Fast Plant Genetics Lab Report
The Genetics of Fast Plants
Kristin Phillips
Dr. Orlando
April 4, 2014
Abstract
The experiment that was being completed was the genetics of plants using Fast Plants. The purpose of this project was to understand Mendel’s concept of a Dihybrid by performing similar crosses and calculated the phenotypes that are displayed. This project was conducted over many weeks by planting F1 seed and waiting for them to grow then cross-pollinating the F1 plants to produce the F2 seeds. Next students were to plant the F2 seeds and record the phenotypes. After all the data was collected we ran a statistical test called a Chi Squared to see if the hypothesis put foreword was correct. After running the Chi Squared Test the Group and Class hypothesis was accepted. So in conclusion the fast plants resemble Mendel’s genetic experiments. …show more content…
The project that was conducted was the genetics of plants, this project was meant to introduce Mendel’s concept of Dihybrid. Mendel was a geneticist that studied patterns of inheritance through true breading pea plant. One pattern that Mendel noticed was called the Dihybrid cross, this is when there are two traits in a single cross. Mendel crossed a true breading round and yellow pea plant with a true breading wrinkled green pea plant and the results were that the F1s’ only displayed the dominant phenotypes. Then Mendel crossed the F1’s and he created the F2’s, the results were that there was a particular ratio of 9:3:3:1 and all phenotypes were displayed. In this project we will perform a Dihybrid cross on fast plants the phenotypes are non-purple stem, green leaf and purple stem, green leaf. We will plant and grow a F1 then cross-pollinate and plant the F2’s and get a ratio. At the end we will conduct a chi squared to see if the ratio is statistically significant to Mendel’s findings.
The plants that we will be studying are Wisconsin Fast Plant; this is a strain of the rapid-cycling Brassica Rapa. The reason we are using Wisconsin Fast Plant is because the life cycle is very short. During the first 1-2 days of the life cycle the seed are germinating, then after the third day the stem pushes through the soil. After 5-8 days into the life cycle the plant is above the soil and is growing the leaves and stem. After 9 days into the life cycle the plant starts growing flowers. After 14-17 days of the life cycle the flowers being to pollinate. Then after 18-20 days of the life cycles the new seeds start to grow. And lastly from days 21-48 the plant begin to wilt and the new seeds can be harvested. The particular phenotypes that were are studying are first non-purple, green stem and the second one is purple stem, green leaf. The possible stem colors for this fast plant are: purple and green. The possible leaf colors for this fast plant are: yellow-green and green.
Figure 1
The life cycle of a Fast Plant is very short after 28 days the plant will start to produce seeds. The purpose of this lab is to use the principles of Dihybrid crosses and predict the expected ratio of phenotypes and then compare the expected ratio of phenotypes to the observed ratio of phenotypes. In this lab we want to determine the dominant and recessive phenotypes. And lastly we want to see if our hypothesis is correct by performing a Chi Squared. The first week of the experiment each group was assigned to plant one quad containing 12 seeds.
Each group labeled the quad they planted as F1. Our Professor had the Parental Fast Plants on the counter so everyone could see the phenotypes. The phenotypes of the Parental were non-purple stem, green leaf and purple stem, green leaf. Over the next week we observed our F1 Fast Plant and recoded the phenotypes. The third week is when we started pollinating the flowers. We used a bee stick to cross-pollinate the flowers. The reason we use a bee stick to pollinate is that fast plants do not self-pollinate, so bee sticks are used to successfully reproduce. Over the next week we had to pollinate the flowers two more times so the flowers could achieve maximum fertilization. In the following week we had to complete the pollination and pinch off any unopened flower buds. Then from weeks 5-7 we monitored the seed formation. And on week 7 we took the fast plants off their watering system and let them dry for five days. Then in week 8 we collected our F2 seeds and planted them in a quad. Then lastly in week 9 we observed our F2
phenotypes. We found that for the group data that the F1 generation plants all had the phenotypes purple stem and green leaves. This data lead me to believe that the purple stem and green leafs are the dominant phenotypes while the green stem and yellow-green leaf is recessive. In table #1 the values number of plants in each phenotype can be found. There were 9 plants with the phenotypes purple stem, green leaf. There was 1 plant with the phenotypes purple stem, yellow green leaf. There were 2 plants with the phenotypes non-purple stem, green leaf. There were 0 plants with the phenotypes non-purple stem, yellow-green leaf. In graph #1 there is a visual of this data.
