Transcript Document

LECTURE PRESENTATIONS
For CAMPBELL BIOLOGY, NINTH EDITION
Jane B. Reece, Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Robert B. Jackson
Chapter 14
Mendel and the Gene Idea
Lectures by
Erin Barley
Kathleen Fitzpatrick
© 2011 Pearson Education, Inc.
Genetics is Like Drawing from the Deck of
Genes.
The BLENDING hypothesis:
• idea that genetic material from the two parents
blends together (like blue and yellow paint blend
to make green)
© 2011 Pearson Education, Inc.
The “Particulate” hypothesis
• the idea that parents pass on discrete
heritable units (genes)
• can explain the reappearance of traits after
several generations
• Mendel!!!
http://www.youtube.com/watch?v=YxKFdQo10r
E
© 2011 Pearson Education, Inc.
Figure 14.1
Mendel used the scientific approach to
identify two laws of inheritance
• Used garden peas
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Why peas??
The advantages…
• many varieties with distinct
heritable features
• Mating can be controlled
– Each flower produces
sperm and egg
– Can be Cross-pollinated
(by dusting)
© 2011 Pearson Education, Inc.
Technique:
Figure 14.2a
TECHNIQUE
1
2
Parental
generation
(P)
Stamens
3
Carpel
4
Figure 14.2b
RESULTS
First filial
generation
offspring
(F1)
5
• Mendel also used true-breeding plants
– plants that produce offspring of the same variety
when they self-pollinate
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Figure 14.3-1
EXPERIMENT
P Generation
(true-breeding
parents)
Purple
flowers
White
flowers
Figure 14.3-2
EXPERIMENT
P Generation
(true-breeding
parents)
F1 Generation
(hybrids)
Purple
flowers
White
flowers
All plants had purple flowers
Self- or cross-pollination
Figure 14.3-3
EXPERIMENT
P Generation
(true-breeding
parents)
Purple
flowers
White
flowers
F1 Generation
(hybrids)
All plants had purple flowers
Self- or cross-pollination
F2 Generation
705 purpleflowered
plants
224 white
flowered
plants
Which color was dominant and which recessive?
• The factor for white flowers didn’t disappear,
because it reappeared in the F2 generation
• “heritable factor” = what we now call a gene
© 2011 Pearson Education, Inc.
Table 14.1
Mendel’s Model
• Mendel noticed a 3:1 (ratio) inheritance pattern
in his breeding experiments in F2 offspring
(F2 = 2nd generation)
• He developed a hypothesis to explain it
4 concepts make up this model
© 2011 Pearson Education, Inc.
1. Different versions of genes exist
• alternative versions of a gene = alleles
• they each have a locus (specific location on
a chromosome)
© 2011 Pearson Education, Inc.
Figure 14.4
Allele for purple flowers
Locus for flower-color gene
Pair of
homologous
chromosomes
Allele for white flowers
2. Organisms inherit two alleles for each
character
• one from each parent
• 2 alleles may be identical or may be different
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3. Dominance exists
• if the two alleles at a locus differ…
– dominant allele - determines the organism’s
appearance
– recessive allele - has no noticeable effect on
appearance
© 2011 Pearson Education, Inc.
4. Law of Segregation (one of Mendel’s Laws)
• the two alleles for a heritable character
separate (segregate) during gamete
formation
– end up in different gametes
– egg or a sperm gets only one of the two alleles
that are present in the organism
• meiosis!!
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Punnett Squares - used to predict
results of genetic crosses between
individuals known genetic makeup
• capital letter = dominant allele
• lowercase letter = recessive allele
© 2011 Pearson Education, Inc.
Figure 14.5-1
P Generation
Appearance:
Purple flowers White flowers
Genetic makeup:
pp
PP
p
Gametes:
P
Figure 14.5-2
P Generation
Appearance:
Purple flowers White flowers
Genetic makeup:
pp
PP
p
Gametes:
P
F1 Generation
Appearance:
Genetic makeup:
Gametes:
Purple flowers
Pp
1/
1/
2 p
2 P
Figure 14.5-3
P Generation
Appearance:
Purple flowers White flowers
Genetic makeup:
pp
PP
p
Gametes:
P
F1 Generation
Appearance:
Genetic makeup:
Gametes:
Purple flowers
Pp
1/
1/
2 p
2 P
Sperm from F1 (Pp) plant
F2 Generation
P
Eggs from
F1 (Pp) plant
p
3
P
p
PP
Pp
Pp
pp
:1
Useful Genetic Vocabulary
• homozygous = two identical alleles for a gene
(bb or BB)
• heterozygous = two different alleles for a gene
(Bb)
• heterozygotes are not true-breeding
• Homozygotes are true-breeding
© 2011 Pearson Education, Inc.
