Lab 2 Lab Report BIOL 2020 PDF

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BIOL 2020U - Labor Laboratory atory # 2 Introd Introducti ucti uction on to Mendelian Gene Genettics Labor Laboratory atory Report # #2 2 (22 marks) Name: K Kylie ylie Jenner

Your laboratory report is due AT THE END OF YOUR LAB PERIOD and must be submitted electronically as an a ass ss ssignment ignment in Can Canvas. vas. Per Perform form the exerci exercises ses and calculations, fill iin n the Tables and answer the questions as described below below.. You need to upload a pdf, doc or docx file as an assignment submission in Canvas. Here are the steps to upload a file as an assignment submission in Canvas. You can upload more than one file for the submission or you can put everything in one file. 1. 2. 3. 4. 5.

Open Assignments. In Course Navigation, click the Assignments link. Select Assignment. Click the title of the assignment. Submit Assignment. Click the Submit Assignment button. Add File. ... Add Another File. ... View Submission

Determining the genot genotype ype of the unknown Indian corn cobs cobs.. You will be given digital photos of 4 ears of Indian corn of unknown heredity. You will make a hypothesis about the genotypes of the plant that produced the ear of corn. You will collect data and submit the data to a chi-square analysis to see if the data support your hypothesis. Every kernel of the Indian corn is a seed, the result of a fusion of a male gamete (pollen nucleus) carrying one allele and a female gamete (ovum) carrying one allele.

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BIOL 2020U Lab 2 Introduction and Protocol

Unknown corn cob # 1: 1. Record the number of phenotypes (color and/or texture texture)) present There are two phenotypes present. Yellow and smooth and purple and smooth.

2. Count the number of kernels with each of the phenotypes present in five rows and determine the proportion of each phenotype Kernel phenotypes (fill in the phenotypes)

Numbers

Yellow & Smooth

Purple & smooth

Total

62

61

123

Proportions 62/123 61/123 Note: you may not need all the columns

123

3. Write down your hypothesis: To which predicted value does your proportion more closely relate to? According to the test cross, it is hypothesised that there is a 1:1 ratio. Therefore, the offspring would be ½ purple smooth kernel and ½ yellow smooth kernel. The yellow smooth kernel has about a 0.504 probability while purple smooth kernel has a probability of 0.495. Therefore, the yellow smooth kernel is most closely related to the 0.5 proportion. Per Perform form a chi-square analysis (see below) between the observed data and the predicted ones, to see if your hypothesis supports the observed data (use P = 0.05). Phenotype

Observed

Expected

(O - E)

(O - E)

(O - E) / E

Yellow & Smooth

62

123 x ½ = 62

62-23 = 0

(0)2 = 0

0/62 = 0

Purple & Smooth

61

123 x ½ = 62

61-62= -1

(-1)2 = 1

1/62 = 0.016

2

2

2

BIOL 2020U Lab 2 Introduction and Protocol

 = 0.016 + 0 = 0.02 Df = 2-1 = 1 P = 0.90

=

=

=

o What was the parental F1 cross that gave rise to your F2 corn? Indicate the genotypes of these parents. The parental F1 cross had to have one parent carrying one heterozygous trait and one homozygous (dominant) trait and the other parent carrying two homozygous (dominant and recessive) traits. The genotypes are, PpSS and ppSS Which gametes are produced by each F1 parent? Gamete for parent 1: PS (purple smooth) and pS (yellow smooth) Gamete for parent 2: pS (yellow Smooth) o Draw the Punnett explaining the data observed and list the phenotypes and ratios

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BIOL 2020U Lab 2 Introduction and Protocol

o Which allele is dominant/rec dominant/recess ess essive? ive? The purple (P) and Smooth (S) allele are dominant while yellow (p) is recessive. 4. Conclusion; Are the results suppo support rt rting/not ing/not supporting your hypothesis? Briefly explain the genetic principle behind the results ob principless obser ser served. ved. The hypothesis was that the phenotypic ratio should be 50% purple smooth and 50% yellow smooth. The results show that out of the five rows, there were 62 yellow smooth kernels and 61 purple smooth kernels out 123 kernels in total. Both had a ratio equivalent to 50% which supports the hypothesis. The chi square also supports the hypothesis. According to the chi square, the P – value is 0.90 which is greater than 0.05. Therefore, the difference between the expected and observed value is not significantly different and we therefore accept the hypothesis.

