Calculating Gene Frequencies
Studying Gene Frequencies in a Population of Domestic Cats
Linda K. Ellis Department of Biology Monmouth University Edison Hall, 400 Cedar Avenue, W. Long Branch, NJ 07764 USA [email protected]
Description: A laboratory activity to calculate allele and genotype frequencies in a population of domestic cats. Keywords: Hardy-Weinberg, gene frequencies, cat coat colors
© 2011 Linda K. Ellis
Introduction
The domestic cat makes an ideal subject for gathering information regarding more than a dozen genotypes, simply by observing coat color and patterning. From the data, the frequency of the alleles at each of the various loci can be calculated and from this information it is possible to determine if a population is in Hardy-Weinberg equilibrium. Using a population of familiar animals as opposed to flipping coins or some of the other "paper labs" that are published on this topic brings the concepts to life and is interesting, motivational and fun. In this exercise, students calculate the frequencies of the alleles of the various genes that act to determine the coat color and patterning in a population of random- bred cats.
Student Outline
One of the most important decisions a scientist makes is the choice of an organism used to study a particular phenomenon. A classic example of a good choice is Mendel’s garden peas, which he used to study inheritance. The plant he chose showed clearly discernable traits; ones that were governed by what we now call “simple Mendelian genetics”: two alleles, one dominant, and the other recessive.
For the same reasons that Mendel’s peas were an ideal subject, the domestic cat is an excellent model to demonstrate how the frequency of alleles for several traits can be calculated. Just by getting a quick look at a cat, one can gather information regarding the genes that have an effect on the coat color and/or patterning of the animal. While there are probably not as many cats in a neighborhood as there are peas in a garden, cats are easy to find. Because it is only necessary to observe them, they do not require care like laboratory animals.
When you begin a study of genetic variability, it is important to establish what is known as the “wild type”. This is the phenotype that is assumed to be the norm for the species. For the domestic cat, Felis cattus, it is the lined or mackerel tabby. This animal has a coat of alternating black and tan stripes. A number of genes at different loci contribute to this phenotype (Wright & Walters, 1980; Elllis, 1993, Gould, 1996, and Trass). They are briefly described below and summarized in Table 1.
The first locus is the Albino series, which is designated as the C locus and has multiple alleles. The protein product of the wild type allele at the C locus is the enzyme tyrosinase, required for the production of melanin. In the complete absence of tyrosinase, the result is albinism. The expression of some mutant alleles is temperature-dependent, resulting in phenotypes associated with specific breeds of cats, such as Siamese and Burmese, but not usually seen in the general population.
Assuming the animal carries the dominant gene for normal tyrosinase activity (C/_), there are two possible pigments that can form in the melanocyte. They are controlled by two additional loci. Eumelanin appears as black (B) or chocolate brown (b). The recessive allele at this locus is seen in rare breeds, where it has been selected for, and its frequency in a domestic or feral population of domestic short hairs is extremely low or perhaps nonexistent.
Pheomelanin is red or orange and is determined by a locus that is carried on the X chromosome known as X-linked or Mutant Orange. O+ results in the expression of the black pigment, while O results in the orange color. Since males carry only one X chromosome, they are either OY (orange) or O+Y (black). Females can be OO, O+O+, or O+O.
Once the choice of pigment has been made, it is modified by a gene at another locus known as the D (dilute) locus. In the wild type, D/_, the pigment is dark (orange or black) while the recessive mutant allele results in buff or gray (d/d).
Calculating Gene Frequencies
Whether the pigmentation along the hair shaft will be uniform or agouti is the result of another pair of alleles at the A locus. Animals that are a/a have a solid colored coat. A/_ cats are striped.
