genetic exp 1 & 2

UDBB 2233

Title : Exp 1-Chi Square Analysis

Exp 2- Mendelian Genetics of Drosophila and Maize

Name : Mohd. Zulfadhli B. Azizan 1004656

Lau Wai Yip 1004923

Siow Zi Kang 1002558

Cheah Jun Zhe 0905802

Ang Yee Von 1007253

Chan Jo Yee 0908033

Group : Practical 1

Lecturer : Mr. Kam Yew Chee

Experiment 1

Title: Chi-Square Analysis

Introduction:

The extent an observed data set fits or differs from the predicted or expectedoccurrences can be evaluated by testing the goodness of fit of the data. When the data donot fit exactly, we will want to find out how much deviation can be allowed before wereject a null hypothesis. One of the simplest statistical tests to assess the goodness of fitof the null hypothesis is chi-square analysis (x2), which tests the difference betweenobserved (O) and expected (E) values. This x2 value is then used to estimate howfrequently the observed deviation can be expected to occur strictly as a result of chance,allowing us to determine whether progeny phenotype ratios fit our assumptions abouttheir genotypes. The formula for calculating the chi-square value is:

x2=∑(O−E)2

Ewhere O = the observed number of individuals in a particular phenotype,

E = the expected number in that phenotype, and∑= the summation of all possible values of (O-E)2/E for the various phenotypic

categories

The degree of freedom (df), which is equal to n – 1, where n is the number ofcategories (the number of phenotypes considered), will then be determined. Degrees offreedom must be taken into account because the greater the number of categories, themore deviation is expected by chance alone. The calculated _2 value can now beinterpreted at a corresponding probability value (p), which can be obtained from _2 table(Appendix) for (n–1) degree of freedom.

The p value can be thought as a percentage. For example, a p value of 0.26indicates that were the same experiment repeated many times, 26% of the trials would beexpected to exhibit chance deviation as great as or greater than that seen in the initialtrial. Conversely, 74% of the repeats would show less deviation than initially observed asa result of chance. The interpretation of the p value reveals that a hypothesis is neverproved or disproved absolutely. Instead, a relative standard must be set to serve as thebasis for either rejecting or failing to reject the hypothesis. This standard is often aprobability value of 0.05. If the calculated _2 value is smaller than the critical value givenin the table at (n–1) df, then we accept the null hypothesis and conclude that thedifference between observed and expected results is not significant, and vice versa.

Objectives:

To 1.Calculate x2 to determine whether a data set approximates a theoretically expected ratio.2.Interpret a calculated x2 value at an appropriate number of degrees of freedom.

Materials:

Non-biological materials: container of equal quantities of colored and white beads, petridishes, two equal-value coins and a calculator.

Procedure :

Activity 1:A beaker contained 150 colored and 150 white beads mixed were given.

1. One petri dish-full random sample of beads was removed from the beaker.2. The beads of the different colors were segregated and counted.3. The data was recorded in table 1.1. The expected numbers were calculated based on the

size of the sample and the known ratio of colored to white beads in the entire population. The table 1.1 was completed and x2 was calculated.

Activity 2:1. Two equal-value coins were together tossed 30 times.2. The data were recorded in table 1.2 and the expected numbers were calculated. The

probability of two or more independent events occurring simultaneously is the product of their individual probabilities were noted.

3. The table 1.2 were completed and x2 were calculated.

Activity 3:A beaker contained 100 beads of two different colors in the ratio of 3:1 . The beads ere thoroughly mixed.

1. One tablespoon-full random sample of beads were removed from the beaker.2. The beads of different colors were segregated and counted.3. The data in were recorded in table 1.3 and then the expected numbers based on the size of

the sample were calculated and the known ratio of colored to white beads in the entire population. The table 1.3 were completed and x2 were calculated.

