Concepts of Genetics Eighth Edition Klug, Cummings, Spencer

1
Concepts of GeneticsEighth EditionKlug,
Cummings, Spencer

• Chapter 5
• Gene mapping (5.1-5)

2
Genes

• More genes in any organism than chromosomes and
  therefore there needs to be many gene loci per
  chromosome.
• These genes are located on chromosomes like beads
  on a string.

3
Genetic maps

• Genetic maps ? determine the order of genes based
  on meiotic recombination.
• Exact locality of gene is not determined but
  relative distance between genes.
• These relative distances are expressed as
  centiMorgan (cM) and do not correspond directly
  to the true distance between genes.

4
Cytogenetic maps

• Cytogenetic (physical) maps identify the precise
  location of gene on chromosome.
• Distance between genes are indicated as base
  pairs ? bp. (basepairs), kb. (1 000 bases
  kilobp) or mb. (1 000 000 basepairs megabp).

5
Gene linkage

• Genes assort independently if they are on
  different chromosomes but show linkage if they
  are on same chromosome.
• Often, mutation causes reduction or loss of
  specific wild-type function ? loss of function
  mutation ? if loss is complete, mutation has
  resulted in what is called null allele.
• If complete linkage exist between two genes
  because of their close proximity and organisms
  heterozygous at both loci are mated ? unique F2
  phenotypic ratio (linkage ratio) results.

6
Gene linkage

• In complete linkage, only parental (noncrossover)
  gametes produced.
• If crossing over between two linked genes occurs
  between two nonsister chromatids, both parental
  and recombinant (crossover) gametes produced
  (Figure 5.1).

7
Gene linkage

• Degree of crossing over between any two loci on
  single chromosome is proportional to distance
  between them ? interlocus distance.
• Genes on same chromosome part of linkage group.
• Number of linkage groups should correspond to
  haploid number of chromosomes.

8
Crosses

• Percentage of offspring resulting from
  recombinant gametes depends on distance between
  two genes on chromosome.
• Synapsed chromosomes in meiosis wrap around each
  other to create chiasmata that are points of
  genetic exchange.
• Two genes located relatively close to each other
  along chromosome ? less likely to have chiasma
  form between them and less likely that crossing
  over will occur.

9
Crosses

• Recombination frequencies between linked genes
  are additive ? frequency of exchange is estimate
  of relative distance between two genes along
  chromosome.
• One map unit (mu) defined as 1 recombination
  between two genes on chromosome ? map units often
  called centimorgans (cM) ? relative distances,
  not exact ones.

10
Crosses

• Single crossover (SCO) alters linkage between two
  genes only if crossover occurs between those two
  genes (Figure 5.5).

11
Crosses

• Percentage of tetrads involved in exchange
  between two genes is twice the percentage of
  recombinant gametes produced.
• When two linked genes are more than 50 mu apart,
  crossover theoretically expected to occur between
  them in 100 of tetrads (Figure 5.6).

12
Gene sequence

• Single crossovers can be used to determine
  distance between two linked genes, but double
  crossovers (DCOs) can be used to determine order
  of three genes on chromosome.
• To study double exchanges ? three pairs of genes
  must be investigated, each heterozygous for two
  alleles (Figure 5.7).

13
Gene sequence

• Expected frequency of double-crossover gametes is
  much lower than that of either single-crossover
  gamete class.
• In three-point mapping ? parent must be
  heterozygous for all three genes under
  consideration.

14
Gene sequence

• Three-point mapping cross (Figure 5.8).
• NCO F2 phenotypes occur in greatest proportion of
  offspring.
• DCO phenotypes occur in smallest proportion.

15
Gene sequence

• Because F2 phenotypes complement each other ?
  i.e., one is wild-type and other is mutant for
  all three genes ? they are called reciprocal
  classes of phenotypes.
• Distance between two genes in three-point cross
  is equal to percentage of all detectable
  exchanges occurring between them and includes all
  single and double crossovers.
• There are two methods for determining gene order
  from three-point cross.

16
Gene sequence

• Three-point cross and mapping of three genes
  involved (Figure 5.10).

17
Gene sequence

• Three-point cross and mapping of three genes
  involved (cont.) (Figure 5.11).