The tabby or T locus exerts its effect on the pattern of stripes in animals that are A/_. The stripes may be arranged in a strikingly symmetrical pattern caused by the dominant allele T. This results in the familiar mackerel or striped tabby. The recessive allele, tb, causes the stripes to appear in whorls or to run together and cause blotches. This (tb/ tb) results in the blotched tabby. Since the tabby markings are only expressed in animals with an agouti coat, the T locus is hypostatic to the non-agouti, a/a phenotype. That is, it is not expressed in a cat with a non-agouti or solid color coat. An exception to this statement applies to the X-linked orange phenotype. Some degree of stripes is always seen in ginger cats. A third allele at this locus, ta, results in spots in place of stripes and is seen in the rare Abyssinian breed.
White patches are the result of areas of the skin without melanocytes. This is due to a gene at the S locus that directs the migration of melanocytes from their site of origin at the neural crest. The homozygous s/s genotype results in a solid colored animal without white patches, while S/_ will result in white patches. These may be limited to the feet, neck and belly, areas that are distal to the neural crest. Some experts state that an animal with white patches on less than half of the body is heterozygous, S/s. Animals with coats that are more than half white are homozygous, S/S.
Another gene (in addition to the C locus) that can produce an all white animal is found at the W locus. The normal function of the gene is to direct the development of neural crest cells. The dominant gene W results in a total, or nearly total, lack of melanocytes. Because neural crest cells differentiate into many different cells, extreme expressivity may result in pleiotropy. Deafness and lack of retinal pigment may also be seen. These animals may have green or blue eyes and often exhibit heterochromia, one blue and one green eye. An animal with green eyes is not an albino, but it is not possible to distinguish an albino from a blue-eyed dominant white, although, if the animal is deaf, the W gene is almost certainly responsible. Variable expressivity at the W locus often results in an animal with pigmentation only on the head and tail.
In addition to coat color and patterning, length of hair, extra toes (polydactyly) and short or absent tails are also phenotypes that can be easily observed. A recessive allele at the L locus is responsible for long hair, but the action of enhancers may result in intermediate phenotypes. The short, stubby or absent tail seen in the Manx cat is an example of a dominant, recessive lethal. A single copy (M/_) results in an abnormal tail length, while the homozygous (M/M) genotype is lethal in utero. Cats with normal tail length are homozygous for the wild-type allele, (m/m). Procedure
Each investigator should gather information on six cats. To document your data, take a photo of each animal and use this to identify the phenotype at each of the loci listed in Table 2. If possible, record the gender of the animal. Based on the phenotypes, record the genotype for each locus in Table 2. When the animal exhibits a dominant phenotype, it may not be possible to determine the other allele. For example, an agouti (striped) animal may be A/A or A/a. For this reason, it should be recorded as A/_.
Results
Analysis of individual data
The Hardy-Weinberg Law states that gene frequencies will remain constant providing the following conditions are met: the population is infinitely large, there are no mutations, there is no immigration or emigration, breeding is random and there are no selection pressures. A population not in Hardy-Weinberg equilibrium is evolving. This is determined by calculating the frequencies in the gene pool of the alleles at a given locus.
Allele frequency can be calculated as follows: Let p equal the frequency of the dominant allele, and q the frequency of the recessive allele. An individual showing a recessive phenotype is homozygous for the recessive allele. Therefore, the fraction of the population showing the recessive phenotype is q2. From this, one can determine q and, since p + q = 1, both p and 2pq can easily be calculated.