Activity 4:The following are the approximate frequencies of the various ABO blood groups in ahypothetical population: 41% A, 9% B, 3% AB, and 47% O. The number of students in the class who have blood group A, B, AB and O were recorded in table 1.4 . The x2 were calculated to 2 decimal places and completed in table 1.4 .

Activity 5:A container contained several thousand beads of 4 different colors in the ratio 9:3:3:1 were given. The beads were thoroughly mixed.

1. One spatula-full random sample of beads were removed from the container.2. The beads of different colors were segregated and counted.3. The data were recorded in table 1.5 and then x2 were calculated.

Experiment 2Title: Mendelian Genetics of Drosophila and Maize

(A) Drosophila Genetics

Introduction:The vinegar or fruit fly, Drosophila melanogaster, has been widely used for

demonstrating the classical Mendelian laws. This small fly undergoes a completemetamorphosis from egg, larva, pupa, through imago in 10 to 14 days at 25oC. Thefertilized eggs normally hatch into white larvae in 22–24 hours after they have been laid.The larvae burrow in the culture medium and feed on it for about four days. They thenpupate on a dry spot, usually at the sides of the container. The pupation lasts for anotherfour to seven days before the adults emerge from the pupa case. The newly emergedadults are pale in colour with unexpanded wings, but mature in about 8–10 hours with acharacteristic darkened body. The females usually do not mate before 8 hours afteremergence and do not lay eggs until they are two days old. In addition to a short lifecycle, Drosophila is highly prolific, and has a lot of genetic variability with manymorphological variants. Names and symbols are used to identify the mutant genes carriedby the flies. Below are some commonly used notations in Drosophila genetics.

a. Wild-type allele of any locus is indicated by the plus (+) symbol.

b. Mutant alleles are shown with italicized letters. A capital letter designates a dominantgene, the lowercase letter represents recessive allele. E.g.,

B represents the dominant ‘Bar’ gene;Bd represents the dominant “Beaded’ gene;b designates the recessive ‘black’ gene;bt designates the recessive ‘bent’ gene.

To represent a wild-type locus, a superscript + symbol is used after the letter. E.g.,

b+ shows that the wild-type gene is dominant over its recessive black gene, b;B+ indicates that the wild-type gene is recessive to its dominant ‘Bar’ gene, B.

c. If more than two alleles are observed for a gene locus, mutant alleles can bedistinguished by small letters in superscript after the alphabet for the genes. E.g.,

w+ represents a wild-type eye-colour gene;w represents a mutant white-eye gene;wa represents a mutant apricot-coloured eye gene

d. The genotype of a gene locus for a diploid organism is represented by two symbols. E.g.

homozygote (wild-type) b+/b+;homozygote (black mutant) b/b;heterozygote (wild-type) b+/b.

e. To show two unlinked loci in a diploid, separate the two pairs of symbols. E.g.

b/b, se/se (black and sepia)

f. If two loci are linked, they are indicated by a slash. E.g.,b vg/b vg (black and vestigial, coupling phase);b vg+/b+ vg (phenotype: wild-type, repulsion phase).

g. Y chromosome is shown as Y, e.g.,

White-eye, miniature winged male can be written as w m/Y, where both w and m are located in X chromosome, or as Xw m.

h. To show a phenotype, a line is written on top of the symbols of particular alleles. E.g.,_B (fly with Bar eye),_ __e vg (ebony and vestigial fly)

Description of phenotype does not indicate homozygosity, heterozygosity, andlinkage group.

(B) Maize Genetics

Introduction:The inheritance of variability in both maize seedlings and aleurone (endosperm)obeys the Mendelian principles. An expected 3:1 ratio has been observed in a recessivemutant allele for albinism (absence of chlorophyll) in the maize seedlings and a recessiveallele for colourless aleurone segregating according the Mendel’s first law.Mendel also postulated that independent assortment of genes occurs when twotraits are considered simultaneously, yielding an expected ratio of 9:3:3:1 among thedihybrid F2 progenies and dihybrid testcross ratio 1:1:1:1. Indeed, the dihybrid F2 kernelsof corns yield a ratio of 9:3:3:1 for the phenotypes starchy (smooth) endosperm, sugary(wrinkled) endosperm, colourless starchy endosperm, and colourless sugary endosperm,respectively.