18
Interference

• Expected frequency of multiple exchanges between
  two genes can be predicted from distance between
  them.
• Interference (I ) reduces expected number of
  multiple crossovers when crossover event in one
  region of chromosome inhibits second event
  nearby.
• Interference is positive if fewer DCO events than
  expected occur and negative if more DCO events
  than expected occur.

19
Coefficient of coincidence

• Coefficient of coincidence (C ) is observed
  number of DCOs divided by expected number of
  DCOs.
• Can quantify interference by using this simple
  equation
• I 1 C
• For example, if I 0.196 ? indicates that 19.6
  fewer DCOs occurred than expected.

20
Gene mapping

• When two genes are close together, accuracy of
  mapping is high.
• As distance between them increases, accuracy of
  mapping decreases.
• Two exchanges between linked genes that are far
  apart on chromosome can involve two, three or all
  four strands ? result in production of different
  percentages of recombinant chromatids (Figure
  5.12).

21
Gene mapping

• These uncommon multiple events even out and two
  genes that are far apart on chromosome
  theoretically yield maximum of 50 recombination
  essential for accurate gene mapping.
• Discrepancy results primarily from multiple
  exchanges predicted to occur between two genes
  but are not recovered during experimental mapping.

22
Gene mapping

• Inaccuracy is result of probability events that
  can be described using Poisson distribution.
• This analysis creates mapping function that
  relates recombination (crossover) frequency (RF)
  to map distance.

23
Example

• Let us assume that the following homozygous lines
  are crossed
• P1 AABB x aabb
• F1 AaBb
• Cross the F1 back to a homozygous recessive line

24
Example

• F2 expect observed
• AB 25 32
• Ab 25 18
• aB 25 15
• ab 25 35

25
Example

• Parental phenotypes (AB ab) are present in more
  than 50 of the offspring, therefore genes A and
  B are linked.
• The parental types are present in 67 of the
  offspring, therefore crossing over between the
  two genes occurred in 33 of the offspring.
• This implies a distance of 33 cM between the
  genes.

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3-point mapping

• Use three genes ? A, B and C.
• Cross two homozygous parents.
• Cross the F1 back to a homozygous recessive for
  all three genes.
• The parental phenotypes should be complementary.
• The numbers of F2s with each phenotype are
  counted.

27
3-point mapping

• There is a positive correlation between the
  number of offspring and the reliability of the
  experiment.
• At least 500 F2s should be used to calculate the
  distances but it is better to use at least 1 000
  F2s.

28
3-point mapping

• Suppose you cross AABBcc with aabbCC
  (complimentary gene compositions).
• Cross the F1s back to the homozygous recessive
  line (aabbcc ) and count the number of F2s in
  each phenotypic class.
• Suppose you observe the following numbers

29
3-point mapping

• Phenotype observed
• ABC 148
• ABc 1828
• AbC 227
• aBc 217
• abC 1779
• abc 139
• Arrange the phenotypes in such a way that
  complimentary phenotypes are together.

30
3-point mapping

• The phenotype with the highest occurrence are
  placed first
• Phenotype observed
• ABc 1828 Parental types
• abC 1779
• ABC 148 Cross over 1
• abc 139
• AbC 227 Cross over 2
• aBc 217
• TOTAL 4338

31
3-point mapping

• Determine whether the genes are linked ? done by
  calculating percentage of parental types.
• Phenotype o /
• ABc 1828 3607 PT
• abC 1779 83.1
• ABC 148 287 CO 1
• abc 139 6.6
• AbC 227 444 CO 2
• aBc 217 10.2
• TOTAL 4338

32
3-point mapping

• Because parental types occur in more than 50 of
  F2s (83.1) ? genes are linked.
• Determine the sequence of the genes (ABC, ACB or
  BAC )
• ABC ACB BAC (theoretical sequences)
• ABc AcB BAc (PT in different TS)
• abC aCb baC (Other PT)
• ABC ACB BAC (Co 1 in TS)
• abc acb bac (Complimentary sequence)
• AbC ACb bAC (CO 2 in TS)
• aBc acB Bac (Complimentary sequence)

