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Programmatic Way to Do a Triple Test Cross Problem |
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The triple test cross has proven to be quite difficult for many students. Every one eventually finds her/his own way to solving these types of problems. My personal approach is outlined below. This approach works for "paper and pencil" questions as well as for laboratory exercises. The problem solving approach will be illustrated with the following example.
Example:
Dr. Alicia Tofuson is constructing linkage maps of Emotion bees. In these bees happiness (h), fear (f) and anger (a) are all recessive and located on the same chromosome. (The Wild type phenotypes are not-happy (H), not-afraid (F) and not-angry (A). Dr. Tofuson crosses a homozygous afraid, happy, not-angry emotion bee to a homozygous an not-afraid, not-happy, angry emotion bee to abtain a triple heterozygote (Fha/fHA). She crosses the triple heterozygote to a complete recessive afraid, happy, angry emotion bee. The results of a triple test cross involving 50,000 emotion bees are given below.
| Phenotype | Number | Genotype | Gamete Type |
| not-afraid, not happy, not angry | 87 | FHA/fha | |
| afraid, not happy, not angry | 3058 | fHA/fha | |
| not-afraid, happy, not angry | 4241 | FhA/fha | |
| not-afraid, not happy, angry | 18229 | FHa/fha | Parental |
| afraid, happy, not angry | 16933 | fhA/fha | Parental |
| afraid, not happy, angry | 4597 | fHa/fha | |
| not-afraid, happy, angry | 2780 | Fha/fha | |
| afraid, happy, angry | 75 | fha/fha |
a. Draw the genetic map (order and distance).
Step 1. Identify the Parental Gametes: Probably the easiest thing to do. If you set up the initial F1 cross, the parental gametes are the same as the phenotypes of the P1 and P2 parents. In this example, that would be FHa (not-afraid, not happy, angry) and fhA (afraid, happy, not angry).
If you did not know the original cross, you could use the definition of statistical linkage to determine the parental gametes. "Statistical linkage is a deviation from independent assortment in the direction of an excess of parental gametes". Since the phenotypes corresponding to FHa and fhA are most frequent, they must be the parental gametes.
Step 2: Determine the double crossover gametes. Probably the second easiest thing to do. Double crossovers are the least frequent gametes. They are so infrequent that often they do not even show up!!! That is, they have a frequency of zero. You must always remember that a triple heterozygote is expected to make 8 different gametes. If you only see 6 phenotypes among the test cross progeny, the missing phenotypes (in most cases) correspond to the double crossovers. In the above example all 8 phenotypes are present and the double recombinant gametes are fha and FHA.
Step 3: Determine the relative order of the genes on the map. The parental pair of gametes is FHa/fhA and the double cross over gametes are FHA/fha. The difference between the two pairs is that the A gene (in red) has switched chromosomes. This mean that the map order is:
The Genetic Map is:
| F--------------- | A--- | -----------H |
Step 4: Now pair up the other gametes: Wherever you have a capital letter in one gamete (e.g., F) you must have a lowercase letter in the complementary gamete (e.g., f)
| Parental Gametes | DC Gametes | Pair X Gametes | Pair Y Gametes |
| FHa | FHA | Fha | fHa |
| fhA | fha | fHA | FhA |
Step 5: Add up the number of each pair and calculate the frequency of each pair of gametes. (Divide the number in each pair by the total of 50,000)
| Parental Gametes | DC Gametes | Pair X Gametes | Pair Y Gametes |
| FHa | FHA | Fha | fHa |
| fhA | fha | fHA | FhA |
| 35162 | 162 | 5838 | 8838 |
| .003 | .117 | .177 |
Step 6: Now select one of of the pairs of gametes that represent a single crossover - for example Pair X (Fha/fHA) and compare this pair to the parental gametes (FHa/fhA). You are looking for the alleles that appear to have switched chromosomes. In this case, the allele pair is H/h. The pair that appeared to have switched is on one end and has crossed over with the gene in the middle of the map (Gene A). The distance between H and A is is the frequency of Pair X gametes and the Double crossovers (as percentages or map units).
Distance H ------ A = 11.7% + .03% = 12% = 12 map units
The Genetic Map is now
| F--------------- | A--- | -----------H |
| 12 map units |
Step 7: Now select the other pair of gametes that represent a single crossover --- that is, Pair Y (fHa/FhA) and compare this pair to the parental gametes (FHa/fhA). You are looking for the alleles that appear to have switched chromosomes. In this case, the allele pair is F/f. The pair that appeared to have switched is on one end and has crossed over with the gene in the middle of the map (Gene A). The distance between F and A is is the frequency of Pair Y gametes and the Double crossovers (as percentages or map units).
Distance H ------ A = 17.7% + .03% = 18% = 18 map units
The Genetic Map is finally:
| F------------- | A--- | -----------H |
| 18 map units | 12 map units |
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Additional Topics |
| Test Cross |
| Test cross for two genes |
| Test cross reveals the number and nature of gametes |
| Test cross can detect linked genes |
| Test cross to detect linkage for RFLP's |
| Triple Test Cross |