Report 9: Population Genetic HARDY WEINBERG

Population genetics: Hardy Weinberg equilibrium.

Early in the twentieth century mathematician Godfrey Hardy and physician Wilhelm Weinberg independently developed a model describing the relationship between the frequency of the dominant and recessive alleles (hereafter, p and q) in a population. They reasoned that the combined frequencies of p and q must equal 1, since together they represent all the alleles for that trait in the population.

Hardy and Weinberg represented random mating in the population as the product (p + q)(p + q),which can be expanded to p² + 2pq + q². This corresponds to the biological fact that, as a result of mating, some new individuals have two p alleles, some one p and one q, and some two q alleles. P² then represents the fraction of the population that is homozygous dominant while 2pq and q² represent the heterozygous and homozygous recessive fractions, respectively.

A significant question in population genetics, therefore, is determining the frequency of the dominant and recessive alleles in a population (for example, the frequency of blood type O allele in the United States), given the frequency of the phenotypes. Note that phenotypic and allelic frequencies are related but are not equal. Heterozygote show the dominant phenotype, but carry a recessive allele. Therefore, the frequency for the recessive allele is higher than the frequency of the recessive phenotype.

            There are 5 assumptions in the Hardy Weinberg equilibrium:

1.            No mutation.

2.            No migration.

3.            No natural selection.

4.            No genetic drift.

5.            Random mating.

Results:

A. PTC taster.

1. Are you a taster of PTC?

 Yes.

2. If so, how did PTC taste? Was it sweet, sour, salty, bitter, or some other combination tastes?

Bitter.

3. You now know your phenotype. Do you know your genotype?

Yes I noe my phenotype and I don’t know my genotype.

1. If not, why?

 It may can be either TT @ Tt.

2. If not, how could you go about finding out what your genotype is?

 find out phenotype of family member. Then, calculate probability or percentage allele frequencies from the family members.

Record your phenotype in the table provided on the chalkboard. After all data are recorded, copy the total class data into the following spaces

Phenotype	Number	Frequencies (decimal fraction)
Tasters	46	46/47 = 0.979
Nontasters	1	1/47 = 0.021
Total	47	1

The frequency of t for the class data is: √0.021 = 0.145.

The frequency for the T allele is thus 0.855.

                  p + q = 1; q = 0.145              p = 1 - 0.145

                                                                     = 0.855 

                                                                  p = T = 0.855

Means the frequency for T (P) = 0.86                  

0.14 + 0.86 = 1 and (0.86)² + 2(0.86)(0.14) + (0.14)² = 1

From Hardy-Weinberg equation, P² of TT = (0.86) ²

                                                                     = 0.7396

The frequency 2pq = 2(0.86) (0.14) = 0.2408                

Chi-square test.

How many degrees of freedom do you have in interpreting X²?

df = n – 1

= 2 – 1

= 1

Does the class seem to be a representative sample of the U.S. white population?

No.

If not, indicate some reasons why the class might not fit this population 

because U.S. population has many nontaster while our sample only has one nontaster.

Phenotype	Observed no.	Expected no.	O – E	(O – E) 2	(O – E) 2/ E
Taster	46	70/100 x 47 = 32.9	13.1	171.61	5.216
Nontaster	1	30/100 x 47 = 14.1	-13.1	171.61	12.171
Totals	47				X2 = 17.387

X² Table = 3.841

X² Calculated = 17.387

X² Calculated > X² Table. So, the U.S. hypothesis is rejected.

B. Blood group.

Phenotype	Genotype	Number	Frequency
A	AA and Aa	9	0.158
B	AA AB and AB a	20	0.351
AB	AAB	6	0.105
O	aa	22	0.386
		Total = 57

1. Frequency of the a allele.

Freq of a allele =     

                         =         2(22) + 9       

                                       2(57)

                         =         0.465

2. Frequency of the A allele.

Freq of A allele = 2(9) + 6

                             2(57)

                      =  0.211

3. Frequency of the A^B allele.

Freq of A^B allele = 20 + 6

                                  114

                             =  0.228

4. What basic assumption are you making about the human population when you calculate allele frequencies in this fashion.
• No mutation.
• No migration.
• No natural selection.
• No genetic drift.
• Random mating.

