BIO Genetics LAB Report 2 PDF

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Genetics of Corn Lab Report...

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Genetics of corn BIO II 11/17/2013

Abstract In this experiment the main goal was to figure out what offspring came from what parent. By looking at two ears of corn the outcome for each four phenotypes were counted: purple smooth, purple wrinkled, yellow smooth, yellow wrinkled. Then they were noted that the outcomes were a 9:3:3:1 ratio. The ratio meant that there was 9/16 of the purple smooth, 3/16 of the purple wrinkled, 3/16 of the yellow smooth, and 1/16 of the yellow wrinkled. The ratios were then put into decimal form and timed by the total outcome of the observed which gave the expected. By subtracting the observed by the expected it was then squared and divided by the original expected number of each. By adding the last column up the result was the final outcome of the chi- square test by which then the degree of freedom formula was used which looked such as: N1. N represented the different number of phenotypes. So the formula was 4-1 =3 which gave the degree of freedom. By finding 3 on the P value scale it then showed if it was greater than or less than 0.05. For example, for corn A the hypothesis was accepted because the value was less than 0.05. However, for corn B the hypothesis was rejected because it was greater than 0.05. For this experiment the hypothesis was that it was believed that the parents of the corn were heterozygous because by looking at the number of outcomes for each phenotype it resembled a 9:3:3:1 ratio. Introduction The dihybrids cross for the parents are heterozygous which means that the cross will look such as: PpSs X PpSs. Phenotypes are the traits that are handed down to each ear of corn it is the realized expression of the genotype the physical appearance or functional expression of a trait (Raven Biology). In this experiment there were four types of phenotypes which were: purple smooth, purple wrinkled, yellow smooth, yellow wrinkled. A punnet square is a chart that is used to find the different combinations for each phenotype. While doing the punnet square we observed that there was a 9:3:3:1 ratio through counting each of the combinations. A genotype is the genetic constitution underlying a single trait or set of traits (Raven Biology). By looking at the results of the chi-square test and looking up the outcomes on the p value scale the hypothesis seemed to be that the parents of the corn were heterozygous. The outcomes followed the 9:3:3:1 ratio. For this experiment, the hypothesis was that it was believed that the parents of the corn were heterozygous because by looking at the number of outcomes for each phenotype it resembled a 9:3:3:1 ratio. Materials & Methods In this experiment you will use two ears of corn to determine if the parents of the corn are heterozygous or homozygous. There are four phenotypes in which you will observe and note how many are in each phenotype which is: Purple Smooth, Purple Wrinkled, Yellow Smooth, Yellow Wrinkled. Place a piece of tape on the row of kernels which are first counted that way the place started will not be lost. Then count each type of phenotype in each row until every row is

counted. Place each observation into a chi-square chart. First place the observed number into the column. The punnet square shows 16 different genotypes from these you get the phenotypic ratio: 9:3:3:1. You take the ratios: 9/16,3/16,3/16,1/16 and put them in decimal form and times them by the total outcome from adding all observed outcomes together. This will give the numbers for the expected outcomes column. The next column on the chi- square chart is used to subtract the observed from the expected. Once those outcomes are calculated each result will be squared and divided by the original expected number for each phenotype. To find the final outcome for the chart the last column is added together in which that outcome will be found on the P value. In which first the degrees of freedom shows that it will be three from the formula: N1. N represents the number of phenotypes. In this case there were four phenotypes. 4-1 is 3 so by finding 3 on the degrees of freedom and going over will show if the final results from the chisquare chart will be less than or greater than 0.05. If the result is less than 0.05 the hypothesis is accepted. However, if the result is greater than 0.05 the hypothesis is rejected. In this experiment there were two ears of corn by which the process must be the same for corn B as corn A.

Results The dihybrids cross for the parents are heterozygous which means that the cross will look such as: PpSs X PpSs. A punnet square is used to determine the phenotypic ratio. PS Ps pS ps PPSS PPss PpSS PpSs PPsS pPSs PpsS Ppss pPSS pPSs ppSS ppSs pPsS pPss ppsS ppss In counting how many combinations of each phenotype will determine the phenotypic ratio in order to do the Chi-square test. In this example above, if you count how many purple smooth combinations the outcome will be 9, the outcome for purple wrinkled will equal to be 3, the outcome for yellow smooth will be 3, the outcome for yellow wrinkled will be 1. This draws the conclusion that the phenotypic ratio for this experiment will be a 9:3:3:1 ratio.