Objectives:

1. To determines the sexes, indentify and differentiate the morphological traits of the 4 known mutant fly stocks .

2. To recognize and interpret maize F2 data and illustrate Mendel’s law of segregation and law of independent assortment.

Materials:Mutant stocks of Drosophila that may be provided are Ebony body colour, Black bodycolour, Vestigial wing, Apterous (wingless), Curly wings, Sepia eye colour, Vermilioneye colour, White eye colour, Apricot (yellow) eye colour, Bar eyes, and Singed bristles, commercially prepared monohybrid, dihybrid and testcross F2 kernels of corns.Procedure:

Activity 1 :The morphological traits of wild and mutant flies were compared. The table 2.1 were completed. The difference seen between the mutant and wild-type flies with appropriate symbol, sketches, or description. The wild-types flies that are considered normal was examined for all traits listed below. A plus sign was placed in the appropriate space to signify the normal condition. The flies from the given mutant stocks were examined carefully and compared with the wild types. The mutant flies are wild type for given traits were placed a plus sign in the appropriate space. The different traits were diagramed.

Activity 2 : 2 possible illustration of endosperm were given in table 2.2 . The number of kernels were counted without removing from the corn ears in each phenotype. The data were recorded in table 2.3 and then hypothesis were formulated to explain the data collected.

Activity 3 : The kernels were groups into 4 different types. The number of kernels were counted and recorded without removing from the corn ears in each of different types of phenotypes in table 2.4

Result :

Experiment 1 Table 1.1 Calculation of x2 for a sample removed from a large population consisting ofequal numbers of coloured and white beads.

Classes(Phenotypes)

Observed(O)

Expected(E)

Deviations(O-E)

(O-E)2 (O−E)2

E

Colored 46 44 2 4111

White 42 44 -2 4111

Total: 88 88 0X2 = 0.18

Degree of freedom :1p value : 0.05 Critical value : 3.84

I accept the hypothesis that the data approximate the expected ratio.

Table 1.2 Calculation of x2 on data from tossing coins.Classes

(Phenotypes)Observed

(O)Expected

(E)Deviations

(O-E)(O-E)2 (O−E)2

EHeads on both coins

4 7.5 -3.5 12.25 1.6333

Head on one, tail on the other coin

14 15 -1 1 0.0666

Tails on both coins

12 7.5 4.5 20.25 2.7

Total : 30 30 0

X2 = 4.3999Degree of freedom : 2

P value : 0.05 Critical value : 5.99I accept the hypothesis that the data approximate the expected ratio.

Table 1.3 Calculation of x2 for a sample removed from a large population consisting of beads of two different colors.

Classes(Phenotypes)

Observed(O)

Expected(E)

Deviations(O-E)

(O-E)2 (O−E)2

EColored 12 11.25 0.75 0.5625 0.05White 3 3.75 -0.75 0.5625 0.15Total : 15 15 0



Table 1.4 Calculation of x2 on blood group data from a hypothetical population.

PhenotypesObserved

(O)Expected

(E)Deviations

(O-E)(O-E)2 (O−E)2

EA 4 10.66 -6.66 44.36 4.16B 10 2.34 7.66 58.68 25.08

AB 2 0.78 1.22 1.49 1.91O 10 12.22 -2.22 4.93 0.40

Total : 26 26 0


P value: 0.05 Critical value : 7.82I reject the hypothesis that the data approximate the expected ratio.

Table 1.5 Calculation of x2 for a sample removed from a large population consisting ofbeads of two different colors.

