DETECTION OF LINKAGE

There are different methods to detect linkages among genes. Some of these are:

  1. Detection of linkage by PD/NPD ratio

The Parental generation (PD) to non-parental generation ratio gives indication about linkage between two genes. Test cross or back cross is performed in order to obtain PD/NPD ratio. If there is ratio of 1 between PD and NPD, then genes are not linked. If this ratio is more than 1, then the genes are linked.

Example 1

Tall pea plant with purple flower was crossed with small plant with white flower. F1 were all tall and purple. The tall and purple of F1 was test crossed. It gives:

Tall purple = 523
Tall white = 528
Sma 1 1 purple = 513
Small white = 502

Parental type (PD)    = 523 + 502 = 1025

Non-parental type NPD = 528 + 5 I 3 = 1041

Ratio of NP to NPD    = 1025: 1041

It gives almost value of 1. Thus the genes of tall/small and purple white are not linked.

Example 2

Purple (PP) flower colour is dominant over red (pp) flower colour in sweat pea. Similarly, long pollen grain (LL) is dominant over round pollen grain (II). They cross pure line of purple long (PPLL) with pure line of red, round (pp11). They obtained purple long in F1 The F1 progeny was test crossed. It gives following progeny:

I. Purple, long          440

  1. Purple, round     14
  2. Red, long           7

. Red, round                 350

Parental type (PD)        = 440+ 350 = 790

Non-parental type NPD = 14 + 7 = 21

Ratio of NP to NPD        = 790 : 21

It gives value for more than I. Thus both these genes are linked.

  1. LOD score method for detection of linkage

The Lod (log of odds) score is used to detect linkage in pedigree arising randomly or by genetic linkage. This test was developed by Newton E. Morton. Following formula is used to measure LOD score:

LOD = log probability of birth sequence with a given linkage value probability of birth sequence with no Linkage

Linkage is present if the LOD score is greater than 3. Linkage is absent if value of LOD score is 1 or less than I.

Example

Two genes, nail-patella and blood type are given in this pedigree.

                                  AO          BO

                                       AB DO


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A cross occurs between an affected man (nail patella) having Blood group A (genotype AO or IAD with a normal woman having blood group B (genotype BO or IRO. They produced an affected son with blood group AB. He was married to normal woman with blood group 0 (genotype 00). Their progeny includes eight children. Four were affected and four were unaffected. All the children have parental

combination like their parent except their son (5 marked as*). All other AO were affected. This 5″ son has blood group AO but it is unaffected. Thus it is a recombinant. Now LOD score of this pedigree is determined as follows:

I. Frequency of recombination in birth sequence = 1/8 = 0.125. It is the probability of linkage in this birth sequence.

  1. Frequency of recombination in this birth sequence, if linkage does not occur = 4/8 = 0.5. It is the probability without linkage

log0.125 =3.54

  1. 1,0D Score =

0.5

  1. As this LOD score is more than 3. Therefore, gene of blood group A and Nail Patella syndrome are linked.
  2. Robust method of detection of linkage

The robust method for detecting linkage was developed by Haseman and Elston. Mathematical integration and regression principles are used in this method. It is used to detect linkage of quantitative genes with other locus.

  1. Modern biotechnological method

These are latest techniques for detecting the gene linkages. In this method modern biotechnological techniques are used. These techniques include REFLP (Restriction fragment length polymorphism), PCR (Polymerase chain reaction) and DNA probes.

  1. DNA is cut in to small fragment called REFLPs. Each fragment has linked genes.
  2. These genes are amplified by PCR.
  3. DNA probes are used to detect the genes present in a fragment.

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