PS Ps pS ps

Phenotype Ratio Purple/Smooth 9/16 = 0.562 Purple/Wrinkled 3/16 = 0.1875 Yellow/Smooth 3/16 = 0.1875 Yellow/Wrinkled 1/16 = 0.0625 By taking this ratio and putting it in decimal form will help to perform the chi-square test. By taking these numbers you will add up the observed number of each phenotype combination and get the total then you will times the total number of observed by each ratio. For example shown below, the observed outcome total was 419 so then you must times 419 by the 9/16(0.562) ratio to get the expected column and you will repeat until expected column is filled in. Then you take the observed and multiply it by the expected number to receive the next column. You will

take the observed minus the expected outcome and square it then divide by the expected number for each phenotype. Which then you will receive the outcome of the chi-square test which leads to know the P value of the data. In finding the P-value one must looks at the outcome and to figure out if it falls greater or equal to 0.05 or less than 0.05. This will tell you if you accept or reject the hypothesis which is the goal here. There are two hypotheses: null hypothesis is there is no difference between the observed/ expected and then there is the alternative hypothesis in which there is a difference between the observed/expected. The null hypothesis is a statistical hypothesis that is only valid for determining whether your data is due to random chance or not. The null hypothesis is not the same as a biological hypothesis. P< 0.05 – Reject null hypothesis P> 0.05 – Accept hypothesis Corn A sample Phenotype

Observed (O)

Degrees of freedom

P= 0.1

0.05

(O-E)2 E Purple/Smooth 280 286 -6 0.13 Purple/ Wrinkled 101 95 6 0.37 Yellow/ Smooth 93 95 -2 0.04 Yellow/Wrinkled 35 32 3 0.28 Total 509 508 1 0.81 The final outcome for corn A sample is that the chi-square test shows us that it is 0.81. The degrees of freedom formula is: N-1 the N represents the number of phenotypes listed. Since there were four phenotypes you subtract that by one which you get 3. Then you will look on the P-value chart and find the 3 on the degrees of freedom column then go over to find where your value lies. If the value lies greater than 0.05 then the hypothesis will be rejected. If the value is less than 0.05 the hypothesis will be accepted. For the corn sample A the hypothesis was accepted because by looking at the chart below the value was less than 0.05.

2.7 1 4.6 1 6.2 5 7.7 8 9.2 4

1 2 3 4 5

Expected (E)

0.01

O-E

3.84

0.02 5 5.02

6.63

0.00 5 7.88

5.99

7.38

9.21

10.6

7.81

9.35

11.34 12.8 4 9.49 11.14 13.2 14.8 8 6 11.07 12.8 15.0 16.7 3 9 5

0.00 1 10.8 3 13.8 2 16.2 7 18.4 7 20.5 2

Corn B sample Phenotype

Observed (O)

Expected(E)

O-E

(O-E)2 E Purple/Smooth 101 236 -135 77.22 Purple/Wrinkled 99 79 20 5.06 Yellow/Smooth 109 79 30 11.39 Yellow/Wrinkled 110 26 84 271.38 Total 419 420 365.05 By looking at the final outcome of the chi-square test the result was 365.05 and with the same degrees of freedom as corn A by using the formula: N-1 which gives you three. The value is greater than 0.05 which means the hypothesis is rejected. Discussions The important procedures in this experiment are the punnet square, chi-square chart, and looking at the P-value scale to figure out if the null hypothesis is accepted or rejected. The punnet square was used to show the ratios for each phenotype and had 16 combinations. The results of the punnet square showed that the ratio was a 9:3:3:1 ratio. There were 9/16 purple smooth, 3/16 purple wrinkled, 3/16 yellow smooth, and 1/16 yellow wrinkled. This was used to do the chisquare test. The chi-square chart was used to organize the data and calculate for the end result. There was a column used for the observed, expected, and the observed subtracted by the expected, and the observed subtracted by the expected squared divided by the original expected of each phenotype. Then the formula to find the degrees of freedom was used which was: N-1. The degree of freedom for this experiment was 3 in which was looked on the P- value to figure out if the hypothesis was accepted or rejected in each sample of corn. For corn A, the hypothesis was accepted and for corn B it was rejected. These procedures helped to be sure that the corn followed the 9:3:3:1 ratio and the parents were heterozygous making the hypothesis correct. Errors that can be found in this experiment are that some kernels of corn fell out prior to observation. The lab could be modified to improve the results by double checking all calculations. By double checking the calculations of the chi-square chart to make sure there were no errors. Also, it would be more accurate if there were a machine to count the kernels texture and color because color is shown differently by every individual. In closing, these procedures were used to figure out if the hypothesis were correct since it showed that the corn sample followed the 9:3:3:1 ratio and the parents were heterozygous....