Unknown corn cob # 2: 1. Record the number of phenotypes (color and/or texture texture)) present 4

BIOL 2020U Lab 2 Introduction and Protocol

There are 2 phenotypes present. Purple and smooth and yellow and smooth 2. Count the number of kernels for each of the phenotypes present in five rows and determine the proportion of each phenotype Kernel phenotypes (fill in the phenotypes) Yellow and Purple and smooth smooth

Total

Numbers

27

85

112

Proportions

27/112

85/112

112

Note: you may not need all the columns 3. Write down your hypothesis: To which predicted value does your proportion more closely relate to? The hypothesis is that there is a 3:1 genotypic ratio with purple smooth having an offspring of ¾ and yellow smooth having an offspring of ¼. Purple and smooth has a proportion of 0.759 while yellow and smooth has a proportion of 0.241. Out of both of them, purple and smooth has the closest predicted value of 0.75. 4. Perform a chi-square analysis (see below) between the observed data and the predicted ones, to see if your hypothesis supports the observed data (use P = 0.05). Phenotype

Observed

Expected

(O - E)

(O - E)

(O - E) / E

Yellow & smooth

27

28

27 – 28 = -1

(-1)2 = 1

1/28 = 0.036

Purple & smooth

85

84

85 – 84 = 1

(1)2 = 1

1/84 = 0.012

 = 0.036 + 0.012 = 0.024 Df = 2-1 = 1 P = 0.90

=

=

2

2

=

o What was the parental F1 cross that gave rise to your F2 corn? Indicate the genotypes of the these se parents.

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BIOL 2020U Lab 2 Introduction and Protocol

Both parents in the F1 generation had one trait that was Heterozygous and the other trait being homozygous dominant. There are three possible genotypes which are, PPSS, PpSS, and ppSS. Both parents have genotype PpSS. o Which gametes are pr produced oduced by each F1 parent? The gametes produced by Parent 1 are, purple smooth (PS) and yellow smooth (pS) Gametes produced by Parent 2 are the same as parent 1 which is purple smooth (PS) and yellow smooth (pS)

o Dr Draw aw the Punnet Punnettt explaining the data observed and list the phenotypes and rati ratios os

o Which allele is dominan dominant/recessive? t/recessive?

The purple (P) and Smooth (S) allele are dominant while yellow (p) is recessive. 6

BIOL 2020U Lab 2 Introduction and Protocol

5. Conclusion; Are the results supporting/not supporting your hypothesis? Briefly explain the genetic principle principless behind the results ob obser ser served ved The hypothesis was that the phenotypic ratio should be 75% purple smooth kernels and 25% yellow smooth kernels. The results show that out of the five rows, there were 85 purple smooth kernels and 27 yellow smooth kernels out 122 kernels in total. The purple smooth kernel ended up having a proportion of about 0.759 which is close to 0.75 and yellow smooth had a proportion of 0.241 which is close to 25%. According to the chi square, the P – value is 0.90 which is greater than 0.05. Therefore, the difference between the expected and observed value is not significantly different and therefore accept the hypothesis. This can be concluded that the results do in fact support the hypothesis.

Unknown corn cob # 3: 1. Record the number of phenotypes (color and/or texture texture)) present There are 4 phenotypes present: purple & smooth, yellow & smooth, purple & shrunken and yellow & shrunken 2. Count the number of kernels for each of the phenotypes present in five rows and determine the proportion of each phenotype Kernel phenotypes (fill in the phenotypes) Purple & smooth

Yellow & smooth

Purple & shrunken

Numbers

34

Proportions 34/155

Yellow & shrunken Total

43

44

34

155

43/155

44/155

34/155

155

Note: you may not need all the columns 3. Write down your hypothesis: To which predicted value does your proportion more closely relate to? The hypothesis was that each phenotype had a ratio of ¼ meaning that each had a proportion of 0.25. 7

BIOL 2020U Lab 2 Introduction and Protocol

Purple & smooth had a proportion of about 0.219, Yellow & smooth had proportion of about 0.277, Purple & shrunken had a proportion of about 0.284, and yellow & shrunken had proportion of 0.219. Out of all four phenotypes, the yellow and smooth kernel closely relates to the hypothesis (proportion of 0.25). 4. Perform a chi-square analysis (see below) between the observed data and the predicted ones, to see if your hypothesis supports the observed data (use P = 0.05). Phenotypes

Observed

Expected

(O - E)

(O - E)

(O - E) / E

Yellow & smooth

43

39

43-39 = 4

(4)2 = 16

16/39 = 0.410

Purple & smooth

34

39

34-39 = -5

(-5)2 = 25

25/39 = 0.641

Yellow & shrunken

34

39

34-39 = -5

(-5)2 = 25

25/39 = 0.641

Purple & shrunken

44

39

44-39 = 5

(5)2 = 25

25/39 = 0.641

 = 0.410 + 0.641 + 0.641 + 0.641= = 2.333 Df = 4-1 = 3 P = 0.50

2

=

2

=

o What was the parental F1 cross that gave rise to your F2 corn? Indicate the genotypes of the these se parents. The F1 generation had one parents have a Heterozygous and homozygous recessive trait Parent 1 has a Ppss genotype and parent 2 has ppSs genotype. o Which gametes are pr produced oduced by each F1 parent? Par Parent ent 1 gamete: Purple shrunken (Ps) and yellow shrunken (ps) Par Parent ent 2 gamete: Yellow smooth (pS) and yellow shrunken (ps)

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BIOL 2020U Lab 2 Introduction and Protocol

o Dr Draw aw the Punnet Punnettt explaining the data observed and list the phenotypes and rati ratios os

o Which allele is dominan dominant/recessive? t/recessive?

