Links
Objectives
- Describe the conditions under which characters fail to assort independently
- Explain a test cross and how one is carried out
- Analyze the results of a test cross and calculate genetic distance
Key points
Background
- sometimes characters don’t assort independently during meiosis
- these characters show a high probability of being inherited together
- characters that travel together during meiosis are linked
Genetic mapping
- linkage can be identified and quantified by careful observation of inheritance patterns
- the basis of genetic mapping is recombination due to crossing over
- characters that are located very close to each other on the chromosome have a lower probability of being separated by crossovers than those that are farther apart
- the frequency of crossovers can be used as a measure of distance between characters
Performing a test cross
- cross a known heterozygous individual with a known homozygous recessive individual
- score the number of offspring having each of the 4 possible phenotypes
Example test cross
- cross a homozygous recessive mutant for body color (black, b) and wing shape (vestigial, vg) with a fly heterozygous for both characters: bb vgvg x b+ vg+ (the + represents the wild-type allele)
- phenotypes of offspring
- parental: black body, vestigial wings (405)
- parental: wild-type body and wings (415)
- recombinant: wild-type body, vestigial wings (92)
- recombinant: black body, wild-type wings (88)
- calculate recombination frequency
- (92 + 88) / 1000 = 0.18 or 18 map units
- notes
- the only way to produce offspring with wild-type bodies and vestigial wings is via recombination
- the only way to produce offspring with black bodies and wild-type wings is via recombination
In-class activities
- work through example test crosses
Questions for Practice
- Describe how you would test two characters for linkage. What predictions would you make about number of offspring for each phenotype if the genes are linked or not?