A hypothesis of inherited traits in rural and suburban cats describing their similarities, differences, and if they are hardy-Weinberg equilibrium at piebald and orange
Abstract
The piebald and orange are traits inherited in rural and suburban cats. The cats are both females and males for the piebald spotting, females only for hardy-Weinberg equilibrium for orange for the whole population and both females and males for orange color in rural and suburban cats. The cats are noted to have a rate at which the traits inherited are relatively the same and the difference is not very high[i]. From this, it is noted that there are no major differences as a result of the area that the cats live in.
Introduction
These traits are phenotypic in the cats in that they are visible and therefore a deeper genetic analysis has to be carried out to determine if the cats carry these alleles in their genes and determine whether they are homozygous or heterozygous. For the piebald spotting, we can see that it occurs in high numbers of cats in both rural and urban areas because the spotting is caused by only one allele. This means that if the cat inherits one allele from one or both parents it will be spotted because only one allele is required for the spotting trait to appear but the cats that have only one allele have less spots that those that have two alleles inherited from both parents.
The paper will aim to analyze the traits of the suburban cats and rural cats. Both of the female and male sex. The analysis will give due consideration to the colour and sex (as mention) among other traits. The analysis focus on the similarity and differences between the suburban and rural cats. The paper will then discuss the results acquired and come up with a valid conclusion about the two cats. The will acknowledge the role played by various group members for the data acquired.
Results
- Rural Cats
Whether cats are in Hardy-weinberg equilibrium for piebald spotting.
Fr(s) p = {(2 * SS) + Ss)}/ 2 * total
= {(2 * 12) + 69)}/ 2 * 131
= 93/262
= 0.355
Fr(s) p = {(2 * ss) + Ss/ 2 * total
= {(2 * 50) + 69)}/ 2 * 131
= 169/262
= 0.645
Freq (SS) = P2 = 0.126
Freq (Ss) = 2 pq = 0.458
Freq (ss) = q2 = 0.416
Piebald SS Ss ss total Observed 12 69 50 131 Expected 16.506 59.998 54.499 131 Chi square 1.23 1.35 0.37 2.95
P2 * 131 = 16.506
2pq * 131 = 59.998
q2 * 131 = 54.499
p < 0.05
- Rural Cats
Calculations on whether the populations are in the Hardy-weinberg equilibrium for orange (females only).
Fr(xO) p = {(2 * OO) + Oo}/ 64 * 2
= (2 * 4) + 21 / 64 * 2
= 29/ 128
= 0.226
Fr (xo) q = (2 * 39) + 21/ 64 * 2
= 99/ 128
= 0.773
Freq xOxO p2 = 0.051
Freq xOxo 2pq = 0.349
Freq xoxo q2 = 0.597
Orange xOxO XoXO xoxo total Observed 4 21 39 64 Expected 3.264 22.336 38.208 64 Chi squar 0.166 0.08 0.016 0.262
P2 * 64 = 3.264
2pq * 64 = 22.336
q2 * 64 = 38.208
- Rural Cats
Calculations on whether cat are in Hardy-weinberg equilibrium for orange colour (males and females)
Fr (xO) p = {12 + (2 * 4) + 21}/ 63 + (2 * 64)
= 41/191
=0.215
Fr (xo) q = {51 + (2 * 39) + 21}/ 63 + (2 * 64)
=150/191
=0.785
Freq (xOxO) p2 = 0.046
Freq (xOxo) 2pq = 0.337
Freq (xoxo) q2 = 0.616
Male/female Orange tortis Non total Observed 16 21 90 127 Expected 5.842 42.799 78.232 127 Chisquar 17.663 11.15 1.77 30.583
P2 * 127 = 5.842
2pq * 127 = 42.799
q2 * 127 = 78.232
- Suburban Cats
Whether cats are in Hardy-weinberg equilibrium for piebald spotting.
Fr (S) p = {(2 * SS) + Ss}/ 2 * total
= (2 * 71) + 163/ 2 * 346
= 305/ 692
= 0.44
Fr (s) q = {(2 * ss) + Ss}/ 2 * total
= (2 * 112) + 163 / 2 * 346
= 387/692
=0.56
Freq (SS) p2 = 0.194
Freq (Ss) 2pq = 0.493
Freq (ss) q2 = 0.314
piebald SS Ss ss total Observed 71 21 90 127 Expected 66.986 170.508 108.506 346 Chisquar 0.24 0.33 0.12 0.69
P2 * 346 = 66.986
2pq * 346 = 107.508
q2 * 346 = 108.506
- Suburban cats
Calculations on whether the popualtions are in Hardy-weinberg equilibrium for orange (females only).
Fr (xO) p = (2 * 13) + 39/ 156 * 2
= 65/ 312
= 0.208
Fr (xo) q = (2 * 104) + 39 / 156 * 2
= 247/312
=0.792
Freq xOxO p2 = 0.043
Freq xOxo 2pq = 0.329
Freq xoxo q2 = 0.627
Orange xOxO xOxo xoxo total Observed 13 39 104 156 Expected 6.749 51.397 97.853 156 Chisquar 2.99 2.99 0.38 9.16
P2 * 156 = 6.749
2pq * 156 = 51.397
q2 * 156 = 97.853
- Suburban cats
Calculations whether cats are in Hardy-weinberg equilibrium for orange colour (males and females).
