Biology Hypothesis
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7.210LMitosisMeiosisandStartMendelianGenetics_2020edit1.pptxLab 7 Mitosis and Meiosis & Start Lab 8 Mendelian Genetics
Terminology
Gene – a discrete unit of hereditary information, such as a segment of DNA that codes for a particular protein or trait (example: a gene that codes for eye color)
Lab Topic 7 contains many bold-faced terms related to genetics and cell division. For future quizzes and lab exams, make sure you can define the bold-faced terms in the Blackboard notes.
More Terminology
Allele – alternate versions of the same gene (example: an allele that codes for blue eye color vs. an allele that codes for brown eye color)
Chromosome – a discrete strand of DNA (and associated organizational proteins) within the nucleus
Chromosome talk
Homologous chromosomes – a pair of chromosomes that carries the same basic set of genes. One homologous chromosome is inherited from the organism’s father, while the other homologous chromosome is inherited from the organism’s mother.
Chromosome talk, continued
Sister chromatids are two copies of a duplicated chromosome that are still physically attached to each other. These chromatids are separated from one another during mitosis and meiosis II; once the chromatids split apart, each one is considered an individual chromosome.
Ploidy
A diploid (2n) cell contains two sets of chromosomes. One set was inherited from the organism’s mother, while the other set was inherited from the organism’s father.
A haploid (n) cell contains one set of chromosomes. In humans, mature egg and sperm cells are haploid.
The Human Life Cycle: Mitosis
Mitosis occurs all over the human body for growth and repair.
During this process, one diploid (2n) parent cell divides into two diploid daughter cells.
Since mitosis is essentially a cloning process, the daughter cells are identical to each other and to the original parent cell.
The Human Life Cycle: Meiosis
Meiosis occurs in the ovaries and testes to produce haploid (n) eggs and sperm.
The chromosome number is cut in half by the end of meiosis, and genetic material also crosses over between homologous chromosomes.
As a result, the daughter cells are not identical to each other nor to the original parent cell.
The Human Life Cycle: Meiosis, continued
During this process, one diploid (2n) parent cell can produce four haploid (n) daughter cells.
The cells are not identical to the parent cell, or to each other because of crossing over
Mitotic Phase of Cell Cycle
During mitosis, sister chromatids are separated from one another and placed into separate nuclei.
Cytokinesis is the process that physically splits one cell into two separate cells. This process occurs at the end of mitosis, meiosis I, and meiosis II.
Cyto-kinesis
& Cytokinesis
Mitosis in Plant Cells
Mitosis in Animal Cells
Prophase Metaphase Anaphase Telophase & Cytokinesis
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Animal cell division
Based on their appearances, be able to differentiate cells in prophase, metaphase, anaphase, and telophase.
Plant cell division
Based on their appearances, be able to differentiate cells in prophase, metaphase, anaphase, and telophase.
Overview of Meiosis I
Homologous chromosomes are ultimately placed into separate nuclei during Meiosis I.
During a process called crossing over, homologous chromosomes exchange genetic material.
Overview of Meiosis II
This process is very similar to mitosis.
One large difference between mitosis & meiosis is that homologous chromosomes do not pair during mitosis.
Sister chromatids are separated from one another and placed into separate nuclei.
Pay special attention to the arrangement of alleles on the chromosomes before and after crossing over
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Sordaria crosses
called ascospores, since they are contained within a sac called an ascus.
Numerous asci are found within a larger structure called the perithecium.
During this experiment, you will examine a fungal species named Sordaria fimicola and look for microscopic evidence of crossing over.
The spores produced by this fungus are
Sordaria crosses, set up
Two fungal strains were crossed that carry different alleles for spore color: tan and black.
When mating occurs between these two strains, the arrangement of colored spores in each ascus indicates whether crossing over occurred during meiosis.
If crossing over DID NOT occur, a 4/4 color arrangement will be produced.
Sordaria crosses, possible outcomes
If crossing over occurred, the arrangement will differ from the 4/4 color arrangement. See diagram below:
Sordaria crossing over: possible outcomes
Crossing over will result in black & tan spores not lying in a 4:4 arrangement
Sordaria cross: procedure
Flame the inoculation loop to sterilize it. This procedure will be explained by your instructor.
Safety note: Use caution while working with the Bunsen burner.
Keep long sleeves, papers, and long hair away from the flame.
Inoculation loop will be extremely hot following incineration. Let it cool for 10-15 seconds before collecting sample.
Sordaria cross: procedure, cont.
Using the sterilized inoculation loop, collect perithecia from one of the hybrid zones.
Minimize the amount of time when lids are off the plate cultures. This will minimize the risk of culture contamination.
Perform Lab 7 Exercises (Omit 7.1 & 7.4)
Lab 8 Mendelian Genetics I: Fast Plants
Unfortunately, we won’t get to perform the final experiments with our fast plants The next few slides will discuss what we WOULD HAVE done, so that you can get a feeling for how genetics experiments work and where the data come from that we’ll use in our examples. NOTE: don’t worry about memorizing the steps of these (or any) procedure!
Plant Seeds for Lab 8
Today, each group would have harvested and planted the seeds produced by their Brassica rapa plants.
Terminology to know for future lab quizzes & exams
Genotype
Phenotype
Wild-type trait
Mutant trait
F1
F2
Dominant trait
Recessive trait
Hybrid
Homozygous
heterozygous
For future quizzes and lab exams, make sure you can define the genetic terms listed in Table 8.1. These terms are defined throughout Lab 8 as well as in the textbook.
