4.1 linkage: basic haploid eukaryotic chromosome mapping
DESCRIPTION
4.1 Linkage: basic haploid eukaryotic chromosome mapping. Haploids organisms. Advantages for genetics studies: There is no dominance or recessivity Only one meiosis in each cross In some fungus and algae, the individual meiosis products stay attached in tetrads Most of them are microbes. - PowerPoint PPT PresentationTRANSCRIPT
4.1 Linkage: basic haploid eukaryotic chromosome mapping
4.1 Linkage: basic haploid eukaryotic chromosome mapping
Haploids organisms
• Advantages for genetics studies:
– There is no dominance or recessivity
– Only one meiosis in each cross
– In some fungus and algae, the individual meiosis products stay attached in tetrads
– Most of them are microbes
The linear meiosis of Neurospora
Allele segregation in ordered tetrads in MI
•The allele segregation takes place during the first meiotic division (MI)
A second-division segregation pattern in a fungal octad
When a crossover between the
centromere and the locus takes
place, segregation occurs in the
second meiotic division (MII)
Four different spindle attachments produce four second-division
segregation patterns
Distance locus-centromere
• Crossover frequency between the locus and the centromere
In this ascus, only half of the chromatids have undergone a crossover
Distance locus- centromere =
1
2 MII Asci
Total100 (map units )
Two loci: a y b
– They are in different arms in the same chromosome
– They are in the same arm• One crossover between the
centromere and a produces the same MII pattern for the two loci: LINKED LOCI
Located in the same chromosome
Two loci
• Crossing a+ b X a b+. ORDERED ASCI
a+ ba+ ba b+
a b+
808
a+ b+
a+ b+
a ba b1
a+ b+
a+ ba b+
a b90
a+ ba b+
a+ ba b+
90
a+ ba b
a+ b+
a b+
5
a+ b+
a b
a+ b+
a b
1
a+ b+
a b
a+ ba b+
5
a - cen =
1
2MII Asci
Total100 =
1
2(5 9015)
1000100 5.05 m.u.
MI MI MI MII MII MII MII MIIMIIMIIMII MIMIMI
b - cen =
1
2(90 90 15)
1000100 9.3 m.u.
• Three possibilities
– Independent Loci : no linkage. Independent segregation. – Most (96) of the a MII asci are also b MII : the third possibility is the correct one. How have the a MII and b MI occurred?a b
a b
a b
a+ ba+ ba b+
a b+
808
a+ b+
a+ b+
a ba b1
a+ b+
a+ ba b+
a b90
a+ ba b+
a+ ba b+
90
a+ ba b
a+ b+
a b+
5
a+ b+
a b
a+ b+
a b
1
a+ b+
a b
a+ ba b+
5
MI MI MI MII MII MII MII MIIMIIMIIMII MIMIMI
Unordered asci
• Saccharomyces cerevisiae • Unable to locate the centromere• a+ b+ X a b cross. Three possible asci types
a+ b+
a+ b+
a ba b
a+ b+
a+ ba b+
a b
a+ ba+ ba b+
a b+
Parentalditypes
no parental ditypes
(recombinante)
Tetratypes
The written order does not mean the ascus order.
Maps with unordered asci
Rec. frequency. =
1
2TT DNP
TOTAL
What happens when there is no linkage?
Two loci in ordered asci
• Cross a+ b X a b+. ORDERED ASCI
a+ ba+ ba b+
a b+
808
a+ b+
a+ b+
a ba b1
a+ b+
a+ ba b+
a b90
a+ ba b+
a+ ba b+
90
a+ ba b
a+ b+
a b+
5
a+ b+
a b
a+ b+
a b
1
a+ b+
a b
a+ ba b+
5
DP DNP DP DNPTT TT TT
RF
1
2TT DNP
Total
1
2(90 5 5) (11)
1000X100 5.2 m.u.
a b5.05 5.2
10.25/9.3