Ch. 5 Linkage, Recombination, and Gene Mapping

genes located on the same chromosome

syntenic genes

gametes whose alleles are in the same combination as in the gametes that gave rise to the previous generation

• progeny generated by parental gametes

parental types/classes

gametes whose alleles are in different combinations than in the gametes that gave rise to the previous generation

• progeny generated by recombinant gametes

  • can also form from independent assortment if the two genes are on non-homologous chromosomes

• gametes or progeny resulting from a meiosis in which a crossover took place between the two genes in question

recombinant types/classes

when two genes are located close enough to each other on the same chromosome so that a doubly heterozygous parents makes more parental type than recombinant type gametes

• when two loci recombine in less that 50% of gametes

linkage

the proportion of the total number of gametes or progeny that are recombinant types

recombination frequency

• RF = recombinants/total

a measure of genetic distance between linked genes

map unit (mu)/centimorgan (cM)

• 1 mu = 1 cM = 1% RF

a specific location on a chromosome

locus/loci

a statistical device used to measure the likelihood that a set of observed experimental results could have been obtained as chance deviation from the expectations from a hypothesis to be tested (the null hypothesis)

chi square test

experimental results that are highly unlikely to be explained by the null hypothesis

• allows you to reject the null hypothesis

• an insignificant result does not allow you to reject the null hypothesis

• chi square test can never prove a null hypothesis; it can only reject the null hypothesis

significant result

in the chi square test for the goodness of fit, the

df = number of classes - 1

• required for interpreting the chi square test

degrees of freedom (df)

the probability that the observed results could be obtained by chance deviations from the expectations of the null hypothesis

• if low (usually below 0.05 for genetic analysis), you can reject the null hypothesis

p value

formation of new genetic combinations by exchange of parts between homologs

recombination

structure formed at the spot where crossing over occurs between homologs

chiasma

the ratio of observed double crossovers to expected double crossovers

coefficient of coincidence

an ascus containing only two non recombinant kinds of spores

parental ditype

when the two alleles of a gene are segregated into different cells at the first meiotic division

first division segregation

individual composed of cells with different genotypes

mosaic

what are 5 tips for solving two point crosses?

• the minimum requirement for detecting recombination is that one parent must be heterozygous for 2 genes. the recombination events that can be detected are the ones that occur between the 2 genes, giving recombinant gametes instead of parent gametes

• it is easiest to detect the parental versus recombinant gametes if you do a testcross

• if genes are assorting independently, in a test cross of aa+bb+ x aabb, the expected classes and frequencies of progeny are:

  • a+—b+— : 1

  • aabb : 1

  • a+—bb : 1

    • but the genes are genetically linked if you see more parental than recombinant progeny

• recombination frequency (RF) = # recombinant progeny / total # progeny (multiply x 100 to express the RF as %)

  • 1% RF = 1 mu or 1 cM

• genes on the X chromosome can be mapped without a testcross. just use the hemizygous male progeny

what are 5 tips for three point crosses?

• in a three point cross, a parent heterozygous for the three genes generates the progeny.

  • therefore, all classes (parental, recombinant, etc.) will occur as reciprocal pairs of progeny.

  • these RP will be both genetic reciprocals and numerically roughly equivalent.

• designate the different gametes or offspring as no crossover (NCO; parental), single crossover (SCO), or double crossover (DCO).

  • the NCO classes are those classes of progeny which have one of the intact, nonrecombinant (parental) homologs from the parent.

    • if the genes are linked, the NCO class will be represented by the RP with the greatest numbers of offspring.

  • there will be SCOs occurring between the gene on the left and the gene in the middle (two RPs), and SCOs occurring between the gene in the middle and the gene on the right (another two RPs).

  • the DCO classes will be represented by the reciprocal classes with the smallest numbers of offspring

    • thus, in a three point cross there will usually be 8 classes of progeny.

    • however, sometimes one or both double crossover classes are missing because they are rare and an insufficient number of progeny were observed

• by examining the pattern of data seen in a problem, you can often start solving the problem with a basic understanding of the linkage relationships of the genes. some of the more common patterns of data are:

  • 3 unlinked genes give 8 classes of data that occur as 4 genetically RPs, but all classes are seen in a 1:1:1:1:1:1:1:1 ratio

  • 3 linked genes give 8 classes of data that occur as 4 RPs genetically and numerically unless one or both DCO classes are missing, in which case you will see 4 classes (as 3 RPs) or 7 classes (3 RPs plus an additional unpaired class)

  • 2 linked genes plus 1 unlinked gene will yield 8 classes of data. 4 of these classes will be numerically equal. the group of 4 classes with larger numbers will consist of the reciprocal parental classes for the linked genes together with either allele of the unlinked gene. the group of 4 classes with the smaller numbers will be the reciprocal recombinant classes of the linked genes together with either allele of the unlinked gene

• begin the process of mapping the genes by ordering the genes. to figure out which gene is in the middle of a group of three genes, choose one of the DCO classes. compare it to the most similar parental class of progeny where two of the three genes will have the same combination of alleles. the gene that differs is the gene in the middle.

• the last step is to determine the distance between the genes on each end and the gene in the middle. use the formula RF = # recombinants between the 2 genes / total # progeny.

  • remember for each interval that the # recombinants is the number of progeny in the reciprocal SCO classes representing crossovers in that interval, plus the number of recombinants in the reciprocal DCO classes