Biostat 551
Lab1: gametic phase disequilibrium
Introduction: This first lab investigates gametic phase disequilibrium (also known as linkage disequilibrium). Unlike subsequent labs this quarter, this will not be a computer lab. Rather, it is a series of problems designed to get you thinking about gametic phase disequilibrium. The material covered in this sequence of problems is covered in Lynch and Walsh's text on pages 94 - 100.
Problem 1: (Similar to Example 3 in Chapter 5 in Lynch and Walsh) A biologist is interested in two genes (from the same chromosome) affecting a population of Drosophila melanogaster. The first of these genes affects eye color. Flies with genotypes RR or Rr have red eyes and flies with the rr genotype have white eyes. The second gene controls the pattern of veins in the wings. Flies with genotypes W W or W w have normal wings and flies with genotype ww have abnormal wings. The biologist decided to mate male flies from the population of interest with female lab-flies of genotype RRW W . Since male fruit flies do not experience recombination, the paternal haplotypes in the resulting offspring form a random sample of haplotypes from the population of interest. The following data was produced:
| haplotype | number of offspring |
| RW | 52 |
| Rw | 105 |
| rW | 27 |
| rw | 17 |
Calculate DRW (the estimate from our sample of the gametic phase disequilibriium), as well as the variance and standard error for the estimate. Do these loci appear to be in disequilibrium?
Problem 2: This problem consists of four parts. For each part you are to (i) calculate DA1B1 (the coefficient of gametic phase disequilibrium for the given sample), (ii) calculate the variance and standard error of DA1B1, and (iii) indicate whether or not you believe disequilibrium between the two loci exists (and why you hold such a belief).
Part A: The following data refer to the height and flower color of the fictional Zuzububu plant. This plant has one gene that controls height and one that controls flower color. The A1 allele gives the plant height (so A1A1 plants are tall, A2A2 plants are short, and A1A2 are intermediate) and the B1 allele gives red tint (so B1B1 plants have red flowers, B1B2 pink flowers, and B2B2 white flowers). The following chart shows the genotype data for these individuals:
| B1B1 | B1B2 | B2B2 | |
| A1A1 | 6 | 25 | 29 |
| A1A2 | 4 | 17 | 19 |
| A2A2 | 2 | 4 | 4 |
Part B: This data set refers to captive bred Peach-Faced Lovebirds (Agapornis roseicollis). The main gene affecting color in this species has two alleles, G and g. To most people, the G allele appears to be dominant, giving the bird a wild-type green color. Two copies of the g allele result in a blue-green color called "Dutch Blue". Experts can distinguish the GG birds from Gg birds (there's some difference in the coloring of the tail feathers). Another gene that affects color is the "Dark Factor" gene. This gene has two alleles, D and d. A DD bird is very dark colored, almost grey (this is called Olive if the bird is green and Slate if the bird is blue). A Dd bird is of medium darkness (called Jade if the bird is green and Cobalt if the bird is blue). A dd bird has regular bright pigmentation (called Light Green if the bird is green and Dutch Blue if the bird is blue). If you want, you can find pictures on African Lovebird Society's web page. Here are the data (I've filled in a couple of the genotypes for you):
| Color Type | Number | Genotype |
| Olive (homozygous for green) | 1 | GGDD |
| Jade (homozygous for green) | 1 | |
| Light Green (homozygous for green) | 3 | |
| Olive (heterozygous for green) | 1 | GgDD |
| Jade (heterozygous for green) | 6 | |
| LightGreen (heterozygous for green) | 21 | |
| Slate | 2 | |
| Cobalt | 16 | |
| Dutch Blue | 57 | ggdd |
For your analysis, consider G to be the A1 allele and D to be the B1 allele.
Part C: This example refers to color inheritance in Shetland Sheepdogs. The basic coat color gene in Shetland Sheepdogs has two alleles, S and s, where S codes for a sable coat and is mostly dominant to s, which codes for a black coat. An SS dog is sable, an Ss dog is shaded sable (hard to distinguish from SS), and an ss dog is black. In addition to the color gene, there is a separate gene that decides whether or not the dog will have tan-colored markings (this is invisible on a sable dog, but becomes an issue on a black dog). This gene has two alleles, T and t, where the dominant T allele gives tan markings. A black dog with tan markings is called tricolored (the third color is the white markings common to the breed) and a black dog without tan markings is called bicolored. The following chart gives the genotypes of a sample of Shetland Sheepdogs:
| SS | Ss | ss | |
| TT | 12 (sable) | 19 (shady sable) | 8 (tricolored) |
| Tt | 16 (sable) | 25 (shady sable) | 11 (tricolored) |
| tt | 6 (sable) | 9 (shady sable) | 4 (bicolored) |
For your analysis, consider T to be the A1 allele and S to be the B1 allele.
Part D: Fictional Rock lizards live in a very hot and desolate land. Their chief causes of death are predation and sunburn. They spend much of their time resting in the sand (hence the sunburn problem) and protect themselves from predators by scrambling to safety in the rocks. We will be considering two skin genes that have evolved alleles that help prevent sunburn. The first gene of interest controls color. It has two alleles, A1 and A2. Lizards with genotype A1A1 are a sandy color that is difficult for predators to see but is prone to sunburn. A1A2 lizards are somewhat darker and are easier for predators to see but somewhat resistant to sunburn. A2A2 lizards are black, which makes them easy for predators to see, but nearly completely resistant to sunburn. The second gene of interest determines the pliability of the lizards' skin. B1B1 individuals have normal, pliable skin that tends to burn. B1B2 individuals have skin that is thicker and less easily burnt, but less pliable so the lizards can't move as quickly. B2B2 lizards have leathery skin that is impossible to burn but so stiff that these lizards are very slow. The following is the genotypic data for a sample of newborn lizards:
| B1B1 | B1B2 | B2B2 | |
| A1A1 | 31 | 28 | 29 |
| A1A2 | 36 | 24 | 15 |
| A2A2 | 16 | 9 | 2 |
Problem 3: The following questions are designed to answer the question: "If a population practices random mating and two genes segregate independently, is it possible for gametic phase disequilibrium to exist between those two genes even if there is no selective pressure toward such disequilibrium?"
Suppose that there are two populations of garden peas (think Mendel). The first population is homozygous for both the yellow pod color allele, A1, and the purple flower allele, B1. The second population is homozygous for the green pod color allele, A2, and the white flower allele, B2. The pod color gene and the flower color gene are known to segregate independently. If the two populations are brought together and allowed to mate randomly, what will be the Gametic Phase Disequilibrium coefficient in the first offspring generation? If the offspring are then randomly mated, what will the disequilibrium coefficient be in their offspring (the second offspring generation)? You may assume that both initial populations are effectively infinite in size.
Would your answers change if the first offspring generation was formed by crossing the two initial populations instead of allowing them to mate randomly?