted F-f, the statistic chi-square by the formula (1.6.11) is,

χF f2

− = { ( )}

×920 285 3 310 85

3 1600

2+ − +

=20 203 1600

××

= 0.0833

By comparing the value of χF f2

− with χ1 0 052 3 841( . ) . ,= one can easily infer that the genes for

full and shrunken traits segregate in the ratio of 3 : 1.

(iii) Now to perform a test for linkage between the genes for seed colour and shape, we makeuse of the chi-square statistic given by (1.6.12)

χL2 =

{( × ) ( )}×

920 9 85 3 310 2859 1600

2+ − +


=( )

×−100

9 1600

2

= 0.6944

Again the value of statistic is less than the tabulated value of χ1(.05)2 . Hence, we accept the

hypothetical ratio 9 : 3 : 3 : 1. This makes us to believe that there is no linkage between the genesresponsible for seed colour and shape.

Overall chi-square for testing the agreement to the expected ratio 9 : 3 : 3 : 1 in the fourcategories can be worked out as follows :

Expected frequencies in the four classes are 900, 300, 300 and 100 respectively. Thus,

χ2 =( ) ( ) ( ) ( )920 900

900310 300

300285 300

30085 100

100

2 2 2 2− + − + − + −

= 0.4444 + 0.3333 + 0.7500 + 2.2500

= 3.7777 != 3.78

Table value of χ2 for P = 0.05 and 3 d.f. is 7.815, which is greater than 3.78. Hence, it isconcluded that the data supports the expected ratio.

Notabene: It is interesting to verify that the sum of χ2C–c, χ2

F–f and χ2L is also 3.7777. It

affirms that there is no linkage.

1.7 MULTIPLE-GENE HEREDITY OR ADDITIVE GENE EFFECTS1.7 MULTIPLE-GENE HEREDITY OR ADDITIVE GENE EFFECTS1.7 MULTIPLE-GENE HEREDITY OR ADDITIVE GENE EFFECTS1.7 MULTIPLE-GENE HEREDITY OR ADDITIVE GENE EFFECTS1.7 MULTIPLE-GENE HEREDITY OR ADDITIVE GENE EFFECTS

As early as 1909, Nilsson-Ehle conducting breeding experiments on wheat observed thatcrossing of Red and white varieties yielded hybirds of different intensities of red colour dependingon the number of genes involved. He found that more the number of genes, darker the colour ofseeds. He propounded that several genes were involved with only partial dominance and cumulativeeffect but individually indistinguishable so that each gene contributed to the redness of the seed. Ifthe effects of two or more genes are a linear function of the individual gene effects, the effects ofthe genes are called multiple-gene heredity or additive gene effect. The condition is normallyobserved in quantitative traits like size of fruit, colour of seed, milk production etc.

It is difficult to say whether each gene is contributing equally or differently. In the lattercondition, the total effect Z will be a linear function of the gene effects. Consider K genes G1, G2,..., GK located at different loci affecting the trait. Let the effect of the gene Gi be αi (i = 1, 2,..., K). then,

Z = a1α1 + a2α2 + ... + aKαK ...(1.7.1)

If αi = α for all i, then

Z = (a1 + a2 + ... + aK)α ...(1.7.2)

where Σαi is the total number of genes.

1.81.81.81.81.8 INTERACTION OF GENESINTERACTION OF GENESINTERACTION OF GENESINTERACTION OF GENESINTERACTION OF GENESIn the post Mendelian era, the breeding experiments revealed that the characters controlled by

two or more genes located at different chromosomes do not express independently following theMendelian laws of inheritence and dominance. Rather the presence of one type of genes interacts


with expression of other gene producing a new trait entirely different from both parents. Thus, asituation where one gene influences the expression of another gene controlling the samecharacter is known as gene interaction.

A classical example of gene interaction was demonstrated by Batson and Punnett (1906) infowl. They selected fowls of two varieties, one with ‘Rose’ comb and the other with ‘Pea’ comb.The cross between these two varieties gave a new strain of comb called ‘Walnut’ comb in F1. Inter-se cross between F1 yielded progeny in the proportion of 9/16 Walnut, 3/16 rose, 3/16 pea and1/16 single type of comb. This is shown in the chequer board below :

Rose comb Pea comb

RR pp × PP rr

F Rr Pp × Rr Pp

Gametes RP Rp rP rp Result

RP [RRPP] [RRPp] [RrPP] [Rr Pp] [Walnut]— 9/16

Rp [RRPp] (RRpp) [RrPp] [Rrpp] (Red) — 3/16

rP [Rr PP] [Rr Pp] {rr PP} {rr Pp} {Pea} — 3/16

rp [Rr Pp] (Rr pp) {rr Pp} rr pp single — 1/16.

