GENETICS- ASSIGNMENTS COMPILATION PDF

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Human Genetics with CytogeneticsLABORATORY ACTI VITIESAbrico, Ryan S.Afuang, Al Randolf Nurelle P.Banal, Denzel S.Fermin, Aaron Axle Marcus D.Mondejar, Gabriel S.Pablo, Reynald Iñigo M.Rivera, Gio Treb V.xGENES SEGREGATION BY BEAN-OTYPEI. Objectives: At the end of the activity, the students should b...

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HumanGenet i cswi t hCyt ogenet i cs

LABORATORYACTI VI TI ES Abr i c o,Ry anS. Af uang,AlRandol fNur el l eP. Banal ,Denz el S. Fer mi n,Aar onAx l eMar c usD. Mondej ar ,Gabr i el S. Pabl o,Rey nal dI ñi goM. Ri v er a,Gi oTr ebV.

xGENES SEGREGATION BY BEAN-OTYPE

I. Objectives: At the end of the activity, the students should be able to: 1. Simulate the sorting and recombination of genes that occurs during fertilization; 2. Learn to compute and apply Chi-square test on observed phenotype frequency; 3. Derive genotypes based on observed phenotype segregation ratios; and 4. Illustrate symbolically how Mendelian inheritance occurs. II. Discussion: A Mendelian trait is caused by a single gene. Modes of inheritance reveal whether a Mendelian trait is dominant or recessive and whether the gene that contains it is carried on autosome or a sex chromosome. Mendel’s first law, which can predict the probability that a child will inherit a Mendelian trait, applies a new to its child. Mendelian Inheritance of gene applies the Law of Segregation and the Law of Independent Assortment. III. Materials: Colored papers cut into small pieces to represent the gene for a certain trait of garden pea. (60 pieces of the same size and shape; 30 pieces of which is with a different color to the other 30 pieces Colored/Labeled containers (will serve as parent containers) IV. Procedure: 1. Place the 30 pieces of paper cuts (one color) to one container and the other 30 pieces to the other container. Assign which colored paper will represent dominant and recessive genes of a specific trait of garden pea. 2. Label the container as First Parent P1 male and other P1 female. 3. At random, segregate the beans on the Lab table. Assume fertilization occurs and First filial generations (F1) are formed. 4. Categorize the formed First filial generations. 5. Put all the papers in one container. Assume that all first filial generations undergo self-fertilization.

6. Pick-up the papers by two from the container without looking at it and collect the second Filial generation (F2) 7. Categorize the formed Second filial generations into two (2) phenotype, and three (3) genotype classes. 8. Fill up Table 15.1 for the genotypic result, and Table 15.2 for the phenotypic analysis. 9. Compute for the Chi-Square Value and compare it with the Tabular Value. 10.Interpret the results of your Chi-square test. Determine if null hypothesis is accepted or rejected. 11. Diagram the cross showing segregation of genes influencing each phenotype or genotype.

Name _____________________________

Laboratory Instructor _____________

Course/Year/Sec ___________________

Date ___________________________

Activity Sheet _____ GENES SEGREGATION BY BEAN-OTYPE A. Genotypic Analysis 1. The Problem

generations Dominant trait,

Fertilization of a garden pea plant occurs and First filial generations (F1) are formed. Assuming that all first filial generations undergo self-fertilization. Categorize the formed Second filial into 3 genotypes. Note that white color represents brown color represents recessive trait.

2. The Null Hypothesis:

The genotypic ratio of second Filial generation of garden pea is 1 PP : 2 Pp :1 pp (PP = 25%, Pp = 50%, pp = 25%)

Table 1. Distribution and Chi-square test of the Genotypes of the Beans Genotype Class

Observed Frequency (O)

Expected Frequency (E)

Deviation (d = O – E)

d2 / E

PP

7

7.5

- 0.5

0.033

Pp

16

15

1

0.067

pp

7

7.5

- 0.5

0.033

Total

30

χ2 =0.133

B. Phenotypic Analysis 1. The Problem

generations Dominant trait,

Fertilization of a garden pea plant occurs and First filial generations (F1) are formed. Assuming that all first filial generations undergo self-fertilization. Categorize the formed Second filial into 2 Phenotype. Note that white color represents brown color represents recessive trait.

2. The Null Hypothesis: The phenotypic ratio of second Filial generation of garden pea is 3:1 (White Bean = 75%, Brown Bean= 25%) Table 2. Distribution and Chi-square test of the Phenotypes of the Beans Observed Frequency (O)

Expected Frequency (E)

Deviation (d = O – E)

White Bean

23

22.5

0.5

0.01

Brown Bean

7

7.5

-0.5

0.033

Phenotype Class

Total

30

d2 / E

χ2 =0.043

V. Questions: 1.

Interpret the result of your Chi-square test.

