Linkage  Mapping  in  Drosophila  Melanogaster  Genetics:  Fall  2012  

Joshua  Hanau      

Introduction:      

An  experiment  was  performed  in  order  to  determine  the  presence  and  

degree  of  gene  linkage  in  Drosophila  Melanogaster.  Gene  linkage  describes  whether  

or  not  two  or  more  genes  are  located  on  the  same  chromosome  in  a  eukaryotic  

individual.  The  degree  of  gene  linkage  can  describe  the  relative  distance  between  

two  linked  genes  on  a  single  chromosome.        

 

Drosophila  Melanogaster,  which  is  also  known  by  the  common  name  “Fruit  

Fly,”  is  an  ideal  organism  to  use  for  genetic  studies.  Drosophila  proliferates  rapidly,  

are  small,  are  cheap  to  maintain,  and  their  phenotypes  are  easily  observable.  

Drosophila  males  and  females  display  numerous  gender  dimorphisms,  which  enable  

gender  segregation,  and  ultimately,  controlled  mating.  Females  have  a  striped  black  

posterior,  on  the  dorsal  side  of  their  abdomen,  while  males  have  a  solid  black  

posterior,  on  the  dorsal  side  of  their  abdomen.  Females  have  no  external  genitalia,  

on  their  ventral  posterior  end,  while  males  do  have  external  genitalia,  on  their  

ventral  posterior  end.  Females  have  a  larger  elongated  abdomen,  when  compared  to  

males.  Males  have  sex  combs  on  their  anterior  appendages,  while  females  do  not.  

These  dimorphisms  (Figure  1)  enable  females  to  be  separated  from  males  before  

they  are  ready  to  mate,  thus  enabling  controlled  mating.    

 

 

 

 

Figure  1:  Ventral  Gender  Dimorphisms  in  Drosophila  [I]  

 

 

A  three-­‐factor  cross  was  the  method  used  to  determine  whether  or  not  three  

drosophila  genes  were  linked,  and  if  so,  to  what  degree.  The  three  genes  that  were  

analyzed  displayed  different  phenotypes  when  they  were  in  mutant  and  wild  type  

forms.  The  first  of  these  genes  corresponded  to  the  fly’s  eye-­‐color.  While  wild  type  

flies  have  dull  red  colored  eyes,  the  mutants  displayed  brighter  orange  colored  eyes.  

The  second  gene  analyzed  corresponded  to  the  bristles  located  on  the  dorsal  side  of  

the  fly’s  thorax.  Wild  type  flies  had  long  straight  bristles,  while  the  mutants  had  

short  and  curled  bristled.  The  third  gene  analyzed  corresponded  to  the  body  color  

being  displayed  on  the  dorsal  side  of  the  fly’s  abdominal  section.  The  wild  type  flies  

displayed  a  regular  pattern  of  alternating  black  and  tan  sections,  which  differed  in  

males  and  females  (Figure  2).  The  mutants,  displayed  the  same  basic  pattern,  only  

the  black  sections  were  faded,  and  less  pronounced  than  they  were  in  the  wild  type  

strain.    

 

Figure  2:  Body  Color  Pattern  in  Wild  Type  Drosophila  [II]  

 

   

After  performing  the  three-­‐factor  cross,  the  ratio  of  the  F2  progeny  (Filial  

generation  #2)  phenotypes  could  be  used  to  determine  whether  or  not  the  three  

genes  were  linked,  and  if  they  were,  to  determine  the  degree  of  the  linkage.  This  

data  could  then  be  compared  to  a  map  of  known  drosophila  genes,  and  the  identity  

of  the  genes  could  potentially  be  inferred  from  the  literature.      