We found that for the group data that the F2 generation plants all had the phenotypes non-purple stem and green leaf. In table #2 the values number of plants in each phenotype can be found. There were 3 plants with the phenotypes purple stem, green leaf. There was 0 plants with the phenotypes purple stem, yellow green leaf. There were 5 plants with the phenotypes non-purple stem, green leaf. There were 3 plants with the phenotypes non-purple stem, yellow-green leaf. In graph #2 there is a visual of this data.
Table #1 - Group Data
The phenotypes from both the F1 and F2 crosses
Phenotype
F1
F2
Purple stem, green leaf
9
3
Purple stem, yellow green leaf
1
0
Non-purple stem, green leaf
2
5
Non-purple stem,
Yellow green leaf
0
3
Table #2 - Class Data
The phenotypes from all the F2 crosses
Phenotype
F2
Purple stem, green leaf
63
Purple stem, yellow green leaf
15
Non-purple stem, green leaf
54
Non-purple stem,
Yellow green leaf
17
Graph #1 – Group Data
The F2 Group Data
Graph #2
The F2 Data from Class Data
In this cross I believe that the mechanism of inheritance is complete dominance. The expected phenotypes for group data can be seen in table #3. There were 6 plants that were expected to have the phenotypes purple stem, green leaf. There were 2 plants that were expected to have the phenotypes purple stem, yellow green leaf. There were 2 plants that were expected to have the phenotypes non-purple stem, green leaf. There was 1 plant that was expected to have the phenotypes non-purple stem, yellow-green leaf.
The expected phenotypes for the class data can be seen in table #4. There were 84 plants that were expected to have the phenotypes purple stem, green leaf. There were 28 plants that were expected to have the phenotypes purple stem, yellow green leaf. There were 28 plants that were expected to have the phenotypes non-purple stem, green leaf. There were 9 plants that were expected to have the phenotypes non-purple stem, yellow-green leaf.
Table # 3
Expected Phenotypes Counts for the Group Data
Phenotype
F2
Purple stem, green leaf
4.125
Purple stem, yellow green leaf
1.357
Non-purple stem, green leaf
4.125
Non-purple stem,
Yellow green leaf
1.357
Table # 4
Expected Phenotypes Counts for the Class Data
Phenotype
F2
Purple stem, green leaf
55
Purple stem, yellow green leaf
18
Non-purple stem, green leaf
55
Non-purple stem,
Yellow green leaf
18
Table # 5 - Chi Squared Analysis of Group Data
Accept the hypothesis
Phenotype
Observed
Expected
O-E
(O-E)2
(O-E)2/E
Purple stem, green leaf
3
4.125
-1.125
1.27
.31
Purple stem, yellow green leaf
0
1.375
-1.375
1.89
1.375
Non-purple stem, green leaf
5
4.125
.875
.77
.18
Non-purple stem,
Yellow green leaf
3
1.375
1.625
2.64
1.92
Total = 11
Total = 11
Degrees of Freedom = 3
Critical Value = 7.815
X2 = 3.785
Table #6 - Chi Squared Analysis of Class Data
Accept the Hypothesis
Phenotype
Observed
Expected
O-E
(O-E)2
(O-E)2/E
Purple stem, green leaf
63
55
8
64
1.16
Purple stem, yellow green leaf
15
18
-3
9
.5
Non-purple stem, green leaf
54
55
-1
1
.018
Non-purple stem,
Yellow green leaf
17
18
-1
1
.018
Total = 149
Total = 149
Degrees of Freedom = 3
Critical Value = 7.815
X2 = 1.696
The Chi Squared that I conducted does reflect my prediction that the Fast Plants are completely dominant because I accepted both the group data and class data. The first Chi Squared that I conducted was on my group data and can be seen in table #5. The X2 number that I calculated was 3.785 and I compared that number to the critical value number of 7.815. Because 3.785 is smaller than the critical value of 7.815 I accepted the hypothesis meaning that my data is statistically significant. The next Chi Squared that I conducted was on the class data and can be seen in table #6. The X2 number that I calculated was 1.696 and I compared that number to the critical value number of 7.815. Because 1.696 is smaller than the critical value of 7.815 I accepted the hypothesis meaning that my data is statistically significant.