Earlier results had shown that either rose or pea when crossed with single comb fowl segregatedindependently in the ratio of 3:1 showing complete dominance over homozygous recessives. But thepresence of even one dominant gene for red comb (R) and pea comb (P) together interacted witheach other producing a new type called ‘Walnut’ comb. However, double recessive (rrpp) expresseditself as single comb. In fact, the results do not defy the laws of inheritance but clearly exhibit theinteraction between two dominant alleles and two recessive alleles.

EpistasisEpistasisEpistasisEpistasisEpistasis

This is another form of gene interaction. When two nonallelic genes affecting the same traitare present in an organism; the presence of one pair of genes masks or covers up the manifestationof other gene, then the phenomenon is called epistasis. The gene covering the expression of othergene is said to be epistatic while the gene masked is called hypotatic.

In a particular strain of dog two different nonallelic genes say A and B were found to affectthe colour of hairs. Gene A was responsible for black colour and presence of at least one B maskedthe colour resulting in white hair. Absence of A i.e., aa alongwith bb gave brown colour due to theabsence of A and B. Thus, mating of AA BB (white) with aabb (brown) gave all white Aa Bboffsprings in F1 and a ratio of 12 : 3 : 1 for white, black and brown in F2 respectively as depictedin the crosses below.


Gametes AB Ab aB ab

AB AABB AABb AaBB AaBb

White White White White

Ab AABb AAbb AaBb Aabb

White Black White Black

aB AaBB AaBb aaBB aaBb

White White White White

ab Aa Bb Aa bb aaBb aabb

White Black White Brown

F2 White Black Brown

12 3 1

The presence of one B has masked the colour of hair. So B is epistatic to A and A is hypostatic.This is an example of dominant espistasis.

Similarly there are several types of epistatic effects such as recessive epistasis giving a ratioof 9 : 3 : 4, incomplete duplicate recessive yielding a ratio of 9 : 6 : 1, duplicate recessive epistasisresulting in the ratio of 9 : 7, duplicate dominant epistasis yielding the ratio of 15 : 1 and dominantand recessive epistasis showing the ratio of 13 : 3 in F2. All these can be verified by the readers ifneeded at all.

QUESTIONS AND EXERCISESQUESTIONS AND EXERCISESQUESTIONS AND EXERCISESQUESTIONS AND EXERCISESQUESTIONS AND EXERCISES

1. What is the constitution of a cell and its function?

2. Differentiate between somatic and gametic cells.

3. Write the conditions that are known as diploid, triploid, haploid and polyploids.

4. Classify homogametic and heterogametic conditions.

5. What do you understand by a gene and its alleles?

6. What kind of activity takes place under mitosis and meiosis?

7. Name the scientist who is called as ‘father of genetics’ and why?

8. Enunciate Mendle’s laws of genetics.

9. Distinguish between genotypic and phenotypic variations.

10. How does laws of probability help in heredity determinations?

11. What is hybridization and how one can ascertain the hybrid ratios theoretically?

12. Explain chi-square test for multiple ratios.

13. What is meant by crossing over in genetic theory?

14. Define and explain the term recombination fraction.

15. Explicate the terms epistatic and hypostatic.

16. An experiment on chlorophyll inheritance in maize consisted of 1092 heterozygous self fertilizedgreen plants which segregated into 869 dominated green plants (DG) and 223 recessive yellowplants (dg). The theoretical ratio of DG to dg is 3 : 1. Test statistically whether the experimentaldata conform the theoretical ratio?


17. Following phenotypic frequencies were obtained in an experiment from a F2 population.

Phenotypes : AB Ab aB ab

Frequency : 475 122 128 75

The expected ratio of frequencies in the given four classes is 9 : 3 : 3 : 1.

(i) Test whether the observed frequencies are in agreement with the expected ratio at 1 per centprobability.

(ii) Test in case of F2 under back cross between two factors (A, a) and (B, b) for linkage andsegregation.

(iii) Test for segregation and linkage between two factors (A, a) and (B, b) in case of back crossin F1.

18. The following data were observed for hybrids of Datura.

Flowers violet, fruits prickly 47

Flowers violet, fruits smooth 12

Flowers white, fruits prickly 21

Flowers white, fruits smooth 3

Using chi-square test, find the association between colour of flowers and character of fruit at5 per cent level of significance.

19. What are autosomes? Write the number of autosomes in human beings.

20. What do you understand by linkage and crossing over?

21. Define genic interaction and give its example.

22. How can we say that additive gene effect exists in a progeny?

23. What is the chemical composition of a chromosome and a gene?

24. Name one most important difference between somatic and gametic cells.

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