The purple and smooth allele is dominant, while the yellow and shrunken allele is recessive.

5. Conclusion; Are the results supporting/not supporti supportin ng your hypothesis? Briefly explain the genetic principle principless behind the results ob obser ser served. ved. The hypothesis was that the each phenotype had a ratio of ¼ (25%). Yellow & smooth had a proportion of about 0.277, yellow & shrunken has a proportion of 0.219, Purple and shrunken has a proportion of, 0.284 and purple & smooth had a proportion of about, 0.219. All of these proportion are somewhat close to 25%. According to the chi square, the probability is 0.50 which is greater than 0.05. Therefore, the difference between the expected and observed value is not significantly different and 9

BIOL 2020U Lab 2 Introduction and Protocol

the hypothesis can be accepted. Therefore, it can be concluded that the results do support the hypothesis.

Unknown corn cob # 4:

1. Record the number of phenotypes (color and/or texture texture)) present There are 4 phenotypes present: Purple & smooth, yellow & smooth, purple & shrunken and yellow & shrunken 2. Count the number of kernels for each of the phenotypes present in five rows and determine the proportion of each phenotype Kernel phenotypes (fill in the phenotypes) Purple & smooth

Purple

yellow

yellow

& shrunken

& smooth

& shrunken

Total

Numbers

100

30

24

11

165

Proportions

100/165

30/165

24/165

11/165

Note: you may not need all the columns 3. Write down your hypothesis: To which predicted value does your proportion more closely relate to? It is hypothesized that there will be a 9:3:3:1 ratio. Purple & Smooth had a proportion of 0.61, purple & shrunken had a proportion of 0.18, yellow & smooth had a proportion of, 0.15, and yellow & shrunken had a proportion of 0.07. Purple & shrunken was closely related to the predicted ratio of 3/16 (0.1875) while yellow and shrunken is closely related to the 1/16 ratio (0.0625) 4. Perform a chi-square analysis (see below) between the observed data and the predicted ones, to see if if your hypothesis supports the observed data (use P = 0.05). Phenotypes

Observed

Expected

(O - E)

(O - E)

(O - E) / E

Yellow & smooth

24

31

24-31 = -7

(-7)2 = 49

49/31 = 1.581

Purple & smooth

100

93

100-93 = 7

(7)2 = 49

49/93 = 0.527

Yellow & shrunken

11

10

11-10 = 1

(1)2 = 1

1/10 = 0.100

10

2

2

BIOL 2020U Lab 2 Introduction and Protocol

Purple & shrunken

30

 = 1.581 + 0.527 + 0.100 + 0.032= 2.24 = Df = 4-1 = 3 P = 0.50

31

30-31 = -1

=

(-1)2 = 1

1/31 = 0.032

=

o What was the parental F1 cross that gave rise to your F2 corn? Indicate the genotypes of the these se parents. Both parent have two heterozygous traits. The genotypes of these parents are, PpSs x PpSs o Which gametes are pr produced oduced by each F1 parent?

Par Parent ent 1 & 2 gametes: Purple & Smooth (PS), Purple & Shrunken (Ps), Yellow & Smooth (pS) and Yellow & Shrunken

o Dr Draw aw the Punnet Punnettt explaining the data observed and list the phenotypes and rati ratios os

11

BIOL 2020U Lab 2 Introduction and Protocol

o Which allele is dominan dominant/recessive? t/recessive? 12

BIOL 2020U Lab 2 Introduction and Protocol

The purple and smooth allele is dominant, while the yellow and shrunken allele is recessive.

5. Conclusion; Are the results supporting/not supporti supportin ng your hypothesis? Briefly explain the genetic principle principless behind the results ob obser ser served. ved. The hypothesis was that the phenotype had a 9:3:3:1 ratio. Purple & Smooth had a proportion of 0.61, purple & shrunken had a proportion of 0.18, yellow & smooth had a proportion of, 0.15, and yellow & shrunken had a proportion of 0.07. These results are somewhat close to the actual proportions. According to the chi square, the probability is 0.50 which is greater than 0.05. Therefore, the difference between the expected and observed value is not significantly different and the hypothesis can be accepted. Therefore, it can be concluded that the results do support the hypothesis.

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BIOL 2020U Lab 2 Introduction and Protocol...