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PRACTICAL 2: FROM DNA TO PHENOTYPE
In this prac, we will:
• Analyse data from a genetic cross involving one autosomal gene locus in Arabidopsis
• Visualise the DNA in cells undergoing meiosis in Lily anthers and grasshopper testes
• Observe bacterial growth on minimal media to determine which supplements are required for the growth of different auxotrophs
BEFORE THE PRAC
• Download and read these prac notes
• Watch the following videos on the LMS:
o BioByte: The Chi Square Test (ESSENTIAL)
o TechTip: Spreading an Agar Plate (if you missed Prac 1)
o TechTip: Setting up the auxotroph experiment (if you missed Prac 1)
• Complete the Pre-prac quiz – Closes on Friday, 16th August at 10am
ASSESSMENT (7.5% subject grade)
• Pre-prac quiz (1.5%)
• Post-prac test (6%) opens AFTER prac (17:00 on Friday and closes 17:00 on Monday, 19th August)
IN THE PRAC
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Time |
Activity |
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10 mins |
Welcome & introduction to Practical 2 |
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50 mins |
Activity 1: An autosomal gene (Arabidopsis) Practical Task 1.1: Observing parental and F1 phenotypes, and F2 cross data Practical Task 1.2: Genetic analysis and statistical testing Practical Task 1.3: Modelling meiosis in the Arabidopsis cross |
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20 mins |
Activity 2: Visualising chromosomes, part 2 Practical Task 2.1: Meiosis in lily anthers Practical Task 2.2: Meiosis in grasshopper testis |
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30 mins |
Activity 3: Characterising auxotrophs Practical Task 3.1: Experimental set-up Practical Task 3.2: Observations and data analysis |
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Your safety in the laboratory is very important: |
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• At all times wear a lab coat, suitable shoes with enclosed heel and toe and safety glasses. • Always work to ensure your safety and the safety of those around you. • Immediately report any injuries or spills to a demonstrator. • Microorganisms will be used in this class, avoid touching things to your mouth and wash hands carefully after the class. A risk assessment has been carried out for the practical classes and identified risks minimised. |
ACTIVITY 1: AN AUTOSOMAL GENE IN ARABIDOPSIS THALIANA 50 MINS
Arabidopsis thaliana is a commonly used model organism in genetics. In this section, you will investigate the inheritance of a gene involved in producing the trichomes on the leaves of A. thaliana. Wildtype plants produce trichomes on their leaves, but mutant plants do not produce these trichomes (Fig. 1). This mutant phenotype is known as glabrous, which means smooth.
Fig. 1 Wildtype (left) and glabrous (smooth) mutant (right) A. thaliana plants
Q1.1a. What characteristics would make an organism like A. thaliana a useful model organism in genetics?
We crossed pure-breeding (homozygous) wildtype plants with pure-breeding smooth (glabrous) mutants to obtain first-filial generation (F1 ) offspring.
Q1.1b. Without looking at the offspring phenotypes, what can we already deduce about the genotype of the F1 plants, given the way the cross was set up?
Familiarise yourself with the parental phenotypes and observe the phenotype of the F1.
Q1.1c. What is the phenotype of the F1 plants?
Q1.1d. What does this observation indicate about the inheritance pattern of the mutant?
The F1 progeny were allowed to self-fertilise to produce second-filial generation (F2 ) offspring. Working as a pair, observe the F2 sample you have been allocated.
Q1.1e. Score your group sample of F2 plants and record the data in Table 1 below.
Table 1: Data for F2
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Number of wildtype (hairy) plants |
Number of mutant (smooth) plants |
Total |
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Group data |
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Class data |
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Q1.1f. Using the standard allelic notation for this gene, Gl for the wildtype allele and gl for the smooth allele, propose a genetic hypothesis to account for these observations by filling in the gaps below.
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Parental phenotypes |
Wildtype (hairy) |
Mutant (smooth) |
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Parental genotypes |
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F1 genotype |
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F1 phenotype |
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Punnet square for working out F2 genotypes:
Q1.1g. The expected F2 genotypic ratio is …
The expected F2 phenotypic ratio is …
Practical Task 1.2: Forming and testing a genetic hypothesis using F2 data
A genetic hypothesis is a statement explaining what you think is happening at the genetic level. You should include the following information in the genetic hypothesis:
• The number of gene loci involved.
• How many alleles are at each gene locus and the names of the alleles.
• The inheritance pattern of the mutant phenotype (dominant or recessive, autosomal or sex- linked).
• The expected phenotypic ratio for the cross (in this case, the F2 generation).
Q1.2a. Write a genetic hypothesis for the F2 generation by completing the following.
There is/are gene locus/loci being considered in this cross with allele/s. The dominant phenotype is . The F2 of this cross will produce a phenotypic ratio of .
After writing our genetic hypothesis, we need to determine if the results obtained from experimental observations support our hypothesis. We are unlikely to obtain results identical to what we predict, so we need to determine whether the difference between our observed results and expected results is due to sampling error or an error in our hypothesis. The more likely we think it is due to an error in our hypothesis, the more likely we will reject the hypothesis and come up with a new one.
A basic method of comparing observed and expected ratios is the c2 (chi-squared) test. Follow the template below to complete the test.
Q1.2b. State the null hypothesis (refer to Appendix “Procedures for Genetic Analysis” for full statement):
Q1.2c. Calculate the chi-square (c2 ) value for the combined class data by completing Table 2 below. Give all the decimal answers to 4 decimal places.