The mechanism of inheritance for the Fast Plants is complete dominance. The reason for the Fast Plants being completely dominant is that there are no intermediate phenotypes displayed. This means that the cross contains dominant and recessive alleles. The parental alleles for this cross are PpGg and ppGg. The F1 alleles for this cross are PpGG. The dominant alleles for this cross are: P_ and G_. The F2 generation Punnett Square can be seen in table #7. The punnett square depicts a ratio that is 6:2:6:2.
Table #7 - Punnett Square
Parental Generation: PpGg and ppGg
F1 Generation: PpGG
F2 Generation ratio: 6:2:6:2
PG
Pg
pG pg pG
PpGG
PpGg ppGG ppGg pg PpGg
Ppgg
ppGg ppgg pG
PpGG
PpGg ppGG ppGg pg PpGg
Ppgg
ppGg
Ppgg
There are many reasons why a hypothesis is not accepted. One reason is that they hypothesis chosen was not correct or if the hypothesis that was chosen correctly and still rejected this is because the population was to small to see the results. That being said if 2 data sets were very different that one that would be correct is the one that was accepted by the Chi Squared. Because the purpose of the Chi squared was to inform a person if their hypothesis was correct.
References
Lauffer H. B., Lauffer, D., Williams, P., (2012). Wisconsin Fast Plants® Program. Retrieved (March 31, 2014), from Wisconsin Fast Plants® Program Web site: www.fastplants.org.
Kristin Phillips
Dr. Orlando
April 4, 2014
Abstract
The experiment that was being completed was the genetics of plants using Fast Plants. The purpose of this project was to understand Mendel’s concept of a Dihybrid by performing similar crosses and calculated the phenotypes that are displayed. This project was conducted over many weeks by planting F1 seed and waiting for them to grow then cross-pollinating the F1 plants to produce the F2 seeds. Next students were to plant the F2 seeds and record the phenotypes. After all the data was collected we ran a statistical test called a Chi Squared to see if the hypothesis put foreword was correct. After running the Chi Squared Test the Group and Class hypothesis was accepted. So in conclusion the fast plants resemble Mendel’s genetic experiments. …show more content…
The project that was conducted was the genetics of plants, this project was meant to introduce Mendel’s concept of Dihybrid. Mendel was a geneticist that studied patterns of inheritance through true breading pea plant. One pattern that Mendel noticed was called the Dihybrid cross, this is when there are two traits in a single cross. Mendel crossed a true breading round and yellow pea plant with a true breading wrinkled green pea plant and the results were that the F1s’ only displayed the dominant phenotypes. Then Mendel crossed the F1’s and he created the F2’s, the results were that there was a particular ratio of 9:3:3:1 and all phenotypes were displayed. In this project we will perform a Dihybrid cross on fast plants the phenotypes are non-purple stem, green leaf and purple stem, green leaf. We will plant and grow a F1 then cross-pollinate and plant the F2’s and get a ratio. At the end we will conduct a chi squared to see if the ratio is statistically significant to Mendel’s findings.