• The null hypothesis is accepted if the value of the computed chi-square is less than the tabulated value. • The null hypothesis is rejected if the value of the computed chi-square is greater than the tabulated value.

Table 1 Computed Value Tabulated Value χ2 =0.133 5.991 Degree of Freedom Level of significance 0.05 Df= n-1 =3-1 =2 Accept null hypothesis

Table 2 Computed Tabulated Value χ2 =0.043 3.841 Degree of Freedom Level of significance 0.05 Df= n-1 =2-1 =1 Accept null hypothesis

2.

Diagram the cross showing segregation of genes influencing each character. Determine the phenotypic ratio (P.R.) and the genotypic ratio (G.R.) of the crosses.

P1 to F1

P2 to F2

P1

PP

pp

gametes

P

p

F1

(ALL)

Pp

Genotypic ratio All Pp Phenotypic ratio All White

Documentation

Pp

gametes

½p

½P

F2

P

p

P

PP

Pp

p

Pp

pp

Genotypic ratio 1 PP : 2Pp :1pp Phenotypic ratio 3 white : 1 brown

STATISTICAL CONCEPTS AND TOOLS (Mean, Variance and Standard Deviation)

I. Objectives: At the end of the activity, the students should be able to: 1.

differentiate variance and standard deviation,

2.

learn how to compute variance and standard deviation on observed phenotype frequency, and

3.

explain the advantage of using variance over standard deviation in analyzing measurements of central tendencies of given samples.

II. Discussion The distribution of a trait in a population implies nothing about its inheritance. The distribution is a description of the population in terms of the proportion of individuals that have each of the possible phenotypes. Many traits that are important in plant breeding, animal breeding and medical genetics are influenced by multiple genes as well as the effects of the environment. These are called multifactorial traits or complex traits because each factor that affects the trait contributes, at most, a modest amount to the total variation in the trait observed in the entire population. Displaying a distribution completely, either in tabular form or in graphical form is always helpful but often unnecessary. In many cases a description of the distribution in terms of two major features is sufficient. These features are the mean and the variance. III. Materials/Specimens leaves still attached at the long stem of San Francisco plant Celluloid ruler Calculator Statistics book as reference

IV. Procedure: 1. Select a dicot plant body and identify ten leaves that will be measured. 2. Make sure that the leaf is always the third leaf in a row considering the lowermost leaf as the first. 3. Measure in millimeter (mm) the length and broadest width of the leaf. 4. Record the measured values on Table 14.1 of the Activity Sheet. 5. Obtain the measurements made by other members of the group. 6. Compute for sample mean ( x), sample variance (s deviation (s) by completing Tables 14.2 and 14.3.

2 )

and sample standard

7. Plot the frequency against the measure value to prepare the histograms.

STATISTICAL CONCEPTS AND TOOLS (Mean, Variance and Standard Deviation) A.1.Table 14.1 Record of the Length and the Width of the Selected Dicot Leaves 1.

Leaf No.

Width(mm.) 14

Length(mm.) Leaf No. 20 6.

Width (mm.) 30

Length(mm.) 60

2.

26

40

7.

36

50

3.

30

50

8.

52

80

4.

26

50

9.

14

30

5.

20

30

10.

22

40

2.Distribution Table for Leaf Length Table 14.2 Distribution of Leaf Length of the Selected Dicot Plant Class Interval Column 20 - 32

26

3

78

-19

361

1,083

33- 45

39

2

78

-6

36

72

46-57

51.5

3

154.5

6.5

42.25

126.75

58-69

63.5

1

63.5

18.5

342.25

342.25

70 - 81

75.5

1

75.5

30.5

930.25

930.25

Midpoints (X)

Frequency (f)

∑(f) 10 X = 45

fX

X-X

∑(fX) 449.5 S = 16.85 S 2 = 283.9

(X-X)2

f(X-X)2

3. Distribution Table for Leaf Width Table 14.3 Distribution of Leaf Width of the Selected Dicot Plant Class Interval Column 14 – 21

17.5

3

52.5

-9.5

90.25

270.75

22 – 29

25.5

3

76.5

-1.5

2.25

6.75

30 – 37

33.5

3

100.5

6.5

39

117

38 – 45

41.5

0

0

14.5

210.25

0

46 - 53

49.5

1

49.5

22.5

506.25

506.25

Midpoints (X)

Frequency (f)

fX

∑(f) 10 X = 27

X-X

(X-X)2

f(X-X)2

∑(fX) 279

S = 10 S 2 = 100

4.Histogram for Leaf Length

LENGTH 90 80 70 60 50 40 30 20 10 0

20

40

50

50

30

60

WIDTH 60 50 40 30 20 10 0

50

80

30

40

5.Histogram for Leaf Width

V. Questions: 1. Describe the histograms for the leaf length and the leaf width. The Histograms for both leaf length and lead width should correlate because the sizes of the length and the width of the leaves correlate with each other. Basically the width is the smaller scale of the length’s histogram.