Table 2: Computation of c2
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Class Data |
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Phenotype |
Wildtype (hairy) |
Mutant (smooth) |
Total |
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Observed (O) |
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Expected (E) |
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(O-E) |
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(O-E)2 |
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(O-E)2 E |
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Degrees of freedom =
Use the Chi-square table (refer to the Appendix ‘Procedures for Genetic Analysis’) to find the associated probability that the difference observed is due to chance =
Q1.2d. Conclusion (delete incorrect alternatives, or annotate the correct alternatives):
The null hypothesis is supported / not supported. Therefore, we accept / reject the underlying genetic hypothesis that gave rise to the prediction. The observed results are consistent / not consistent with the genetic hypothesis.
Practical Task 1.3: Modelling meiosis in the Arabidopsis cross
The glabrous gene is located on Chromosome 3 in A. thaliana. On the diagram below, track Chromosome 3 through meiosis in the two parents leading up to the F1. Label an arbitrary position on each chromosome where the gene is located, and clearly indicate which allele is present on each chromosome. Hint: using different colours will make this clearer.
In the parents, you should draw the chromosomes as they would be after DNA replication. But in the F1, draw the chromosomes as they would be just after fertilisation.
ACTIVITY 2: VISUALISING CHROMOSOMES, Part 2 20 MINS
Meiosis occurs in specialised organs called gonads. The cells within these organs that are destined to undergo meiosis are called the germline cells. In this activity, we will explore both meiotic divisions in cells from the lily anther during pollen production. We will also observe the first meiotic division in cells from the testis of a grasshopper species.
Meiosis consists of two successive divisions which serves to halve the chromosome number. But the two divisions of meiosis differ. In the first division, three special events occur that are not seen in the second meiotic division (or in mitosis). These are that the:
• homologous chromosomes pair up as bivalents,
• homologous chromosomes physically swap parts of their chromatids, and
• homologous chromosomes separate from each other (while the sister chromatids stay together).
On the other hand, the steps that occur in the second meiotic division are analogous to those in mitosis.
Practical Task 2.1: Meiosis in the Lily anther
We can examine the stages of both meiotic divisions in slides prepared from lily anther cross sections, which have then been digitised using the Roslin platform. These slides reveal the developmental stages of pollen from early prophase of meiosis I, through to the pollen tetrads which are the final product of meiosis.
Using the computer stations set up for you, observe the digitised high-resolution slides from Lily anthers.
Q2.1a. In the images from the 1st meiotic division and the image from the 2nd meiotic division, identify as many of the different stages as you can. How many can you find? Are there some that have not been captured in these particular images? You may want to take phots or draw the best examples.
Q2.1b. Given that this species has 24 chromosomes (2n = 24), state how many chromosomes you would expect to find in a cell that was in each of the following stages:
(i) Prophase of mitosis
(ii) Metaphase I of meiosis
(iii) Prophase II of meiosis
(iv) Telophase II of meiosis
Practical Task 2.2: Meiosis in the grasshopper testis
In male grasshoppers, meiotic chromosomes are relatively easy to observe. In somatic cells of the plague locust, Locusta migratoria, there are 24 chromosomes in the female and 23 in the male. Each sex has 11 pairs of autosomes, but the females have two X chromosomes whereas the males have only one. There is no Y chromosome. Thus, sex is determined by the number of X chromosomes present, XX animals are female, and XO animals are male.
Observe the high-resolution digitised slides of meiosis in grasshopper testis using Roslin on the provided computers.
Q2.2a. How many bivalents would be present during prophase of meiosis I?
Q2.2b. Sister chromatids are held together at the centromere. Can you identify the centromere in any of these chromosomes?
Q2.2c. How many sperm are produced from each parental diploid cell? How many chromosomes are present in each sperm produced?
ACTIVITY 3: AUXOTROPHS 30 MINS
Your average wildtype bacterium can produce all the organic compounds it requires, such as amino acids and vitamins, from inorganic sources in the environment. This type of bacterium is known as a prototroph. Prototrophs can grow on minimal media agar, which only contains inorganic ingredients.
However, if there is a block in a pathway to manufacture one of these compounds, the bacteria must be provided with the compound in the environment and will not grow on minimal media without it. This strain of bacterium is known as an auxotroph. By adding different compounds to minimal media and seeing which compound allows bacteria to grow, we can establish in which pathway this block occurs.
In this activity, you will determine the requirements of 4 strains of E. coli auxotrophs.
In the last practical, you set up the bacterial plates for this experiment. Remember that each strain has been spread on minimal media. Then, paper tabs, each infused with a different amino acid, were placed on the plate in the arrangement shown in Fig.2 below. The amino acids in the paper tabs diffused out into the agar. These plates were incubated and have been returned to you to be checked for growth.
Fig. 2: Positions of paper tabs impregnated with amino acids on each petri dish
Q3.1a. Without using the word ‘control’, explain why one paper tab is infused with water only.
Q3.1b. Had there been growth of E. coli around the paper tab infused with water, how would this influence the conclusions you could draw from your experiment?
Q3.2a. Record these results in Table 3 below, indicating whether growth occurred around each paper tab with a + symbol, or using – for no growth.
Table 3: Results of auxotroph growth on minimal media
Strain A Strain B Strain C Strain D
Arginine
Leucine
Methionine
Tyrosine
Phenylalanine
Water
Q3.2b. Which amino acid pathway was disrupted in each of the mutant strains?
Strain A
Strain B
Strain C
Strain D
Q3.2c. What was unusual about Strain D? How can you explain this result?
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