Fr (xO) p = {(39 + (2 * 13) + 39) / 175 + (2 *156)
= 104 / 487
= 0.214
Fr (xo) q = {(136 + (2 * 104) + 39) / 175 + (2 *156)
= 383 / 487
= 0.786
Freq (xOxO) p2 = 0.046
Freq (xOxo) 2pq = 0.336
Freq (xoxo) q2 = 0.617
Male/female Orange tortis non total Observed 52 39 240 331 Expected 15.158 111.351 204.491 331 Chisquar 89.546 47.011 6.166 142.723
P2 * 331 = 15.158
2pq * 331 = 111.351
q2 * 331 = 204.491
Discussion
As seen in the investigation, there is a very big difference in the number of orange traits in the rural and urban cats. The numbers are higher in the suburban areas than in the rural. These cats are in hardy-Weinberg equilibrium for orange at relatively low numbers[ii]. The orange gene has two alleles: a non-orange and orange. The non-orange, o, allele is recessive and it allows full expression of the black locus. The dominant orange allele, O, however, influences expression of the black and agouti loci because it substituted the production of phaeomelanin for eumelanin. It masks the effect of the black gene by converting a black or brown coat to orange. This ability of one gene to mask the effect of another gene is called epistasis[iii]. Further, all orange cats are tabbies because they contain the orange allele is epistatic to the non agouti (solid coat), phenotype that is normally produced by aa at the agouti locus. This masking occurs because the orange band of phaeomelanin granules in the hair shaft is not visible against the yellowish background of the hair without melanin granules.
The orange gene is usually carried on the X chromosome which makes it sex-linked. In the male cats, this locus can normally produce only two phenotypes, that is, black or orange. In females, it can produce three phenotypes: black, orange, or tortoiseshell. This is because males are usually XY (heterogametic) and therefore have only one X chromosome. This means that if a male carries the orange allele at all, he will be orange. Females are XX meaning that they have homogametic and if both carry the orange allele, then the cat will be orange but if she is heterozygous, her coat will be have black and orange patches (tortoiseshell)[iv]. In the investigation, there are high numbers of females carrying the allele because of the big number of the tortoiseshell coat for the suburban than the rural cats and the chi square is high for the orange and tortoiseshell coat.
The piebald spotting is very high among the suburban cats and research generally shows that this is very common. The larger areas of white spotting are usually SS and the cat can even be completely white. The spotting varies tremendously, but spotting is said to follow a regular progression. Cats with the least spotting have small spots on the breast and belly[v]. Spotting increases, to cover the entire belly, the neck, chin and front feet. Lastly, cats with most spotting have spots up the sides, over the back and onto the head. With the ss phenotype, there is no spotting and the orange and black are intermingled, mostly without large patches of either orange or black. With either the Ss or SS genotypes, there are white spots and the orange and black occur as distinct patches.
The spotting provided an interesting genetic trait distribution between the suburban and rural population. The p(S) frequencies were higher in the suburban areas. The differences in the inherited genes are attributed to both genetics and environment[vi]. Therefore, the p(S) distributions can be linked to habitat variations in the rural and suburban areas and this is determined through selective hypothesis.
Acknowledgement
I would like to send my appreciations to ——————-, the rest of the class who took part and offered the necessary data set required in the experiment.
Cited References
O’Brien, E. The Genetic Profile of Cat Coats in Northern New Jersey. 2003.
Kerr, S. J. Mutant allele frequencies in Swiss rural cat populations. Journal of Heredity, 1983; 74(5): 349-352.
Ruiz-Garcia, M., & Alvarez, D. Genetic microstructure in two spanish cat populations. I: genic diversity, gene flow and selection. Genes & genetic systems, 2000; 75(5): 269-280.
Vinogradov, A. E. Locally associated alleles of cat coat genes. Journal of Heredity, 2000; 85(2): 86-91.
Christensen, A. C. Cats as an aid to teaching genetics. Genetics, 2000; 155(3): 999-1004.
Jones, E., & Horton, B. J. Gene frequencies and body weights of feral cats, Felis catus (L.), from five Australian localities and from Macquarie Island. Australian journal of zoology, 1984. 32(2): 231-237.
[i] O’Brien, E. The Genetic Profile of Cat Coats in Northern New Jersey. 2003.
[ii] Kerr, S. J. Mutant allele frequencies in Swiss rural cat populations. Journal of Heredity, 1983; 74(5): 349-352.
[iii] Jones, E., & Horton, B. J. Gene frequencies and body weights of feral cats, Felis catus (L.), from five Australian localities and from Macquarie Island. Australian journal of zoology, 1984. 32(2): 231-237.
[iv] Christensen, A. C. Cats as an aid to teaching genetics. Genetics, 2000; 155(3): 999-1004.