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• A B c A c B B A c (Parental type)
• a b C a C b b a C (Compl. PT)
• A cross-over between the first and second gene
  results in
• A b C A C b B a C
• Determine whether these theoretical combinations
  occur in cross-over type 1 or 2
• A b C (2) A C b (2) B a C
• Determine the theoretical implications of a
  cross-over between the second and third genes
• A B c A c B B A c (Parental type)
• a b C a C b b a C (Compl. PT)
• A cross-over between the second and third gene
  results in
• A B C A c b B A C (already excluded
  so ignore this sequence)
• Determine whether this theoretical combinations
  occur in cross-over types 1 or 2
• A B C (1) A c b (does not occur in either
  1 or 2, therefore it is
  an incorrect sequence)

34

• The correct sequence is consequently ABC (because
  both cross over types are theoretically
  possible).
• Rewrite the table with the correct sequence
• Phenotype observed /
• ABc 1828 3607 Parental types
• abC 1779 83.1
• ABC 148 287 Cross over 1
• abc 139 6.6
• AbC 227 444 Cross over 2
• aBc 217 10.2
• TOTAL 4338

35

• Make a schematic representation of the DNA with
  the genes indicated on it
• A B C
• Calculate the distance between the genes.
• A and B occur together in parental types.
• Cross over between A and B number of cross
  over that occurred between them (i.e. cases where
  Ab or aB occur together)
• 10.2 ? 10.2cM.
• B and C ? 6.6cM.
• A 10.2cM B 6.6cM C

36

• In this example some phenotypic classes are
  absent (aBC and Abc are absent).
• They represent DCO events ? i.e., cross over
  between A and B, as well as a cross over between
  B and C.
• The frequency of DCO is very low.
• Calculations in cases where DCO occur, is done in
  a similar fashion, with the exception that only
  the DCO is used to determine the sequence of the
  genus.

37

• Suppose we obtain the following results from gene
  mapping
• A 17.2cM B 11.6cM C
• Crossover between A and B occur in 17.2 of cases
  ? probability 17.2/100 0.172.
• Probability of DCO probability of crossover
  between A and B, as well as between B and C ?
  therefore 0.172 x 0.116 0.02.
• Sometimes synapses on one point of chromosome
  influence synapses on another point ? to
  determine degree of chiasma interference
• C (observed frequency of DCO) / (expected
  frequency of DCO)
• I 1 C
• where C equals the coefficient of similarity and
  I represents the chiasma interference.

38
Problem

• Use the following linkage data to construct a
  chromosome map of these loci
• Loci Map units
• a-b 10
• a-c 16
• a-d 3
• a-e 8
• b-c 6
• b-d 13
• b-e 2

39
Problem

• A cross between individuals with Lg N phenotype
  lg n phenotype produced F1 with Lg N phenotype.
  F1 crossed with individuals with lg n phenotypes
  and following results observed
• Progeny Number
• Lg N 180
• Lg n 50
• lg N 60
• lg n 190
• Total 480
• a) Are these genes linked?
• b) If they are linked, what were the genotypes
  of the original parental individuals?

40
Problem

• Several female Drosophila heterozygous for four
  mutations were test-crossed. The mutations and
  their map positions were engrailed (en ) map
  position 62.0, scabrous (sca ) 66.7, droopy (dr )
  71.2, and curved (cu ) 75.5.
• a) If the genotype of the original heterozygotes
  was en sca dr cu / , how many different
  phenotypic classes would you expect to find in
  the progeny?
• b) If 25,000 progeny were examined, how many
  progeny with an engrailed, droopy phenotype
  would you expect to find, if there is no
  interference?

41
Problem

• This map shows the distances between three loci
  (v, b and lg ) in corn. If individuals with a v
  b lg / genotype were test-crossed, what
  classes of phenotypes would you expect to find in
  the progeny, and how may of each would you expect
  if you examined 1000 progeny?
• v_________b________lg
• 18 28

42
Problem

• A tomato plant heterozygous for alleles at three
  loci (Aa Bb Cc ) was test-crossed to an aa bb cc
  plant. From the progeny listed here, which loci
  would you say are linked?
• Phenotype Number
• ABC 2250
• ABc 2225
• AbC 240
• Abc 255
• aBC 260
• aBc 250
• abC 2250
• abc 2275

43

• A wild-type Drosophila with gray body and
  straight wings was crossed with a black-bodied,
  curled winged fly. F1 all gray bodied with
  straight wings. F1 back-crossed to black bodied,
  curled-winged flies. Progeny as shown here
• Cross Progeny Number
• F1 female x black curled male
• Black curled 762
• Black straight 210
• Gray curled 234
• Gray straight 784
• F1 male x black curled female
• Black curled 1013
• Black straight 0
• Gray curled 0
• Gray straight 987
• a) What is the map distance between these loci?
• b) What do you conclude about crossing over in
  male Drosophila?