The plants that we will be studying are Wisconsin Fast Plant; this is a strain of the rapid-cycling Brassica Rapa. The reason we are using Wisconsin Fast Plant is because the life cycle is very short. During the first 1-2 days of the life cycle the seed are germinating, then after the third day the stem pushes through the soil. After 5-8 days into the life cycle the plant is above the soil and is growing the leaves and stem. After 9 days into the life cycle the plant starts growing flowers. After 14-17 days of the life cycle the flowers being to pollinate. Then after 18-20 days of the life cycles the new seeds start to grow. And lastly from days 21-48 the plant begin to wilt and the new seeds can be harvested. The particular phenotypes that were are studying are first non-purple, green stem and the second one is purple stem, green leaf. The possible stem colors for this fast plant are: purple and green. The possible leaf colors for this fast plant are: yellow-green and green.
Figure 1
The life cycle of a Fast Plant is very short after 28 days the plant will start to produce seeds. The purpose of this lab is to use the principles of Dihybrid crosses and predict the expected ratio of phenotypes and then compare the expected ratio of phenotypes to the observed ratio of phenotypes. In this lab we want to determine the dominant and recessive phenotypes. And lastly we want to see if our hypothesis is correct by performing a Chi Squared. The first week of the experiment each group was assigned to plant one quad containing 12 seeds.
Each group labeled the quad they planted as F1. Our Professor had the Parental Fast Plants on the counter so everyone could see the phenotypes. The phenotypes of the Parental were non-purple stem, green leaf and purple stem, green leaf. Over the next week we observed our F1 Fast Plant and recoded the phenotypes. The third week is when we started pollinating the flowers. We used a bee stick to cross-pollinate the flowers. The reason we use a bee stick to pollinate is that fast plants do not self-pollinate, so bee sticks are used to successfully reproduce. Over the next week we had to pollinate the flowers two more times so the flowers could achieve maximum fertilization. In the following week we had to complete the pollination and pinch off any unopened flower buds. Then from weeks 5-7 we monitored the seed formation. And on week 7 we took the fast plants off their watering system and let them dry for five days. Then in week 8 we collected our F2 seeds and planted them in a quad. Then lastly in week 9 we observed our F2
phenotypes. We found that for the group data that the F1 generation plants all had the phenotypes purple stem and green leaves. This data lead me to believe that the purple stem and green leafs are the dominant phenotypes while the green stem and yellow-green leaf is recessive. In table #1 the values number of plants in each phenotype can be found. There were 9 plants with the phenotypes purple stem, green leaf. There was 1 plant with the phenotypes purple stem, yellow green leaf. There were 2 plants with the phenotypes non-purple stem, green leaf. There were 0 plants with the phenotypes non-purple stem, yellow-green leaf. In graph #1 there is a visual of this data.
We found that for the group data that the F2 generation plants all had the phenotypes non-purple stem and green leaf. In table #2 the values number of plants in each phenotype can be found. There were 3 plants with the phenotypes purple stem, green leaf. There was 0 plants with the phenotypes purple stem, yellow green leaf. There were 5 plants with the phenotypes non-purple stem, green leaf. There were 3 plants with the phenotypes non-purple stem, yellow-green leaf. In graph #2 there is a visual of this data.
Table #1 - Group Data
The phenotypes from both the F1 and F2 crosses
Phenotype
F1
F2
Purple stem, green leaf
9
3
Purple stem, yellow green leaf
1
0
Non-purple stem, green leaf
2
5
Non-purple stem,
Yellow green leaf
0
3
Table #2 - Class Data
The phenotypes from all the F2 crosses
Phenotype
F2
Purple stem, green leaf
63
Purple stem, yellow green leaf
15
Non-purple stem, green leaf
54
Non-purple stem,
Yellow green leaf
17
Graph #1 – Group Data
The F2 Group Data
Graph #2
The F2 Data from Class Data
In this cross I believe that the mechanism of inheritance is complete dominance. The expected phenotypes for group data can be seen in table #3. There were 6 plants that were expected to have the phenotypes purple stem, green leaf. There were 2 plants that were expected to have the phenotypes purple stem, yellow green leaf. There were 2 plants that were expected to have the phenotypes non-purple stem, green leaf. There was 1 plant that was expected to have the phenotypes non-purple stem, yellow-green leaf.