 

 

Methods:  

  At  the  start  of  the  experiment,  two  strains  of  flies  were  crossed.  The  first  

strain  was  a  true-­‐breeding  wild  type  stock,  which  proves  that  it  was  homozygous  for  

all  three  wild  type  alleles,  or    +/+  ;  +/+  ;  +/+,  where  “+”  indicates  the  wild  type  allele  

for  any  given  gene.  The  second  strain  was  a  true  breeding  triple  mutant,  which  

proves  it  was  homozygous  for  all  three  mutant  alleles,  or  e/e  ;  br/br  ;  bc/bc,  where  

“e”  indicates  the  unknown  mutant  eye  color  allele,  “br”  indicates  the  unknown  

mutant  bristle  allele,  and  “bc”  indicates  the  unknown  mutant  body  color  allele.    

 

  To  perform  a  cross,  virgin  females  of  one  strain  must  be  placed  in  a  vial  with  

males  of  the  other  strain.  Virgin  females  can  be  identified  by  their  pale  coloring,  

their  folded  wings,  their  enlarged  abdomens,  and  by  the  presence  of  a  meconium  

(Figure  3).  Not  all  four  of  these  signs  are  required,  but  the  presence  of  a  meconium  

is  the  most  important  sign  of  recent  eclosion  (emergence  from  the  pupa  casing),  and  

the  best  guarantee  of  a  female  fly  being  a  virgin,  since  drosophila  females  do  not  

mate  for  the  first  few  hours  (around  8  on  average)  after  eclosion.  Using  virgin  

females  in  all  crosses  is  crucial.  Female  drosophila  can  only  mate  with  one  male,  and  

they  then  store  that  male’s  sperm  for  the  rest  of  their  adult  life  cycle.  Therefore,  in  

order  to  control  mating,  females  must  be  isolated  soon  after  they  eclose  from  their  

pupa  casing,  while  they  still  exhibit  the  features  listed  above,  and  have  a  guaranteed  

virgin  status.    

 

     

Figure  3:  Virgin  Female  Drosophila  Image  [III]  

 

 

  The  general  pattern  of  a  cross  was  as  follows  (Figure  4).  Once  4-­‐5  virgin  

females  were  isolated,  they  were  placed  in  a  vial  with  8-­‐10  males  from  the  other  

cross  strain.  The  flies  would  mate,  and  the  females  began  laying  eggs  shortly  

thereafter.  After  the  eggs  were  laid,  the  larva,  or  1st  instars,  would  emerge  within  a  

day  or  so.  The  larva  would  continue  to  grow,  and  pass  through  the  2nd  and  3rd  instar  

phases.  During  the  3rd  instar  phase,  the  larva  would  climb  up  the  wall  of  the  vial  and  

form  a  pupa  case.  Around  four  days  later,  and  around  10  days  after  the  cross  was  

started,  a  new  generation  of  adult  flies  would  emerge.  The  parental  generation  was  

removed  from  the  vial  by  this  point,  to  ensure  that  the  two  generations  were  kept  

separate.        

 

Figure  4:  Drosophila  Life  Cycle  [IV]  

 

 

  Assuming  that  none  of  the  three  genes  are  sex  linked,  the  progeny  of  this  

initial  cross  would  all  display  wild  type  phenotypes,  since  the  wild  type  is  

presumably  dominant,  and  all  the  offspring  would  be  heterozygous  for  all  three  

genes  (+/e  ;  +/br  ;  +/bc).  However,  if  one  or  more  genes  are  sex  linked,  then  there  is  

a  possibility  that  the  male  offspring,  who  are  hemizygous  for  x  linked  genes,  would  

display  the  mutant  phenotype.  To  test  for  sex  linkage,  reciprocal  crosses  were  

performed.    This  means  that  there  were  two  crosses  performed  between  the  two  

strains,  with  the  genders  of  each  strain  alternated  in  the  two  crosses.  One  cross  was  

between  wild  type  males  and  female  mutants,  while  the  other  was  between  male  

mutants  and  wild  type  females.  If  one  or  more  genes  were  sex  linked,  then  the  male  

F1  generation  (Filial  generation  #1)  offspring  of  the  reciprocal  crosses  would  always  

have  the  same  phenotype  as  their  mothers  for  that  one  or  more  genes,  instead  of  

simply  having  the  wild  type  phenotype  that  would  be  expected  for  all  F1  flies.    