5. Is this a reasonable assumption for the trait of blood types? Justify the answer.

No. Because genotype frequencies in the next generation are different from the first generation.

DISCUSSION:

            An understanding of evolution depends upon knowledge of population genetics. One of the more difficult concepts to understand when studying population genetics is Hardy-Weinberg Equilibrium. The Hardy-Weinberg equilibrium is the statement that allele frequencies in a population remain constant over time, in the absence of forces to change them. Its name derives from Godfrey Hardy, an English mathematician, and Wilhelm Weinberg, a German physician, who independently formulated it in the early twentieth century. The statement and the set of assumptions and mathematical tools that accompany it are used by population geneticists to analyze the occurrence of, and reasons for, changes in allele frequency. Evolution in a population is often defined as a change in allele frequency over time. The Hardy-Weinberg equilibrium, therefore, can be used to test whether evolution is occurring in populations.

p + q = 1
p= frequency the dominant allele A
q= frequency the recessive allele a

Then the second equation are came out:

p² + 2pq + q² = 1 
 p²= frequency for the genotype homozygous dominant (AA)
 2pq= frequency for the genotype heterozygous (Aa)
q²= for the genotype homozygous recessive (aa)

The chi-square test can determine if observed and expected data are in agreement. Therefore, we can use a chi-square test to determine whether a population exhibits Hardy Weinberg equilibrium for a particular gene. To do so, it necessary to distinguish between homozygotes and heterozygotes, either phenotypically or at the molecular level. This is necessary so that we can determine both allele and genotype frequencies. In this experiment, for the PTC tester for the class the result for X² is 17.387. the result was too big then the X² chi-squarewich is 3.8441. from this show that the result was rejcted. It is because the result had very big different then the US tester.The second test the blood type, the result also rejected it is because the result for the class was too big from the expected result. The result  of blood type for our class is  X²=63.5587 and the real result that we need to get is X² chi-square = 7.815

CONCLUSION

As a conclusion, we are able to calculate the gene (allele) frequencies for a population sample in which each of the genotypes AA, Aa, aa has a unit phenotype that regulate the ability to taste PTC. Beside that, we are able to determine whether a population sample in which each of the genotypes AA, Aa and aa has a unique phenotype represents a population in a Hardy-Weinberg equilibrium. Lastly, we can calculate the frequencies of the I^A I^B and i alleles, given a population sample in which the frequencies of individuals having blood types A, B, AB and O are known.

REFERENCES

Brooker. R. J. 2009. Genetics: Analysis & Principles. Third Edition. McGraw-Hill Higher Education.

Eldra P. S. Linda R. B. Diana W. M. 2005. Biology. Seven Edition.

Brooks/cole-Thomson Learning.

Kenneth R.M. and Joseph L. 2004. Prentice Hall Biology. Pearson Prentice Hall.

http://www.talkorigins.org/faqs/genetic-drift.html

http://www.conservapedia.com/Genetic_drift

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/H/HardyWeinberg.html

http://www.tiem.utk.edu/bioed/bealsmodules/hardy-weinberg.html

Report 8: Red Blood Cell

INTRODUCTION

Red blood cells have certain proteins on their surface called antigens. Also, plasma contains antibodies which will attack certain antigens if they are present. There are various types of red blood cell antigens - the ABO and rhesus types are the most important.

These were the first type discovered.

• If we have type A antigens on the surface of our red blood cells, we also have anti-B antibodies in our plasma.
• If we have type B antigens on the surface of our red blood cells, we also have anti-A antibodies in our plasma.
• If we have type A and type B antigens on the surface of our red blood cells, we do not have antibodies to A or B antigens in our plasma.
• If we have neither type A or type B antigens on the surface of our red blood cells, you have anti-A and anti-B antibodies in our plasma.

Most people are 'rhesus positive' as they have rhesus antigens on their red blood cells. But, about 3 in 20 people do not have rhesus antibodies and are said to be 'rhesus negative'.