2. Differentiate mean, variance and standard deviation? Mean is a other term of “average” wherein all given are summed and divide to how many are they. Variance basically show the Precision of a data, it shows if all data are near to the Mean or away from the mean. Standard Deviation is just the same with Variance, they show how the sheet is spread but Standard Deviation is usually used with the set of Groups while variance is used with set of Points. 3. What does a relatively small standard deviation indicate? Small Standard Deviation Indicates that Set of data points is close to the Mean, therefore it also indicates that the group of data are precise when it comes to likeness. High or Low Standard deviation varies from cases to cases to say if it’s a good Standard Deviation or it’s a bad standard deviation.

4. Give the importance of statistical concepts and tools in genetic study? Genetic study basically deals with inheritance of individual, statistical concepts are being used in the study of genetics to know if the Heritability is being followed and if there are “extremes” in the spreadsheet where in indicates that Inheritance is not happening or maybe altered. For example, a Normal bloodline should have a low standard deviation of Genes, these shows that they are close to their mean and indicates that they are almost the same due to their bloodline. Variation is also a key for

the study of genetics that’s why it uses Variance, it usually shows and proves certain point on being related with each other.

STATISTICAL CONCEPTS AND TOOLS (Chi-Square Test)

I. Objectives: At the end of the activity, the students should be able to: 1.

explain the relevance of null hypothesis in the analysis of a problem,

2.

appreciate the relevance of using Chi-square test in a mathematical calculations involving large populations, and

3.

learn to apply Chi-square test on observed phenotypic frequency,

II. Discussion Progeny of crosses are predicted by the binomial probability formula. The addition rule of probability deals with outcomes of genetic cross that are mutually exclusive. On the other hand, the multiplication rule of probability deals with the outcomes of a genetic cross that are independent. For example, in the sequence of births, the sex of anyone child is not affected by the sex distribution of any children born earlier and as no influence whatsoever on the sex distribution of any siblings born later. Each successive birth is independent of all the others. Probability calculations in genetics frequently use the addition and the multiplication rules together. Chi-square test is a statistical test that applies to experiments in which each trial can have a definite number of outcomes and the number of times each outcome occurs is counted. The observed numbers are then compared with the expected number as predicted by the hypothesis tested. Chi-square test is very applicable to mathematical approximations that become increasingly accurate as the observed population become large. III. Materials:

20 Surveyed families with four kids (by partner) Tabular value of Chi-square table Calculator

IV. Procedure: 1. With a partner in class, make a survey of 20 families with four children in a chosen community. 2. Record and tabulate the distribution of sexes of your surveyed results. 3. Before completing the provided Chi- square table, be sure to construct a simple problem referring to the consistency of equal number of boys and girls in your surveyed families. 4. Make a tentative answer of the stated problem in a null form. 5. Justify your tentative answer using the Chi-square test. 6. At the end, make the final decision, that is to accept or reject the stated hypothesis by comparing your computed Chi-square value to a Tabular value.

STATISTICAL CONCEPTS AND TOOLS (Chi-Square Test) A. 1. Statement of the Problem -

This activity aims to test the progeny of crosses probability and its independence by the usage of the Chi-square test. Through the use of the aforementioned test, the sequence of births of children by sex is not affected by the children born before them and has no influence whatsoever, hence proving it to be independent. 2. Statement of the Null hypothesis.

-

The sequence of sex of the children is not affected by the sex distribution by birth order.

Family Number 1. Fermin 2. Legaspi 3. Chua 4. Mandia 5. Delos Santos 6. Revadavia 7. Dagal 8. Salazar 9. Alvarez 10. Principe 11. Martin 12. Bautista 13. Garcia 14. Entila 15. Salamat 16. Kano 17. Lim 18. Sionzon 19. Faustino 20. Baunsit

Male 3 1 2 0 0 2 1 3 1 1 0 2 4 1 2 2 2 2 2 3

Female 1 3 2 4 4 2 3 1 3 3 4 2 0 3 2 2 2 2 2 1

Table 12.1 Distribution and Chi-square test of Sexes in Families of Four Children Observed number of families(O)

Combination

Expected number of families(E)

Deviation (d)

d 2/ E

All boys (b4)

1

1

0

0

3 boys & 1 girl(b 3 g )

3

3

-2

0.8

2 boys & 2 girls ( b2 g2 )

8

8

0

0

1 boy & 3 girls (bg3)

5

5

0

0

All girls (g4)

3

1

2

0.8

Total

20

20

0

1.6

Computed chi-square value _____1.6_____

Tabular chi-square value

___30.144__ at _19_ df and at 0.05 level of Significance

Decision: since that X2(19) = 1.6, p < 0.05, and Crit = 30.144. and X2...