Describe how the relative numbers of PDs and NPDs
can be used to establish linkage
Parental ditype is a type of tetrad consisting of two different genotypes both of that are parental. The arrangement of spores is Ascomycetes which consists of only two ascospores which are non-recombinant type.
Non parental ditype is a tetrad type which consists of only two different genotypes both of which are recombinant. A spore arrangement of Ascomycetes which contains only two are recombinant type ascospores.
The ratio between the types of different segregation that arises after sporulation process is the measure of a linkage between the two genes.
In order to determine the relative frequency between the two genes, we can break open all the spore cases, pool the spores and then analyse them to determine which ones are parental and which are recombinants.
PDS = NPDS for unlinked genes on chromosome:
Let us consider a cross between a haploid strain of yeast of type a which carries the his4 mutation and the wild type of TRP1 gene which is a strain opposite to type a.
Here three types of meiosis process could takes place each producing a different tetrad type. A PD tetrad will result from one of the two random alignments of homologous chromosomes during the process of meiosis I. Here the number of PDS and NPDS will be equal as the two genes are present on different chromosomes. As expected the related frequency is 50% which means that the number of PDS which are parental spores and the number of NPDS in which the recombinant spores are equal.
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