Blood group depends on which antigens occur on the surface of our red blood cells. Genetic make-up which we inherit from our parents determines which antigens occur on our red blood cells. Because of this fact a blood group test is sometimes used to help settle disputes about who is the father of a child.Blood group is said to be:

• A+ (A positive) if you have A and rhesus antigens.
• A– (A negative) if you have A antigens, but not rhesus antigens.
• B+ (B positive) if you have B and rhesus antigens.
• B– (B negative) if you have B antigens, but not rhesus antigens.
• AB+ (AB positive) if you have A, B and rhesus antigens.
• AB– (AB negative) if you have A and B antigens, but not rhesus antigens.
• O+ (O positive) if you have neither A nor B antigens, but you have rhesus antigens.
• O– (O negative) if you have do not have A, B or rhesus antigens.

Basically, a sample of our blood is mixed with different samples of plasma known to contain different antibodies. For example, if plasma which contains anti-A antibodies makes the red cells in your blood clump together, then we have A antigens on our blood cells. Or, if plasma which contains rhesus antibodies makes the red cells in our blood clump together, then we have rhesus antigens on our blood cells. By doing a series of such tests it is possible to determine what antigens are on our red blood cells and therefore determine our blood group.

Routine blood grouping checks for our ABO and rhesus status. Other red cell antigens are tested for in certain other situations.

A blood group test is always done on pregnant women. If the mother is rhesus negative, and the unborn baby is rhesus positive (inherited from a rhesus positive father), then the mother's immune system may produce anti-rhesus antibodies. These may attack and destroy the baby's blood cells. This is rarely a problem in a first pregnancy. However, without treatment, this can become a serious problem in subsequent pregnancies as the mothers immune system will be 'sensitised' after the first pregnancy.

LEARNING OUTCOMES

In this experiment, we are able to determine

• Our own blood type with respect to the A and B antigens.

MATERIALS

• Slide
• Sterilizer
• Lancets
• Swabs
• Toothpicks
• Container for swabs disposal
• Antisera
• 70% ethyl alcohol
• Anti-A (a) serum
• Anti-B (b) serum

METHODS

Two glass slide for each students was prepare. The slide must be clean. The slide was mark at the end with a letter “A” and other end “B”. the other slide was mark with letter “C”

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Carefully we wash our hand by using the soap detergent supplied and then blotted with a dry clean paper tissue. Then left index finger was swab with a sterile cotton swab dipped into 70% alcohols.

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Quickly the sterile disposable lancet was remove and was quickly prick on the tip of the finger. The lancets were disposed.

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The lanced finger was massage and squeeze out 3 drops of blood into each glass slide.

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At the glass slide with letter ”A”, put antisera A

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At the place with letter “B” put antisera B

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At the other slide with letter “C” put the anti-D

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A clean toothpick was use to mix the blood with the antisera. Different toothpick was use for different antisera. Lastly, the toothpick was disposable.

The sera you have used are color coded. What color is the serum containing alpha antibodies?

Yellow

What color is the serum containing the beta antibodies?

Blue

Place the slide on a piece of paper and observe. What do u see?

I have see that my blood become agglutination with anti-B only.

What is your blood type, with respect to the A and B antigen? Did you know your blood type before this test?

My blood type is type B and respect to anti-B only. Yes I know my blood type.

Does your determination agree with what your medical record indicates?

Yes

RESULT

My own blood type      : B

Antigen on RBC’s      : B

RH⁺ test                        : Positive

DISCUSSION

                                From the experiment, we already known our blood type. I have blood B+. means that I have antigen B and antibodies A with rhesus positive. How do I know that my blood is type b? this is because when a drop of my blood was put on the slide, then put a drop of anti a and anti b, my blood will agglutinate with anti b and for the rhesus test, my blood will agglutinate with anti d and it show the positive result..For blood A type, the blood will agglutinate with anti A , while if there are no agglutination means the blood is O type. If both are agglutinate, means that it is AB blood type. Blood type is control by multiple allele. Multiple allele is more than 1 allele ate the same loci.

CONCLUSION

                             As a conclusion, we are able to apply the sterility in the lab before identified the blood type of myself and my friends. Other than that, we also learn how to use the disposable lancets.

REFERENCES

Brooker, R. J. (2009). Genetics Analysis & Principle. Third edition: Mc Graw Hill Companies Inc.

Campbell O.N. (2005). Biology. Seventh Edition: Benjamin Cumming.