 

  The  second  cross  was  a  test  cross.  A  test  cross  is  when  heterozygous  

individuals  (F1  virgin  females)  are  crossed  with  double  mutant  individuals  (Parental  

mutant  males).  This  cross  is  performed  in  order  to  determine  whether  or  not  the  

genes  assort  independently.  If  the  genes  assorted  independently,  then  all  8  potential  

phenotype  combinations  (2  phenotypes  for  3  genes.  23  =  8)  would  ideally  show  up  

with  the  same  frequency,  12.5%,  in  the  F2  generation.  If  the  parental  phenotypes  are  

displayed  the  majority  of  the  time,  or  far  greater  than  the  expected  12.5%  (i.e.  no  

need  for  chi  squared  analysis),  then  it  can  be  assumed  that  the  three  genes  are  

linked.  Since  linked  genes  are  on  the  same  chromosome,  it  makes  sense  that  the  

genes  are  inherited  together  the  majority  of  the  time,  and  it  explains  the  lack  of  

independent  assortment.  Additionally,  any  phenotypes  that  are  not  parental,  in  the  

linkage  scenario,  must  be  the  result  of  crossing  over  during  meiosis,  which  is  also  

known  as  meiotic  recombination.  Thomas  Hunt  Morgan’s  research  showed  that  the  

frequency  of  recombination  between  two  genes  is  directly  proportional  to  the  

physical  distance  between  them.  Thus,  using  recombinant  frequencies,  an  RF  map  of  

the  chromosome,  which  is  similar  to  the  physical  map  (but  not  identical),  can  be  

constructed  using  recombinant  frequencies.  The  progression  of  crosses  performed  

is  shown  below  (Figure  5).      

Figure  5:  Progression  of  Crosses  Performed  

 

 

 

 

 

 

 

 

Results:  

  The  results  of  the  reciprocal  crosses  suggest  that  all  three  genes  are  sex  

linked,  since  the  male  offspring  of  the  cross  with  the  mutant  phenotype  females  

were  mutant  for  all  three  genes.  By  extension,  all  three  genes  are  linked,  since  there  

is  only  one  sex  chromosome.  Therefore,  the  ratio  of  F2  phenotypes  were  expected  to,  

and  in  fact  did,  deviate  from  the  1:1:1:1:1:1:1:1  ratio  that  would  be  expected  in  a  

case  of  independent  assortment.  The  number  and  percentage  of  each  phenotype  is  

displayed  below  (Table  1).  In  addition  to  there  being  eight  possible  phenotypes,  

there  are  eight  possible  genotypes  that  can  result  from  this  type  of  test  cross.  Each  

genotype  corresponds  to  a  phenotype,  where  the  wild  type  allele  (+)  is  always  

P:   ♂ +/+ ; +/+ ; +/+ x ♀ e/e ; br/br ; bc/bc + reciprocal cross        

F1:   ♂ e/e ; br/br; bc/bc x ♀ +/e ; +/br ; +/bc  

F2:   Eight  possible  phenotypes  (See  results  section)    

expressed  over  the  mutant  type.  Eight  testcrosses  were  performed,  resulting  in  a  

total  of  515  progeny.  