At each locus, determine the number of animals exhibiting the recessive phenotype. Show the frequency (q2) as a fraction, (number showing recessive phenotype/number of animals showing the trait) in Table 2. This will make it easier to collate your data with the rest of the class’s data. The number of animals exhibiting a trait may be different for some of the loci. A solid color animal will not express the alleles at the T locus and you will not be able to tell what alleles are carried at several of the loci in a white animal. Note that Hardy-Weinberg calculations cannot be applied to X-linked genes; show this as number of O/ (O+ plus O). Calculating Gene Frequencies
Table 1. Genetics of coat color in the cat. LOCUS INHERITANCE FUNCTION ALLELES C - Albino C is dom. to all… Codes for tyrosinase, C = full pigmentation cs and cb show required for pigment ca = blue-eyed albino incomplete production cs = Siamese dominance and ca cb = Burmese is dom. to c c = pink-eyed albino B - Black Dominant Codes for the pigment B = eumelanin, black melanin b = brown bl = cinnamon O - Mutant X-Linked, Converts eumelanin to O+O+ or O+Y = black Orange expressed when orange pheomelanin OO or OY = orange present O+O = orange & black D - Dilute Recessive “Dilutes” black and D = full colour orange to gray and buff d = diluted colour A - Agouti Dominant Results in agouti pattern A = stripes on hair shaft a = solid coat colour T - Tabby Dominant Affects pattern of stripes T = Mackerel tabby, symmetrical stripes tb = Classic or blotched tabby ta = spots on body S - Pie-balding Dominant Results in white patches S = white patches s = solid colour W - Dominant Dominant Results in all white W = solid white White animal or pigmentation w = full pigmentation only on head & tail I - Silver Dominant No pigment at base of hair I = “silver” appearance shaft i = full pigmentation L - Length Recessive Regulates length of hair L = short hair l = long hair M - Manx Dominant, Regulates length of tail Mm = short tail homozygous lethal mm = normal tail length MM = lethal in utero
Analysis of class data
In Table 3, combine the individual data and record the total number of homozygous recessive animals and the total number of animals expressing the trait for each locus in Table 3. Use these data to calculate the frequency of the homozygous recessive (q2) individuals at each locus. From this value, complete the table by calculating q, p, and 2pq.
Table 2. Individual genotype data. CAT C B O D A T S W L 1
2
3
4
5
6
Frequency * of homozygous recessive animals * Number of X-linked orange /total number of alleles (females have 2, males only 1)
Calculating Gene Frequencies
Discussion Questions
1. Based on the class data for gene frequencies, how many cats in a sample of 500 from this population would you expect to be long-haired? 2. How does the frequency of homozygous recessive animals at each locus in your data compare to the frequency found when the class data was used. 3. How might a population pick up the mutant cs allele at the C locus that is seen in the Siamese cat? 4. Why might you expect the dominant allele at the W locus to a. Increase? b. Decrease? 5. Explain how the “founders’ effect” could explain unusually high or low frequencies of certain alleles within a population. 6. The gene responsible for the short tail seen in the Manx cat is dominant, but lethal when homozygous. How can you account for the continuing presence of the allele in a population? 7. Explain how selection pressures contribute to changes in gene frequencies and therefore to evolution.
Table 3. Analysis of class data. For the X-linked O locus, count the number of O and O+ alleles and calculate frequency based on the number of alleles, not the number of animals. Omit the remaining calculations. C B O* D A T** S W** L
Total number of homozygous recessive animals at each locus
Total number of animals surveyed
Frequency of homozygous recessive animals (q2)
Frequency of recessive allele (q)
Frequency of dominant allele (p)
Calculated frequency of heterozygotes (2pq)
*Calculate p and q using the number of each allele/total number of alleles. Omit the rest of the calculations. **It is not possible to learn the genotype at this locus if the animal is non-agouti. The same is true for an all white animal at several other loci. Calculate frequency based on the number of animals expressing each trait.
Calculating Gene Frequencies
Literature Cited
Coat Colors FAQ: Cat Color Genetics. Cat Fanciers Chat. Web. 17 May 2010.
About the Author
Although Linda Ellis currently teaches at the university level, this activity was developed when she was teaching advanced placement biology in high school and she had the students collect data from neighborhood cats. She has used a modified version using photographs in my university-level genetics course as well. In addition, the background information from this paper in the form of a PowerPoint presentation has been presented to an audience at Acadia University in Nova Scotia and, with some more details on the genetics, to the Student Veterinary Association at the School of Veterinary Medicine at St. George's University in Grenada, West Indies.