Uitwerking alle case studies van principles of human genetics
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Course
Principles of human genetics (GEN21803)
Institution
Wageningen University (WUR)
Dit document bevat alle case studies (week 1-4) van het vak principles of human genetics. Met de case studies is een gemiddeld cijfer van 8.46 behaald.
Case Study 1: CATLAB and the Genetics of Coat Colour in Cats
This case refers to Classical Mendelian Genetics in chapters 1 and 2 as covered in lecture, as well as
chapters 3, 4, and 6, in the textbook by Griffiths et al. 2015.
Report
This file, with your answers, is your report, which you can upload to Brightspace.
Make sure you indicate the names of all group members!
Goal: To understand and experience the processes of scientific reasoning. In the first exercise, you
will get familiar with the CatLab program and analyse the genetic basis of five traits in house cats
using monohybrid crosses. In the second part, you will perform dihybrid crosses to test for gene
interaction. In the third part you reflect on what you have learned.
Method: The computer simulation programme CatLab allows you to act as a virtual cat breeder. You
can simulate crosses in cats and practice scientific reasoning for problem solving. This is also termed
the empirical cycle: Problem definition, experiment design, data production, analysis and
interpretation, hypothesis definition and testing.
About the χ2 (chi square) test (details found in textbook, look in index for where to find the
table):
1. Expected Mendelian ratios for monohybrid crosses are 1:1, 1:2:1, or 3:1. Typically, the
observations deviate a little bit from the expectations by chance.
2. The χ2 test, available in the CatLab program, is a statistical test to see how well the
observed numbers fit the expectations. The outcome is called the chi-square value. The
better the fit, the lower the chi-square value. It is 0 if the observed and expected numbers
are exactly the same. We consider the data a bad fit if they exceed the critical value,
meaning that íf the hypothesis would be correct, the chance that the data would fit this
poorly or even worse, would be less than 5%.
3. The critical value of the chi-square test depends on the number of degrees of freedom
(df), which is the number of numbers that can be filled in independently. You can look it
up in the book or online. For df=1 and probability level=0.05 the critical value is ~3.84.
4. The χ2 test in the CatLab program expects you to fill in the expected ratio and the
observed numbers, although the calculation is done with numbers only! The program uses
the expected ratio to calculate the expected numbers.
5. The χ2 test is not reliable when the numbers in any class are less than 5.
6. Excel also contains a χ2 test formula, but it gives the P value (the chance) as an outcome,
which is higher when the data fit better.
7. You can reject your null hypothesis when the χ 2 value is higher than the critical value, or
when the P value is lower than 0.05. It means the same, as the P value can be calculated
from the χ2 value and vice versa.
,Exercise I – Monohybrid crosses
First take a few minutes to get familiar with the CatLab program. Realise that this is a simulation
program: You choose a cat with a specific phenotype, not a specific genotype. Furthermore, when
you make crosses, realise that all traits may affect the phenotype at the same time. This may be
confusing, so try to focus on one single contrast at a time. In exercise I we will study:
1. Dilute (modifier of pigmentation):
Dense (full colour) or diluted (gray or cream-coloured)
2. Manx:
No tail (manx) or normal tail
3. Pigmentation pattern:
Mackerel (stripes/dots) or Blotched (blotches/bullseye) or non-agouti (no pattern)
4. Orange and black:
Orange or black or both (calico or tortoise shell),
5. Piebald spotting (white spots on the body):
No/little white or some white (<50%) or lots of white (>50%)
1
,Problem I-1
The dilute gene affects the amount of pigment produced by melanocytes, whether this is the
black/brown pigment (eumelanin) or the yellow/red pigment (pheomelanin). Grey is the dilute
version of black; cream-coloured is the dilute version of orange.
First do some observations (see above in the empirical cycle) and state a hypothesis (e.g. “dilute is
a dominant monogenic trait”), then think about the data you need to test this hypothesis. Then
choose the desired parents and simulate the breeding data in CatLab. If the data do not fit, you
have to try another hypothesis. You may have to use a trial-and-error approach before you have
the most likely explanation.
Hint:
When you think you have solved the problem, make crosses between cats of known genotype,
and obtain data to test your conclusions both qualitatively (what progeny types are expected) and
quantitatively (what progeny ratio is expected).
a. Your final hypothesis with presumed genotypes and phenotypes of parents and offspring (max.
1 sentence):
We expect grey (diluted gen) to be recessive, and black to be dominant.
b. Which crosses were performed to test your hypothesis, and what progeny was obtained (also
indicate the number)?
1. grey female x grey male -> grey cats (aa x aa -> aa)
2. black female x black male -> grey and black cats (Aa x Aa -> AA or Aa or Aa or aa)
c. Statistical support for your hypothesis (χ2 test)? Note that the program works with expected
ratio rather than numbers.
χ2 = 0.40. this is lower than the critical value (3.84), so it can be concluded that the observed data
fits the assigned ratio of 3:1 well.
d. Conclusion: How is the dilute trait inherited (max. 1 sentence)?
The dilute trait (grey) is recessive (aa) inherited.
2
, Problem I-2
Manx cats are tailless (Manx is a cat breed from the Isle of Man, UK) and have a defect in spine
formation.
a. Try to create a true-breeding Manx cat in CatLab in maximum 10 crosses. Did you succeed?
Explain your approach. (max. 1 sentence)
By crossing two Manx cats, the offspring consists of both Manx cats and non-Manx cats. We would
not succeed in creating a true-breeding of Manx cats, probably because the Manx allele is
dominant heterozygous (Aa).
b. In reality cat breeders have so far failed to produce true-breeding Manx cats. What could be the
reason?
The Manx allele is dominant heterozygous (Aa), because both Manx cats and non-Manx cats are
created by two Manx cats. When crossing a heterozygous with a heterozygous (Aa x Aa), it is
possible to create a homozygous (AA) cat. However, a dominant homozygous Manx cat will not
survive, so a true-breeding manx cat is impossible.
3
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