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Genetics 371B, Autumn 2000 Answer key, Problem set 2 |
| 1. | (a) | There are 1000 progeny total, so if we are dealing with a simple Mendelian trait, we expect 3/4 of the progeny to show the dominant trait -- i.e., 750 non-smashing, 250 smashing.
These being the expected progeny numbers, we can calculate the chi-square values:
There are two independent categories, so df = 1. For chi-square = 8.112 and df = 1,
Since the P value is less than 0.05, we can reject the null hypothesis (that there is no significant difference between the expected and observed values); the data do not support the initial hypothesis of a simple Mendelian trait. (The chi-squared value we have obtained here is outside the range of values covered in the chi-squares table in your lecture notes. You can set up a chi-squares test in a spreadsheet, e.g., in Microsoft Excel, or use an online chi-squares calculator, such as the one here. Using these other means, the P value is found to be 0.0044.) |
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| (b) | The obvious explanation is that our expectation was based on a false premise -- if the trait does is follow simple Mendelian inheritance, then we have no basis to form predictions of what progeny numbers to get. But even if our initial hypothesis was correct, there are several reasons why our expected numbers mightnot match the observed --
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| 2. | (a) | The genotype of the male parents indicates that the genes are not X-linked. If they were X-linked, we'd expect to see only one copy of the gene in males, and the genotype (e.g., sm/sm) indicates that there are two copies (i.e., if the traits were X-linked, we'd expect to see a genotype such as XsmY for males -- using sm as an example, the same holds for the other two genes also). It is formally possible that one of these genes is in the pseudautosomal region, but in fact, the pseudoautosomal region in Drosophila contains few functional genes. | |||||||||||||||||||||||||||
| (b) | Looking at the progeny numbers, we can conclude that for the first cross, the genes are linked; wildtype and sm bw must be the parental types (because they are the most abundant). The heterozygous parent must be + +/sm bw and the recombinant progeny types are + bw and sm +. The recombinant types add up to 130 of the 1000 progeny, giving a map distance of 13 cM:
The second cross also shows evidence of linkage. Here the progeny numbers indicate that the genes in the heterozygous parent are in the trans configuration, + sp/sm +. The recombinant types make up 160 of the 1000 progeny, so the map distance here is 16 cM:
Since we do not know the distance between sp and bw, we are left with two possible maps. In one, bw lies between sm and sp; in the other map, sm lies between sp and bw: Map 1
Map 2
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| (c) | The ambiguity can be resolved by mapping the bw - sp interval. One parent in the cross would be +/bw +/sp and the other would be fully homozygous recessive. If Map #1 is correct, the bw-sp interval should be 3 cM, while if Map 2 is correct, the interval should be 29 cM. | ||||||||||||||||||||||||||||
| (d) | since the most abundant progeny (the parental types) are wildtype and fully recessive, the dominant alleles must be in cis in the heterozygous parent. The double crossover (DCO) types are + sm sp and bw + +. The middle gene must be bw:
Therefore, Map 1 is correct; the gene order is sm - bw - sp |
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| (e) | The expected progeny phenotypes and numbers are:
How these numbers were obtained-- Based on the map distances, we expect:
However, with interference being a factor, the actual number of DCO progeny will be lower. The actual DCO number can be calculated from the equation:
The coefficient of coincidence (COC) = (1 - interference) = 0.67. Since COC = actual DCO/expected DCO, we can solve for observed DCO --
The two DCO classes together should add up to 26; each one should be 13. The sm-bw map distance is 13 cM, so recombinants in this interval (single crossover progeny and DCO) account for (0.13 x 10000) = 1300 progeny. Since this number includes the DCO class, the number of SCO in the sm-bw interval = 1300 - 26 = 1274; each member of this class will have 1274/2 = 637 flies. Likewise, there are expected to be (0.03 x 10000) = 300 recombinants in the bw-sp interval; again, this number includes DCO progeny, so excluding the DCO, we get 274 progeny resulting from SCO in the bw-sp interval. The remainder make up the two parental types. |
| 3. | (a) |
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| (b) | The probe to use is #2. Probe #1 would hybridize to a 2 kb fragment for both alleles A and C, so it wouldn't work; Probe #3 would hybridize to a 3 kb fragment in alleles A and B. With Probe #2, we should get hybridization to a 5 kb fragment from Allele A, a 7 kb fragment from Allele B, and an 8 kb fragment from allele C, allowing us to distinguish between these three alleles. |
| 4. | (a) | If the allele for the dominant phenotype is represented by G and the recessive phenotype by g, the mother (I-1) has genotype Gg {12,16} -- where the numbers in brackets represent alleles of the polymorphic site. There are four possible gamete genotypes that this individual could produce:
To tell if there is linkage between these to loci or not, we need to look at the progeny and ask if the four progeny types are represented in equal proportions or not -- i.e., what is the frequency with which I-1 has transmitted the four gamete genotypes. If all four are represented equally, then the loci do not show evidence of linkage; if two of the genotypes are over-represented, then the loci may be linked. Looking at the progeny genotypes, we can tally the number of progeny inheriting each of these gamete types.
Two of the gamete types (G 12 and g 16) outnumber the other two types, which would seem to indicate that there may be linkage between the two loci; the putative recombinant progeny (inheriting G 16 and g 12 gametes) would be II -3, II-9, and II-10. NOTE, however, that if this were a real experiment, we wouldn't be happy with the small sample size; we really wouldn't conclude that there was linkage. In fact, the sample size is too small to do even a chi-square analysis -- a rule of thumb that many people follow is that in order for chi-square analysis to be meaningful, each category must have at least 5 members. (As for your answer to this question, all we are looking for is whether you know how to go about looking for evidence of linkage.) |
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| (b) | Keeping in mind the caution above (that the sample size is too small to draw real conclusions), if we were to project from the available numbers, we'd say that the map distance was 30 cM (3 of 10 progeny are presumed to be recombinant).
An important point to note: This pedigree has been set up to resemble a Drosophila cross, where we can look at meiosis in just one parent (I-1) and ignore the other parent (because recombinant gametes and non-recombinant gametes from I-2 are indistinguishable). So all we are counting are recombinant gametes from one parent. In many real pedigrees, we could get recombinant gametes from either parent. If we counted recombinant gametes from BOTH parents, we'd have to divide the # of recombinant progeny by TWICE the total number of progeny to get the recombination frequency, because each child has two ways of being recombinant -- recombination in the mom, or recombination in the dad. |
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| (c) | As discussed above, if the loci are unlinked the probability that I-1 will transmit any one genotype is 1/4. The father (I-2) can transmit either g 15 or g 20; the probability of each is 1/2 assuming independent assortment. Therefore, the probability that each child will inherit any given genotype = 1/8. (If you include the probability of the child's sex, the overall probability drops to 1/16.)
Therefore, the probability of the offspring in this pedigree is (1/8)10 (not counting the sex) or (1/16)10 (counting the sex). These values are pretty small; what they are saying is that there are multitudes of possible outcomes of progeny from parents with the given genotypes (where the loci are assorting independently); the probability of this particular outcome is pretty low compared to all possible outcomes. |