Mendelian Genetics: Fast Plants
At the beginning of the semester, F1 seeds were planted for this experiment; these seeds were heterozygous for the traits of interest.
Two weeks later, the phenotypes of these plants were recorded.
The F1 plants were then pollinated to produce F2 seeds, which are now mature in the pods.
Mendelian Genetics: Fast Plants
Today, the F2 seeds would have been harvested and planted.
Next week, the phenotypes of the F2 seedlings will be observed, and Chi-square statistical analysis will be performed to see if the phenotypic ratios follow the laws of Mendelian inheritance.
Let’s talk Monohybrid Crosses
1. P (parental) generation starts with 2 homozygous parents.
2. When parents are crossed, all offspring (F1 generation) will exhibit the dominant phenotype
3. When two F1 individuals are crossed, offspring (F2 generation) will exhibit a 3:1 phenotypic ratio (3 dominant trait:1 recessive trait)
Now for a Dihybrid Cross
1. P (parental) generation AGAIN starts with 2 homozygous parents.
2. When parents are crossed, all offspring (F1 generation) will exhibit the dominant phenotype
Now for a Dihybrid Cross
3. When two F1 individuals are crossed, offspring (F2 generation) will exhibit a 9:3:3:1 phenotypic ratio
(9 dominant /dominant,
3 dominant /recessive,
3 dominant /recessive,
1 recessive /recessive trait)
Mendelian Genetics
Exercise 8.1 Inheritance of Anthocyanin Gene (Monohybrid Cross) (AKA Fast Plants!)
F1 plants (which were heterozygous for stem color (1 purple allele, 1 green allele)) were crossed
Phenotype of F1 plants = majority recorded as purple
Hypothesis for this experiment?
How do we expect these genes to be inherited?
Prediction for this experiment?
What do you expect to see in the F2 generation?
Mendelian Genetics
Exercise 8.2 Inheritance of Plant Color: Green, Yellow-Green, and Purple (Dihybrid Cross)
F1 plants were heterozygous for stem color (1 purple allele, 1 green allele) and leaf color (1 green allele, 1 yellow-green allele)
Phenotypes of F1 plants = majority recorded as purple stem with green leaves
Hypothesis for this experiment?
Prediction for this experiment?
Chi-Square Analysis
χ2 is the symbol for Chi-square, Σ means “sum of”, o is observed value, e is expected value, n is sample size
Chi-Square Analysis, continued
χ2 cannot be accurately calculated if the expected value for any category is less than 5
Once χ2 has been calculated, you must find its associated p-value using the χ2 table (page 807).
The degrees of freedom and calculated χ2 value must be determined to find the appropriate p-value. The χ2 value means nothing on its own.
Degrees of Freedom and Probability
Degrees of freedom (df) = number of categories – 1
In 8.1, there are two phenotypic categories: purple and green. Therefore, df = 2-1 = 1.
In 8.2, there are four phenotypic categories: purple stem and green leaves, purple stem and yellow-green leaves, green stem and green leaves, and green stem and yellow-green leaves. Therefore, df = 4-1 = 3.
Probability (p value) tells us how likely it is that our conclusions are reliable based on our data.
p values of 0.05 are commonly used to analyze Chi-square results.
We will use p = 0.05 for this course.
Interpretation of χ2 Results, option 1
If the p-value is greater than 0.05, the small differences between observed and expected values are not statistically significant and you cannot reject the null hypothesis. In other words, the observed data are consistent with the hypothesis.
In relation to our experiment, this means the phenotypic ratios did follow the patterns of Mendelian inheritance.
p-value > 0.05 = cannot reject the null hypothesis
AKA – p-value is the likelihood of a false conclusion from these data. That we would see these same numbers, with large differences between observed and expected, but that these genes are actually inherited following mendel’s rules.
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Interpretation of χ2 Results, option 2
If the p-value is less than 0.05, the relatively large differences between observed and expected values are considered statistically significant and the null hypothesis should be rejected. In other words, the observed data are not consistent with the hypothesis.
In our experiment, this means the phenotypic ratios did not follow the patterns of Mendelian inheritance.
p-value < 0.05 = reject the null hypothesis
Sample Problem setup
An experiment was set up in order to see whether or not the fast growing plant Brassica rapa shows the general pattern of Mendelian inheritance.
The researcher counts the F2 generation of the plants which has the following phenotypes: 83 plants are purple and 22 plants are green.
From this data, the researcher would like test the goodness to fit of the experiment using χ2 analysis.
Let’s do the numbers
Purple
O = 83
E = 105 x (3/4) = 78.75
O-E = 83-78.75 = 4.25
O-E2 = 18.0625
O-E2/E = 0.2294
Green
O = 22
E = 105 x (1/4) = 26.25
O-E = 22-26.25 = -4.25
O-E2 = 18.0625
O-E2/E = 0.6881
n = 105
Σ = 0.9175, D.F = 2-1=1,
p-value ≈ 0.3, fail to reject null hypothesis
The thing that MOST students miss is WHY we multiply the total by ¼ or ¾. It’s because of the expected ratios for each phenotype. We expect ¾ to be purple, and 1/4 to be green, based on the 3:1 ratio from our Punnett square. Yeah?!?
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Figure 12.UN02
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Figure 12.UN02 Test Your Understanding, question 7
Figure 12.UN05
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Vocab Review
Gene
Allele
Chromosome
Homologous Chromosome
Sister Chromatid
Diploid
Haploid
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