[v] Ruiz-Garcia, M., & Alvarez, D. Genetic microstructure in two spanish cat populations. I: genic diversity, gene flow and selection. Genes & genetic systems, 2000; 75(5): 269-280.
[vi] Vinogradov, A. E. Locally associated alleles of cat coat genes. Journal of Heredity, 2000; 85(2): 86-91.
APPENDIX
Appendix 1: Rural Cats
Whether cats are in Hardy-weinberg equilibrium for piebald spotting.
Fr(s) p = {(2 * SS) + Ss)}/ 2 * total
= {(2 * 12) + 69)}/ 2 * 131
= 93/262
= 0.355
Fr(s) p = {(2 * ss) + Ss/ 2 * total
= {(2 * 50) + 69)}/ 2 * 131
= 169/262
= 0.645
Freq (SS) = P2 = 0.126
Freq (Ss) = 2 pq = 0.458
Freq (ss) = q2 = 0.416
Piebald SS Ss ss total Observed 12 69 50 131 Expected 16.506 59.998 54.499 131 Chi square 1.23 1.35 0.37 2.95
P2 * 131 = 16.506
2pq * 131 = 59.998
q2 * 131 = 54.499
p < 0.05
Appendix 2: Rural Cats
Calculations on whether the populations are in the Hardy-weinberg equilibrium for orange (females only).
Fr(xO) p = {(2 * OO) + Oo}/ 64 * 2
= (2 * 4) + 21 / 64 * 2
= 29/ 128
= 0.226
Fr (xo) q = (2 * 39) + 21/ 64 * 2
= 99/ 128
= 0.773
Freq xOxO p2 = 0.051
Freq xOxo 2pq = 0.349
Freq xoxo q2 = 0.597
Orange xOxO XoXO xoxo total Observed 4 21 39 64 Expected 3.264 22.336 38.208 64 Chi squar 0.166 0.08 0.016 0.262
P2 * 64 = 3.264
2pq * 64 = 22.336
q2 * 64 = 38.208
Appendix 3: Rural Cats
Calculations on whether cat are in Hardy-weinberg equilibrium for orange colour (males and females)
Fr (xO) p = {12 + (2 * 4) + 21}/ 63 + (2 * 64)
= 41/191
=0.215
Fr (xo) q = {51 + (2 * 39) + 21}/ 63 + (2 * 64)
=150/191
=0.785
Freq (xOxO) p2 = 0.046
Freq (xOxo) 2pq = 0.337
Freq (xoxo) q2 = 0.616
Male/female Orange tortis Non total Observed 16 21 90 127 Expected 5.842 42.799 78.232 127 Chisquar 17.663 11.15 1.77 30.583
P2 * 127 = 5.842
2pq * 127 = 42.799
q2 * 127 = 78.232
Appendix 4: Suburban Cats
Whether cats are in Hardy-weinberg equilibrium for piebald spotting.
Fr (S) p = {(2 * SS) + Ss}/ 2 * total
= (2 * 71) + 163/ 2 * 346
= 305/ 692
= 0.44
Fr (s) q = {(2 * ss) + Ss}/ 2 * total
= (2 * 112) + 163 / 2 * 346
= 387/692
=0.56
Freq (SS) p2 = 0.194
Freq (Ss) 2pq = 0.493
Freq (ss) q2 = 0.314
Piebald SS Ss ss total Observed 71 21 90 127 Expected 66.986 170.508 108.506 346 Chisquar 0.24 0.33 0.12 0.69
P2 * 346 = 66.986
2pq * 346 = 107.508
q2 * 346 = 108.506
Appendix 5: Suburban cats
Calculations on whether the popualtions are in Hardy-weinberg equilibrium for orange (females only).
Fr (xO) p = (2 * 13) + 39/ 156 * 2
= 65/ 312
= 0.208
Fr (xo) q = (2 * 104) + 39 / 156 * 2
= 247/312
=0.792
Freq xOxO p2 = 0.043
Freq xOxo 2pq = 0.329
Freq xoxo q2 = 0.627
Orange xOxO xOxo xoxo total Observed 13 39 104 156 Expected 6.749 51.397 97.853 156 Chisquar 2.99 2.99 0.38 9.16
P2 * 156 = 6.749
2pq * 156 = 51.397
q2 * 156 = 97.853
Appendix 6: Suburban cats
Calculations whether cats are in Hardy-weinberg equilibrium for orange colour (males and females).
Fr (xO) p = {(39 + (2 * 13) + 39) / 175 + (2 *156)
= 104 / 487
= 0.214
Fr (xo) q = {(136 + (2 * 104) + 39) / 175 + (2 *156)
= 383 / 487
= 0.786
Freq (xOxO) p2 = 0.046
Freq (xOxo) 2pq = 0.336
Freq (xoxo) q2 = 0.617
Male/female Orange tortis non total Observed 52 39 240 331 Expected 15.158 111.351 204.491 331 Chisquar 89.546 47.011 6.166 142.723
P2 * 331 = 15.158
2pq * 331 = 111.351
q2 * 331 = 204.491