PhenotypesObserved

(O)Expected

(E)Deviations

(O-E)(O-E)2 (O−E)2

EWhite 50 46.31 3.69 13.61 0.29Yellow 127 138.94 -11.94 142.56 1.03Red 432 416.81 15.19 230.74 0.55Blue 132 138.94 -6.94 48.16 0.35Totals : 741 741 23.88



Experiment 2 :

Table 2.1 Comparison of wild-type and mutant Drosophila.Trait Wild-type Mutant 1 Mutant 2 Mutant 3 Mutant 4

Body Color Yellow Black

Black line Yellow Black


Yellow Black


Eye Color Red White Red Red RedEye Shape

Square shape with a blunt edge

Round shapeSquare shape with a blunt edge



Wing shape Normal Normal Normal Normal Normal

Bristle and Shape Large Small Small Very small small

Table 2.2 Record of F2 data for maize (species 6500) endosperm trait.Phenotypes Number of

kernels observed

Number of kernels expected

Genotypes of the kernels

Deviations(O-E)

(O-E)2 (O−E)2

E

Purple 446 443 P_ 3 9 0.02

Yellow 145 148 pp -3 9 0.06Totals : 591 591 0.08

x2 = 0.08Degree of freedom : 1


The diagram of monohybrid cross and test cross of the species of 6500 maize

Parent : PP x pp

Gametes : P p

F1 : Pp x Pp

Gametes : P,p P,p

F2 : PP,Pp,Pp,pp

Hence, we can determine that the parent genes of PP and pp will produce the F2 offspring with a ratio of 3:1

Table 2.3 Record of F2 data for maize (species 6502) endosperm trait.Phenotypes Number of

kernels observed

Number of kernels expected

Genotypes of the kernels

Deviations(O-E)

(O-E)2 (O−E)2

E

Purple 291 291 P_ 0 0 0Yellow 291 291 pp 0 0 0Totals : 582 582 0

x2 = 0Degree of freedom= 1

P value= 0.05 Critical value= 3.84



Parent : Pp x pp

Gametes : P,p p

F1 : Pp x pp

Gametes : P,p p

F2 : Pp,pp

Hence, we can determine that the parent genes of Pp and pp will produce the F2 offspring with a ratio of 1:1

Activity 3:Let the gene “P” represents the purple traits while gene “p” represents the yellow traits and the gene P is dominant over the gene p. Let the gene “S” represents the smooth traits while gene “s” represents the wrinkled traits and the gene S is dominant over the gene s.

Table 2.5 – Data of independent assortment in maize and calculation of x2 for the maize “6600”

x2= 7.775

Degree of freedom= 3



Classes (phenotype)

Genotypes Observed (O)

Expected (E)

Deviations(O-E)

(O-E)2 (O−E)2

EPurple smooth kernels

P_S_339 316.125 22.875 523.266 1.655

Yellow smooth kernels

ppS_96 105.375 -9.375 87.89 0.834

Purple wrinkled kernels

P_ss85 105.375 -20.375 415.14 3.940

Yellow wrinkled kernels

ppss42 35.125 6.875 47.266 1.346

Total: 562 562 0


Parent : PPSS x ppss

Gametes : PS ps

F1 : PpSs x PpSs

Gametes : PS,Ps,pS,ps PS,Ps,pS,ps

F2 : PPSS,PPSs,PpSS,PpSs,PPSs,PPss,PpSs,Ppss,PpSS,PpSs,ppSS,ppSs,PpSs,Ppss,ppSs,ppss

Hence, we can determine that the parent genes of PPSS and ppss will produce the F2 offspring with a ratio of 9:3:3:1

Table 2.6 – Data of independent assortment in maize and calculation of x2 for the maize “6602”Classes

(phenotype)Genotype

sObserved (O)

Expected (E)

Deviations(O-E)