44

• From a three-point test-cross mapping experiment,
  the following gamete genotype frequencies were
  obtained
• XYZ 365 xyz 367
• xYz 110 XyZ 105
• xYZ 3 Xyz 4
• XYz 25 xyZ 21
• TOTAL 1000
• From these data, what can be said of the genes?
• a) All are unlinked.
• b) XYZ is the gene order.
• c) YZX is the gene order.
• d) YXZ is the gene order.
• e) Two of the genes are linked, one is
  independently assorting.

45

• Stern Bridges crossed stock carrying dominant
  eye mutation star on the 2nd chromosome to stock
  homozygous for the 2nd chromosome recessive
  mutations aristaless and dumpy. The F1 star
  females then back-crossed to homozygous
  aristaless dumpy males and following phenotypes
  observed
• Phenotype Number
• aristaless dumpy 918
• star 956
• aristaless star 7
• dumpy 5
• aristaless 132
• star dumpy 100
• a) What are the recombination distances and the
  order of loci for these three genes?
• b) What classes of phenotypes are missing and
  why?

46

• In maize recessive mutant genes bm (brown
  midrib), v (virescent seedling) and pr (purple
  aleurone) are known. A plant heterozygous for
  all these genes are crossed with homozygous
  recessive one for these genes and following
  results observed
• Phenotype Number
• 4
• bm 182
• v 1200
• pr 2237
• pr v 79
• pr bm 1195
• v bm 2230
• pr v bm 3
• Determine whether these genes are linked. If
  they are linked, what is the order of the genes
  and the distances between them? Does chiasma
  interference occur? What are the genotypes of
  the parental types?

47

• In the nematode Caenorhabditis elegans, d
  (dumpy), u (uncoordinated) and k (knobby) are all
  recessive genes located on same chromosome.
  Females heterozygous normal for all three traits
  mated with dumpy, uncoordinated, knobby males and
  following progeny observed
• Phenotypes Number
• dumpy uncoordinated knobby 3
• uncoordinated knobby 392
• knobby 34
• uncoordinated 61
• dumpy uncoordinated 32
• dumpy knobby 65
• dumpy 410
• wild-type 3
• Total 1000

48
Problem

• In the heterozygous normal females, what is the
  cis/trans arrangement of the following pairs of
  genes d u, d k, u k ?
• Determine whether these genes are linked. If
  they are linked, what is the order of the genes
  and the distances between them?

49

• In Drosophila a dominant mutation, dichaete (D ),
  and two recessive mutations, pink (p ) and ebony
  (e ) are possibly linked. The F1s obtained from
  cross of dichaete line with pink and ebony line
  are crossed to homozygous recessive line and
  following results were obtained
• Phenotype Number
• ebony 8
• pink 187
• dichaete 8240
• pink, ebony 8781
• dichaete, ebony 214
• dichaete, pink 4
• dichaete, pink, ebony 28
• wild type 31
• Determine whether these genes are linked. If
  they are linked, what is the order of the genes
  and the distances between them? Does chiasma
  interference occur?

50

• Yellow versus gray body colour in Drosophila is
  determined by the alleles y and , vermilion
  versus wild-type eyes by the alleles v and , and
  singed versus straight bristles by the alleles sn
  and . When females heterozygous for each of
  these X-linked genes were test-crossed with
  yellow, vermilion, singed males, the following
  classes and numbers of progeny were obtained
• Phenotypes Number
• yellow, vermilion, singed 53
• yellow, vermilion 108
• yellow, singed 331
• yellow 5
• vermilion, singed 3
• vermilion 342
• singed 95
• wild-type 63 (cont.)

51

• What is the order of the three genes (y, v, sn )?
  
• Construct a linkage map with the genes in their
  correct order, and indicate the map distance
  between the genes.
• How does the frequency of double crossovers
  observed in this experiment compare with the
  frequency expected if crossing-over occurs
  independently in the two chromosome regions?
• Determine the coefficient of coincidence and the
  interference.