Table  1:  F2  Progeny  Ratios  

Genotype/Phenotype   Number  of  F2  Progeny   %  of  Total  Progeny  

1:  +/e  ;  +/br  ;  +/bc   181   35.1%  

2:  e/e  ;  br/br  ;  bc/bc   186   36.1%  

3:  +/e  ;  +/br  ;  bc/bc   12   2.3%  

4:  e/e  ;  br/br  ;  +/bc   21   4.1%  

5:  +/e  ;  br/br  ;  +/bc   48   9.3%  

6:  e/e  ;  +/br  ;  bc/bc   63   12.2%  

7:  e/e  ;  +/br  ;  +/bc   2   0.4%  

8:  +/e  ;  br/br  ;  bc/bc   2   0.4%  

 

Using  these  ratios  a  recombination  map  of  the  X  chromosome  (because  they  

are  sex  linked)  can  be  constructed.  For  example,  in  the  parental  phenotypes  (#’s  

1+2)  the  gene  for  eye  color  and  bristles  are  the  same  type  of  allele  (both  wild,  or  

both  mutant).  Thus,  any  offspring  where  they  display  different  types  of  alleles  on  the  

maternal  homologue  must  be  a  case  where  recombination  took  place  between  the  

two  genes.  Thus,  recombination  took  place  between  these  two  genes  in  progeny  

numbers  5,6,7,  and  8.  By  adding  up  the  percentages  of  recombinants,  the  total  

recombination  frequency  between  the  two  genes  can  be  calculated  to  be  22.3%.  

Therefore,  the  genes  for  eye  color  and  bristles  should  be  located  around  22.3  map  

units  (map  units)  apart  on  an  RF  map.  The  distance  between  the  other  sets  of  genes  

can  be  calculated  in  the  same  way  (Table  2).  The  one  small  point  to  note  is  that  in  

between  “br”  and  bc,”  the  percentage  of  double  recombinants  needs  to  be  counted  

twice,  since  two  crossover  events  took  place  between  these  two  genes  in  those  

scenarios.    

Table  2:  Recombinant  Frequencies  

Alleles  of  Interest   Calculation   Recombinant  Frequency  

“e”  and  “br”   9.3  +  12.2  +  0.4  +  0.4   22.3%  (m.u.)  

“e”  and  “bc”   2.3  +  4.1  +  0.4  +  0.4   7.2%  (m.u.)  

“br”  and  “bc”   2.3  +  4.1  +  9.3  +  12.2  +  

2(0.4)  +  2(0.4)  

29.5%  (m.u.)  

 

Using  the  tabulated  data,  an  RF  map  for  these  three  genes  can  be  constructed  

(Figure  6).  A  useful  trick  for  constructing  a  recombinant  map  is  to  look  for  the  

“flipped  allele”  in  the  least  common  phenotype,  which  is  always  the  result  of  the  

double  crossover.    This  flipped  allele  in  the  double  recombinant  will  always  be  

located  between  the  other  two  genes  [V].  Since  the  allele  for  eye  color  differs  from  

the  two  parental  strains  in  the  double  recombinant,  the  gene  for  eye  color  must  be  

located  between  the  other  two  genes  on  the  recombinant  map.    

Figure  6:  RF  Map  of  Three  Genes  of  Interest  

 

 

Conclusions:    

  By  comparing  these  results  to  a  map  of  known  genes  on  the  drosophila  X  

chromosome  (Figure  7),  the  identity  of  the  three  genes  can  be  inferred.  The  three  

genes  must  correspond  to  eye  color,  bristle  shape,  and  body  color.  The  distances  

between  the  three  known  genes  must  also  be  similar  to  the  distances  calculated  

between  the  three  unknown  genes  in  this  lab  report.    