The expected phenotypes for the class data can be seen in table #4. There were 84 plants that were expected to have the phenotypes purple stem, green leaf. There were 28 plants that were expected to have the phenotypes purple stem, yellow green leaf. There were 28 plants that were expected to have the phenotypes non-purple stem, green leaf. There were 9 plants that were expected to have the phenotypes non-purple stem, yellow-green leaf.
Table # 3
Expected Phenotypes Counts for the Group Data
Phenotype
F2
Purple stem, green leaf
4.125
Purple stem, yellow green leaf
1.357
Non-purple stem, green leaf
4.125
Non-purple stem,
Yellow green leaf
1.357
Table # 4
Expected Phenotypes Counts for the Class Data
Phenotype
F2
Purple stem, green leaf
55
Purple stem, yellow green leaf
18
Non-purple stem, green leaf
55
Non-purple stem,
Yellow green leaf
18
Table # 5 - Chi Squared Analysis of Group Data
Accept the hypothesis
Phenotype
Observed
Expected
O-E
(O-E)2
(O-E)2/E
Purple stem, green leaf
3
4.125
-1.125
1.27
.31
Purple stem, yellow green leaf
0
1.375
-1.375
1.89
1.375
Non-purple stem, green leaf
5
4.125
.875
.77
.18
Non-purple stem,
Yellow green leaf
3
1.375
1.625
2.64
1.92
Total = 11
Total = 11
Degrees of Freedom = 3
Critical Value = 7.815
X2 = 3.785
Table #6 - Chi Squared Analysis of Class Data
Accept the Hypothesis
Phenotype
Observed
Expected
O-E
(O-E)2
(O-E)2/E
Purple stem, green leaf
63
55
8
64
1.16
Purple stem, yellow green leaf
15
18
-3
9
.5
Non-purple stem, green leaf
54
55
-1
1
.018
Non-purple stem,
Yellow green leaf
17
18
-1
1
.018
Total = 149
Total = 149
Degrees of Freedom = 3
Critical Value = 7.815
X2 = 1.696
The Chi Squared that I conducted does reflect my prediction that the Fast Plants are completely dominant because I accepted both the group data and class data. The first Chi Squared that I conducted was on my group data and can be seen in table #5. The X2 number that I calculated was 3.785 and I compared that number to the critical value number of 7.815. Because 3.785 is smaller than the critical value of 7.815 I accepted the hypothesis meaning that my data is statistically significant. The next Chi Squared that I conducted was on the class data and can be seen in table #6. The X2 number that I calculated was 1.696 and I compared that number to the critical value number of 7.815. Because 1.696 is smaller than the critical value of 7.815 I accepted the hypothesis meaning that my data is statistically significant.
The mechanism of inheritance for the Fast Plants is complete dominance. The reason for the Fast Plants being completely dominant is that there are no intermediate phenotypes displayed. This means that the cross contains dominant and recessive alleles. The parental alleles for this cross are PpGg and ppGg. The F1 alleles for this cross are PpGG. The dominant alleles for this cross are: P_ and G_. The F2 generation Punnett Square can be seen in table #7. The punnett square depicts a ratio that is 6:2:6:2.
Table #7 - Punnett Square
Parental Generation: PpGg and ppGg
F1 Generation: PpGG
F2 Generation ratio: 6:2:6:2
PG
Pg
pG pg pG
PpGG
PpGg ppGG ppGg pg PpGg
Ppgg
ppGg ppgg pG
PpGG
PpGg ppGG ppGg pg PpGg
Ppgg
ppGg
Ppgg
There are many reasons why a hypothesis is not accepted. One reason is that they hypothesis chosen was not correct or if the hypothesis that was chosen correctly and still rejected this is because the population was to small to see the results. That being said if 2 data sets were very different that one that would be correct is the one that was accepted by the Chi Squared. Because the purpose of the Chi squared was to inform a person if their hypothesis was correct.
References
Lauffer H. B., Lauffer, D., Williams, P., (2012). Wisconsin Fast Plants® Program. Retrieved (March 31, 2014), from Wisconsin Fast Plants® Program Web site: www.fastplants.org.