Eldra P. S., Linda R. B. and  Diana W.M. (2005). Biology. Third Edition: Thomson Brooks/Cole.

Kenneth R. M.,and Joseph L. (2004). Prentice Hall Biology. Teacher's Edition: Pearson Publishing.

Report 4: Mitosis

INTRODUCTION:

Mitosis is the process that facilitates the equal partitioning of replicated chromosomes into two identical groups. Before partitioning can occur, the chromosomes must become aligned so that the separation process can occur in an orderly fashion. The alignment of replicated chromosomes and their separation into two groups is a process that can be observed in virtually all eukaryotic cells.

Both the alignment and separation processes are the consequence of the chromosomes interacting with filamentous structures, known as microtubules. The microtubules become organized into a biconical array known as a spindle, which forms early in mitosis, and then disassembles as mitosis nears completion. Mitotic spindles are visible in living cells with the polarizing light microscope. Some of the spindle microtubules become attached to the chromosomes at sites known as kinetochores. The kinetochores cannot be seen with the light microscope, but they reside near the place on the chromosome known as its Centromere, which can be observed with the light microscope. There are two kinetochores on each replicated chromosome (one on each chromatid), and when the replicated chromosome splits apart at its Centromere at the onset of anaphase, each daughter chromosome possesses one Centromere and one kinetochore. The linkages between kinetochores and microtubules are thought to be central in controlling both the positioning of the replicated chromosome at the central portion of the spindle during the alignment phase, and in moving the daughter chromosomes apart after they split at their centromeres. The separation of daughter cells from each other is a process known as cytokinesis, and is separate from mitosis. In cytokinesis, animal and plant cells differ considerably from each other. These differences are the consequence of having or not having a cell wall. Cytokinesis in fungi reveals.

LEARNING OUTCOMES:

            After this experiment, we are able to outline the procedure for preparing acetocarmine squash of a root tip to demonstrate the process of mitosis and able to distinguish and examine;

1.    Every stage of mitosis in plant cell including :

a.    Prophase

b.    Metaphase

c.    Anaphase

d.    Telophase

2.    The process and changes that occurred in every stage of mitosis.

MATERIALS:

1.    1 molar (M) hydrochloric acid (HCL) in dropping bottle

2.    70% ethyl alcohol

3.    Absolute ethyl alcohol

4.    Acetocarmine stain in dropping bottle (dissolve 1.0g in 100mL of 45% acetic acid, boil 5 min, cool, decant and filter)

5.    Alcohol lamp or slide warmer

6.    Beaker in which to let onion sprout

7.    Cover slip

8.    Dissecting needles

9.    Forceps

10. Glass rod for crushing root tips

11. Microscope

12. Scalpel or razor blade

13. Slides

14. Supply of onion bulbs

15. Watch glass

METHOD:

Mitosis in onion root tips

1.    Firstly, we soften the root so that the cells can be separated and flattened, thus making it possible to see the chromosome, nuclei, spindle and other cell parts.

2.    Several drops of 1 molar HCL was placed in a watch glass safely because don’t want the acid get to skin and cloth.

3.    The terminal 3 or 4 mm of the 1 cm long onion root was placed into the acid.

4.    The root tips will feel soft when touched with dissecting needles in short times.

5.    The softened root tips was pick up using the forceps and transferred it to a drop of acetocarmine stain on a clean slide and wait about 15-30 minutes.

6.    The root tip most 1mm was cut and retains by using the razor blade or scalpel. The rest of root was discarded. The remaining root tips were chopped into many pieces with the razor blade and crush the material with glass rod.

7.    A clean cover was applying to the slide and gently heated over the Bunsen burner.

8.    The slide was examined under the low power and high power.

Mitosis in prepared slide

1.    By using the prepared slide, we compared the slide that we did and the prepared slide.

Question and answers:

1.    Most of the cells are in what stages?

In interphase stage.

2.    Can you determine the number of chromosome in any cells?

No

3.    Can you see the spindle in any cells?

Yes

4.    At late telophase, can you see the cell plate?

Yes

5.    Do you observe more cells in mitotic stages than you observed in the untreated root tips?

yes

6.    How does the prepared slide differ from ones you prepare?

It not much differ but slightly differ because do not crashed the onion into many pieces

7.    What advantage do your slides have over the commercial ones?

Slide is fresh prepared compared to the commercial ones might have some damages.

