Chapter 7 genetics PDF

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Chapter 7 • Linked Genes Do Not Assort Independently • Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them • A Three-Point Testcross Can Be Used To Map Three Linked Genes • Physical-Mapping Methods Are Used To Determine the Physical Positions of Genes on Particular Chromosomes • Recombination Rates Exhibit Extensive Variation • • • •

• • • • • •

• • • •

Chromosome Theory of Inheritance states that genes are located on chromosomes (for the most part that’s true) Genes are arranged along the length of the chromosome in a specific order The site on the chromosome where a gene resides is referred to as the genetic locus The human genome is estimated to contain ~ 25,000 genes arranged along 23 chromosomes Chromosomes are numbered according to size of the autosomes, with chromosome #1 being the largest and chromosome #22 the smallest. The 23rd chromosome is the sex chromosome Genes that are located on the same chromosome are said to be physically linked The full complement of genes on a given chromosome constitutes a linkage group Because genes on a given chromosome are linked, they tend to be transmitted to offspring as a unit or group However, eukaryotes are diploid organisms, and each chromosome possesses a homolog that it pairs with during Meiosis I Parental type - allele configuration in the parents in a cross Recombinant type - allele combinations different from those in the parental chromosomes Frequency of recombination - percent value for recombinant chromosomes Syntenic - refers to genes in the same chromosome regardless of whether they show independent assortment or linkage

Nonindependent assortment of flower color and pollen shape in sweet peas

Crossing over takes place in meiosis and is responsible for recombination

1.For single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because: each crossover takes place between only two of the four chromatids of a homologous pair. Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them. • Coupling and Repulsion Configuration of Linked Genes. – Repulsion (trans configuration): wild type allele and mutant allele are found on the same chromosome. • Testing For Independent Assortment

Genes on different chromosomes assort independently • A double heterozygote produces four possible types of gametes in equal proportions

The physical basis of the independent assortment of genes in different chromosomes is that nonhomologous chromosomes orient independently at metaphase I of meiosis. Crossing-over produces recombinant gametes

Linked Genes Segregate Together and Crossing Over Produces Recombination Between Them. • Calculating Recombination Frequency – Recombination frequency = (No. recombinant progeny/Total No. of progeny) X 100% • Coupling and Repulsion Configuration of Linked Genes. – Coupling (cis configuration): wild type alleles are found on one chromosome; mutant alleles are found on the other chromosome.

The arrangement (coupling or repulsion) of linked genes on a chromosome affects the results of a testcross. Linked loci in the Australian blowfly, Lucilia cuprina, determine the color of the thorax and that of the puparium.

In this case, the recombination frequency is 16% (0.16) You have two possible recombinant gametes. Therefore, you divide the percentage of recombination frequency between the two possible Gametes. . If the recomb. freq. = 0.16 and you have two recombinant gametes, then the frequency for each recombinant gamete will be 0.08 (8%).

Establishing Linkage

P1 A B X a A B

F1

a

Linked Genes

b

Testcross X a b

A B a

b

b

a AB

ab A B a b

Coupling Phase (cis)

b

ab a b a b

All Parental

A

a

B

b

Parental Gametes

P1 A b X a A b

F1

a

A b a

B

Linked Genes

B

X

B Ab ab A b a b

Testcross a b a

b

aB a B a b

Repulsion Phase (trans)

A

a

b

B

Parental Gametes

All Parental

A chi-square test of independence can be used to determine if genes at two loci are assorting independently.

A chi-square test of independence can be used to determine if genes at two loci are assorting independently.

A chi-square test of independence can be used to determine if genes at two loci are assorting independently.

Linked Genes • In reality, however, testcrosses involving linked genes may also generate both parental and recombinant type offspring • BUT, the ratio of parental to recombinant progeny is usually not in the 1:1:1:1 ratio observed with unlinked genes • Why? Recombinant chromosomes are produced by crossover events occurring during meiosis and end up in some gametes • •

• • •

Genes with recombination frequencies < 50% are linked and are on the same chromosome Genes that have a recombination frequency of 50% are unlinked – these genes may be on non-homologous chromosomes or are located far apart in the same chromosome These crossover events are relatively rare Thus, with linked genes, the majority of the progeny will be of parental type and a minority will be represented by the recombinant type If recombination does not occur, the progeny will be 50% parent 1 and 50% parent 2

Alleles of linked genes can be in cis or trans configuration • Recombination occurs at the same frequency whether the alleles are in cis or trans

Possible configurations of the mutant alleles in a genotype heterozygous for two mutations General rules of recombination • Recombination frequency is a characteristic of a particular pair of genes • Recombination frequencies are the same in cis and trans heterozygotes Recombinant gametes are produced by crossing-over between two genes Here the crossing-over did not occur between the two genes. As a result, all four gametes are non-recombinant

Crossing-over outside of the region between two genes is not detectable through recombination

Results of two crossovers between a pair of genes • Four nonrecombinant gametes are produced

When two crossovers take place between marker genes, and both involve the same pair of chromatids, neither crossover is detectable Genetic distance between two genes can be represented four ways • Frequency of recombination • Percent recombination • Map distance in map units • Map distance in centimorgans (cM) One map unit represents 1% cross-over frequency

Chromosomal configurations in 50 meiotic cells, in which 1 cell has a cross-over between two genes Map distance map distance = 1/2 x the average crossovers between two genes per meiotic cell x 100

Thomas Hunt Morgan Did it Again!! • The concept that genes on the same chromosome can be inherited as blocks, came from T.H. Morgans results with Drosophila – pr+(red-eye) is dominant over pr (purple-eye) – vg+(normal wing) is dominant over vg(vestigial wing)

pr+pr+ vg+vg+

X

prpr vgvg

(Red-eye; normal wing) (purple-eye, vestigial wing)

pr+pr vg+vg (red-eye; normal wing)

Two-Point Testcross • Crossovers occur at random along the length of a linear chromosome • The frequency of recombination between two genes is directly proportional to the distance between the two genes • By counting the number of recombinant progeny produced from a dihybrid testcross, a direct estimate of the frequency of crossovers that occurred during gamete formation can be determined • • •

The frequency is directly proportional to the distance between the 2 genes in question A large number of recombinant progeny indicates that the genes are far apart By measuring the frequency of recombination, geneticists can establish a linkage map of relative distances between genes on a chromosome

Recombination Frequency (R) • R=

the # of recombinant testcross progeny the total number of testcross progeny

• An R-value can be converted into percent recombination simply by multiplying by 100 – R = 0.01 = 1% recombination

• By definition, 1% recombination equals a map distance of 1 map unit or 1 centiMorgan (cM) which is equivalent to ~ 1,000,000 bp • The map unit represents the percentage of chromosomes that have undergone a crossover event between the 2 genes of interest

Two-Point Testcross • Example: Drosophila body color and wing length – b+(wt body color) is dominant to b(black body color) – vg+(wt wing length) is dominant to vg(vestigial wings) b+ b vg+vg

X

b b vgvg

Phenotype wt, wt

Gametes b + vg+

Observed 975

Expected 600

(O-E) 2/E 234

wt, vestigial

b + vg

217

600

245

Black, wt Black, vestigial

b vg b vg

+

236 972

600 600

221 231

2400

Parental gametes?

2400

Linkage phase?

c 2 = 931

•

•

Gametes are not in a 1:1:1:1 ratio – The genes for body color and wing length are not assorting independently; most likely LINKED! c2 = 931, with 3 degrees of freedom, gives a p value...