 

Figure  7:  RF  Map  of  Drosophila  “X”  Chromosome  [VI]  

 

     

This  small  segment  of  the  Drosophila  X  chromosome  contains  the  most  likely  

candidates  for  our  three  unknown  genes.  The  calculated  distance  between  our  

bristle  and  body  color  mutations  is  29.5  map  units,  while  in  this  map,  the  distance  

between  “forked  bristles”  and  “tan  bodies”  is  29.2  maps  units.  Since  both  of  these  

mutations  sound  like  they  could  correspond  with  our  mutant  phenotypes,  it  is  

reasonable  to  conclude  that  the  two  genes  that  contain  these  two  mutant  alleles  are  

the  same  as  our  first  unknown  genes.  Our  eye  color  mutation  is  located  7.2  maps  

units  away  from  our  body  color  mutation,  and  22.3  map  units  away  from  our  bristle  

mutation.  On  this  map  of  known  genes,  the  mutation  for  “Vermillion  eyes”  is  located  

5.5  maps  units  away  from  the  “tan  body”  mutation,  and  23.7  map  units  away  from  

the  “forked  bristles”  mutation.  These  results  are  similar  enough  to  conclude  that  our  

eye  color  mutation  might  be  caused  by  a  mutant  “vermillion”  allele.  Thus,  it  is  

reasonable  to  conclude  that  our  three  unknown  genes,  correspond  with  the  genes  

whose  mutated  forms  result  in  the  “tan  body,”  “Forked  bristles,”  and  “vermillion  

eyes”  mutations.      

 

This  inference  was  much  easier  to  make  due  to  the  use  of  a  reciprocal  cross.  

The  reciprocal  crosses  showed  that  all  three  genes  were  sex  linked,  which  led  to  the  

conclusion  that  all  three  genes  were  located  on  the  X  chromosome.  This  narrowed  

down  the  search  for  the  identity  of  the  unknown  alleles  considerably.    

 

Although  the  results  were  pretty  close  (assuming  these  are  the  correct  

genes),  there  was  some  error,  which  has  to  be  explained.  The  most  obvious  cause  is  

human  error.  It  was  not  always  easy  to  distinguish  the  mutant  body  color  from  the  

wild  type.  This  may  have  resulted  in  incorrect  phenotype  tallying,  and  therefore,  

faulty  recombinant  frequencies.  The  same  is  true  for  eye  color,  which  was  not  

always  100%  discernable.  The  bristle  mutation  was  the  only  one  that  was  

completely  obvious  in  all  cases.    

 

Another  source  of  error  may  have  been  due  to  a  failed  separation  of  

generations.  In  one  case,  flies  were  scored  from  a  vial  22  days  after  the  cross  was  set  

up.  It  is  possible  that  some  of  the  flies  that  were  counted  belonged  to  the  F3  

generation,  and  may  have  interfered  with  the  results.    

 

To  verify  the  identity  of  the  genes  a  complementation  test  could  be  

performed.  Flies  that  are  known  to  be  heterozygous  for  the  three  genes  mentioned  

above  could  be  crossed  with  the  unknown  mutant  parental  strain.  If  the  result  were  

a  wild  type  phenotype  (or  phenotypes),  then  the  hypothesis  of  the  genes  identity  of  

that  gene  (or  genes)  must  be  incorrect,  since  complementation  suggests  the  

mutations  are  on  different  genes.    If  mutant  phenotype  flies  were  prevalent  in  50%  

of  the  offspring,  for  one  or  all  three  genes,  then  the  identity  of  that  mutant  allele,  or  

alleles,  will  be  confirmed.    

 

 

Citations:  

[I]:  http://flymove.uni-­‐

muenster.de/Genetics/Flies/MaleFemale/MaleFemalepage.html?http&&&flymove.u

ni-­‐muenster.de/Genetics/Flies/MaleFemale/MaleFemaleTxt.html  

[II]:  http://www.ableweb.org/volumes/vol-­‐10/15-­‐pollock.pdf  

[III]:  See  [I]  

[IV]:  See  [II]  

[V]:  Griffiths,  Anthony;  Wessler,  Susan;  Carroll,  Sean;  Doebley,  John  (2012).  

Introduction  to  Genetic  Analysis-­‐  10th  Edition.  New  York  City:  W.H.  Freeman  and  

Company.  Pages  134-­‐135.      

[VI]:  http://www.blacksage.com/APBiology/Handouts/DrosophilaMap.htm