8.    What are the advantages of the commercial slides?

Can use many time and last long period.

9.    What is acetocarmine? What is the function of this solution?

Acetocarmine is a staining that colors the chromosome to make the chromosome more visible through microscope.

DISCUSSIONS:

            Mitotic division consists of two, meiosis and mitosis. The onion root tips occurs the mitosis process. In mitosis process there are 5 stages, Interphase, Prophase, Metaphase, Anaphase and Telophase. Why do we observe the root tips of the onions? This is because in the root tips there were actively divided. The onions bulbs supported by tripod of toothpicks in beaker of water.

            Most of the cells are spent in the interphase stages. Interphase consist of G₁ (growth phase or gap), S (synthesis phase) and G₂ (second growth phase or gap 2) then followed by mitosis. In G₁  volume of cytoplasm increase and the protein start to synthesis. Increase the number of organelles such as mitochondria, endoplasmic reticulum and Golgi apparatus. Also synthesis of carbohydrate lipid and ribonucleic acid (mRNA, rRNA and tRNA). S phase is DNA synthesis phase. The cell DNA replicate and chromosome duplicate. In this stage in which genetic materials are synthesize. In G₂ the energy stored are increase. The formation of microtubules (kinetochore, aster and polar. Increase in cell organelles (mitochondrion, Golgi body and endoplasmic reticulum) also increase in size of nucleus. 

            Mitosis begins with Prophase. In Prophase, the chromatin fibers more tightly coiled, condensing into discrete chromosome. In the same time, the nucleoli were disappeared. Each duplicated chromosome appears as two identical sister chromatids joined together. The mitotic spindles begin to form. It is composed of the centrosomes and microtubules that extend from them to form spindle. The radial arrays of shorter microtubules that extend from the centrosomes are called asters. The centrosomes move away from each other, apparently propelled by the lengthening microtubules between them.

            In metaphase, the centrosomes are now at opposite ends of cell. The centrosomes convene on the metaphase plate. The chromosomes, centromeres lie on the metaphase plate. For each chromosome, the kinetochore of the sister chromatids are attached to the kinetochore microtubules coming from opposite poles. The entire apparatus of microtubules is called the spindle.

            In anaphase, it begin when 2 sister chromatids of each pair suddenly apart. When the microtubules shorten, the centromeres splits and sister chromatids are pulled to different poles. By the end of anaphase, the 2 ends of the cell have equivalent and complete collection of chromosomes.

            In Telophase, nucleolus is forming. 2 daughter nuclei begin to form in the cell. Nuclear envelope arises from the fragment of the parent cell nuclear envelope and the chromosome less condense.

            Cytokinesis is the division of the cytoplasm is usually well underway by late telophase, so the daughter cell appears shortly after the ends of mitosis. In the onion root tips cells, it has the cell wall, so during the telophase, the vesicle derived from the Golgi apparatus which contain carbo-filled vesicle move along the microtubules to the middle of the cells producing the cell plate. The cell plates enlarge until its surrounding membrane fuses with the plasma membrane along perimeter of the cell. Two daughter cell result, each with its own plasma membrane. A new cell wall arising from the contents of the plate has performed between the daughter cells.

The mitotic index:

Interphase: 

Prophase:

Metaphase:

Anaphase

Telophase

Conclusion:

            As a conclusion in the root tips cell that I observed are all active divided because of the mitotic percentage of each cells are more than 0.05%. We also able to distinguish each stage in mitosis process (interphase, prophase, metaphase, anaphase, telophase). We also know the function of the acetocarmine and the function of HCL.

REFERENCES:

Brooker, R. J. (2009). Genetics Analysis & Principle. Third edition: Mc Graw Hill international edition.

Eldra P. Solomon, Linda R. Berg, Diana W.Martin. (2005). Biology. Third Edition: Thomson Brooks/Cole.

Kenneth R. Miller, joseph Levine. (2004). Prentice Hall Biology. Teacher's Edition: Pearson Publishing.

Neil, C. O. (2005). Biology. Seventh Edition: Benjamin Cumming.