(O-E)2 (O−E)2

EPurple smooth kernels

P_S_ 151 133.5 17.5 306.25 2.294

Yellow smooth kernels

ppS_ 128 133.5 -5.5 30.25 0.227

Purple wrinkled kernels

P_ss123 133.5 -10.5 110.25 0.826

Yellow wrinkled kernels

ppss 132 133.5 -1.5 2.25 0.017

Total: 534 534 0X2= 3.364

Degree of freedom= 3




Parent : PpSs x ppss

Gametes : PS,Ps,pS,ps ps

F1 : PpSs x ppss

Gametes : PS,Ps,pS,ps ps

F2 : PpSs,Ppss,ppSs,ppss

Hence, we can determine that the parent genes of PpSs and ppss will produce the F2 offspring with a ratio of 1:1:1:1

Discussion :

In experiment 1, all x2 analysis are accepted except the activity 4. The rejection of the hypothesis is due to the small sample of 26 students cannot represents the whole population on the Earth. A sample must reach a minimum numbers to represents a populations, in this case, we reject the hypothesis due to this error. In contrary, others x2 analysis can be accepted and obeys the mendel’s law.

In experiment 2 activity 1, the fly characteristics are determine. We seen a great deviation of characteristics between the fly species. This is due to the variations of genes in fly. Hence, we can know that different genes can produce different types of traits in fly. In this experiment, the comparison of the morphological trait in fly show the variations.

In experiment 2 activity 2, it is the study of the Mendel’s Law of segregation. The law states that allele pairs segregate during gamete formation, and randomly unite at fertilization. There are 4 main concepts in this law :

1)A gene can exists in more than one form

2)Organisms inherit two alleles for each traits

3)When gametes are formed through meiosis, allele pairs separate leaving each cell with a single allele for each traits

4)When the two alleles of a pair are different, one is dominant and the other is recessive.

As an example, for the maize 6500 species, parents cells consist of homozygous dominant gene (PP) and homozygous recessive gene (pp). The F1 generation will inherit the both of the dominant gene and recessive gene which resulting a heterozygous genes. When F1 mate with each other, they will produce a F2 generations with a phenotypes of ratio 3:1 (refer to result in 6500 maize species).

Written only the offspring that will produce wanted F2

But in the maize species of 6502, when a heterozygous dominant gene parent mates with a homozygous recessive genes, this mating will produce an offspring of heterozygous gene that is dominant and homozygous gene which is recessive. Hence, the F2 generations will show a phenotype with a ratio 1:1 . (refer to results in 6502 maize species)

In experiment 2 activity 3, the experiments are carried out to test the Mendel’s Second Law which is the law of independent assortment which states that the segregation of alleles of one allelic pair is independent of the segregation of the alleles of another allelic pair during gamete formation which involves dihybrid crosses.

As examples is the maize species 6600 and 6602. For species 6600, it is similar to maize 6500 but only introducing one more genes which is known as dihybrid. When a dominant dihybrid homozygous (PPSS) mates with recessive dihybrid homozygous (ppss), it will produce a dominant dihybrid heterozygous. When they mates with each other, they will produce f2 generations that obey the Mendel’s Law Second Rule with a ratio of 9:3:3:1.

For the maize species 6602, parents of dominant heterozygous dihybrid genes mates with homozygous dihybrid recessive gene. The F1 generations will mates with each other that have different genes (PpSs and ppss)only will produce the F2 generations with the ratio of 1:1:1:1 .

Conclusion :

1. Chi square analysis used to determined whether the data approximating the expected ratio.2. A Chi square analysis were carried out using a appropriate degree of freedom to compare

with the critical value.3. The fly were identified by observing their morphological traits.4. F2 generations of maize can be used to illustrate the Mendel’s First and Second Law.

References :

Bowler, P. J. (1989). The Mendelian Revolution : The EMergence of Hereditarian.

Mendel, G. J. (2009). The Lost and Found Genius of Gregor Mendel, the Father of Modern Genetics. The Monk in the Garden .

genetic exp 1 & 2

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