Chapter 13 problems:

• 2, 4, 6, 9, 10, 12,19, 20, 22, 24, 32, 36, 38

Tetrad analysis

• In some haploid fungi and algae, the meiotic products stay together in 'tetrads'
• Because you see all products of meiosis, you can do more detailed analyses
• • like mapping the centromere
  • distinguishing various kinds of crossovers
• In many ways, everything is the same except
• • Meiotic tetrads are counted, not individuals
• Tetrads can be either ordered (e.g. Neurospora) or unordered (yeast)
  

Mapping the centromere

• Centromeres always segregate in Meiosis I
• • AAaa or aaAA
  • This is known as a "first division segregation pattern".
• If a gene segregates in meiosis II, then there must have been a crossover between it and the centromere.
• • AaAa or AaaA
  • this is "second division segregation"

If that doesn't make sense, draw out several chromosomes in the process of meiosis, with and without crossovers, and convince yourself.

Note that when a single crossover event occurs, only 2 of the 4 stands are involved, so only half of the gametes show crossovers.

Tetrad example

• Here is an example that we worked in class, showing meiotic tetrad types from a heterozygous Aa individual.
• What is the distance from A to the centromere?

AAaa	36
aaAA	44
AaAa	4
aAaA	6
AaaA	3
aAAa	7
Total: 100

• This is the same basic principle that we have always used. The percentage of recombined gametes is a measure of map distance, in centimorgans.
• To do that problem, first figure out which tetrads indicate that a crossover occurred, and which were produced without crossing over. Then, the distance to the centromere will be 0.5 * fraction of total tetrads that were products of crossovers
• In Neurospora, each of these gametes would be doubled, producing ascii with 8 spores. Here I'm just showing the four meiotic products.

Mapping two loci with tetrad analysis

Parental ditype:	AB,AB,ab,ab
Non-parental ditype:	Ab,Ab,aB,aB
Tetratype:	AB,aB,Ab,ab

How do you distinguish the types? Ditypes have two kinds of gametes per tetrad; tetratypes have four different gametes. Parental ditypes have alleles in the same arrangement as they were in the parents.

If the genes are linked,

• Single crossovers produce tetratypes
• Double crossovers can produce all three kinds, depending on which chromatids are involved
• NPD can be produced only by double crossovers.

If A and B are on different chromosomes, so they assort independently, what gamete types do you expect? Draw it out to see.

To map genes in tetrads,

Use 1/2 TT + NPD as the number of recombinants

Or, for more accurate map that takes account of all double crossovers,

1/2 (TT + 6 NPD)

Although I didn't derive these formulae in class, the mapping principle is the same as always: (recombinants / total gametes) gives the map distance.

Example

• AB x ab
• Produces the following tetrads:

AB AB ab ab 60

Ab Ab aB aB 5

Ab AB ab aB 16

AB aB Ab ab 19

Total=100

• Are they linked?
• How far apart are they?
• How many single vs double crossovers?

• To do this, identify the parental ditypes, the nonparental ditypes, and the tetratypes.
• If PD and NPD are equal, then the genes are probably not linked.
• If NPD << PD, then they are linked.
• Calculate the number of recombinant tetrads using the formulae above (e.g. 1/2TT + NPD) and divide by the total number of tetrads.

Tetrad Analysis in Arabidopsis

Usually, tetrad analysis can only be done in fungi that keep the products of meiosis together. Most higher organisms produce individual gametes. The Quartet mutant keeps pollen together in tetrads, which allows you to do tetrad analysis.

I showed you data from a paper by Copenhaver et al. 1998. PNAS 95:247

• They used those unordered tetrads to map the centromeres (which is very difficult to do any other way)
• Determined details of crossover frequency; in particular they showed the frequency of crossovers was not entirely random. There was less variation than expected, as if there had been some kind of selective tradeoff over evolutionary time to keep recombination rate approximately 2 per chromosome.
• They showed that conversion is rare: there were no tetrads with 3:1 or 4:0 ratios.

Mapping in Humans

Here is a pedigree that shows blood types and "nail patella syndome"

NPS is a rare genetic disease occurring in only about 2/100,000 births, characterized by underdeveloped nails and kneecaps.

Human pedigree mapping

1. Determine genotypes in the pedigree
2. Find informative progeny (e.g. test crosses)
3. What fraction of offspring are recombinant?

You might want to check out these web sites for two ways to analyze that same pedigree:

as we did in class > and likelihood method

The basic difficulty with mapping genes in humans is that it is hard to get big enough pedigrees, with enough informative families.

• Other strategies include:
• Statistical approaches that try to consolidate information from all of the individuals in a pedigree. They work best when you have lots of markers (e.g. molecular markers).
• Human-rodent somatic cell hybrids can localize a gene to a particular chromosome.
• We now have the whole physical map of the human genome

Mapping a hair loss gene Allopecia universalis

Ahmad et al., Science, 1998

• Found a large family in Pakistan with many affected individuals
• Looked for correspondence between molecular markers and the trait, using 300 highly polymorphic markers
• Searched for markers that were homozygous in affected individuals and heterozygous otherwise (because the disease is recessive, so closely linked markers in affected individuals must be homozygous too).
• One such marker was found on chromosome 8p12

Finding the actual gene

• In other work, they had also mapped the "hairless" gene (hr) using somatic cell hybrid mapping
• Used the mouse hr gene to amplify the human gene sequence
• Fused bits of human chromosomes to hamster cells, to find the fragments that corresponded to the mouse hairless gene
• It also mapped to 8p12
• The hr gene was a good candidate, so they sequenced the gene from each individual in this pedigree.
• All affected individuals had an allele of hr that differed in a single base A to G, which changed a threonine to alanine in the protein

Another example: Renal Fanconi syndrome

Lichter-Konecki et al, 2001 Am J Hum Gen

• RFS is a dominant disorder
• Again they used a big pedigree from Wisconsin and a lot of polymorphic molecular markers
• They found a set of markers on Chromosome 15 that were associated with the disorder
• Rare recombination events between closely linked markers helped them localize the disorder to a short stretch of the chromosome