Sometimes, traits don’t reveal genetics
• Phenotype = physical appearance
• Genotype = genetic makeup
• For example, in pea plants,
– PP and Pp plants have the same phenotype (purple)
but different genotypes
© 2011 Pearson Education, Inc.
Figure 14.6
3
Phenotype
Genotype
Purple
PP
(homozygous)
Purple
Pp
(heterozygous)
1
2
1
Purple
Pp
(heterozygous)
White
pp
(homozygous)
Ratio 3:1
Ratio 1:2:1
1
The Testcross = what we do to try and
determine the genotype of a mystery
individual
• How is a test cross performed?
– breed the mystery individual with a homozygous
recessive individual
– If any offspring display the recessive phenotype,
the mystery parent must be heterozygous
© 2011 Pearson Education, Inc.
Figure 14.7
TECHNIQUE
Dominant phenotype,
unknown genotype:
PP or Pp?
Predictions
If purple-flowered
parent is PP
Sperm
p
p
Recessive phenotype,
known genotype:
pp
or
If purple-flowered
parent is Pp
Sperm
p
p
P
Pp
Eggs
P
Pp
Eggs
P
p
Pp
Pp
Pp
Pp
pp
pp
RESULTS
or
All offspring purple
1/
2
offspring purple and
1/ offspring white
2
Law of Independent Assortment:
(another one of Mendel’s Laws)
• each pair of alleles segregates independently
of each other pair of alleles during gamete
formation
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• Mendel identified his second law of inheritance
by following two characters at the same time
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• Mendel produced dihybrids (BbFf) by
crossing two true-breeding parents differing in
two characters (bbff x BBFF)
He then performed a dihybrid cross between
these individuals.
© 2011 Pearson Education, Inc.
What can a dihybrid cross show us
about the genetics of the individuals?
• can determine whether two characters are
transmitted to offspring as a package or
independently
Figure 14.8
EXPERIMENT
YYRR
P Generation
yyrr
yr
Gametes YR
F1 Generation
Predictions
YyRr
Hypothesis of
dependent assortment
Hypothesis of
independent assortment
Sperm
or
Predicted
offspring of
F2 generation
1/
Sperm
1/
2
YR
1/
2
2
YR
YyRr
YYRR
Eggs
1/
2
1/
4
YR
4
Yr
4
yR
4
yr
Eggs
yr
YyRr
3/
yyrr
1/
4
YR
1/
4
1/
Yr
4
yR
1/
4
yr
yr
1/
1/
4
1/
YYRR
YYRr
YyRR
YyRr
YYRr
YYrr
YyRr
Yyrr
YyRR
YyRr
yyRR
yyRr
YyRr
Yyrr
yyRr
yyrr
4
Phenotypic ratio 3:1
1/
9/
16
3/
16
3/
16
1/
16
Phenotypic ratio 9:3:3:1
RESULTS
315
108
101
32
Phenotypic ratio approximately 9:3:3:1
The Phenotypic Ratios he found when he
performed monohybrid and dihybrid
crosses:
• Monohybrid cross (Rr x Rr)
–3:1
• Dihybrid Cross (YyRr x YyRr)
–9:3:3:1
These become expected ratios
for hybrid crosses
Probability is KING when it comes to
Mendel’s genetics.
• Mendel’s laws reflect the rules of probability
• When tossing a coin, the outcome of one toss
has no impact on the outcome of the next toss
– Just like how the alleles of one gene segregate
into gametes independently of another gene’s
alleles
© 2011 Pearson Education, Inc.
The Multiplication and Addition Rules and
Monohybrid Crosses
• multiplication rule - probability that two or more
independent events will occur together is the
product of their individual probabilities
• addition rule - probability that any one of two or
more exclusive events will occur is calculated by
adding together their individual probabilities
© 2011 Pearson Education, Inc.
Figure 14.9
Rr
Segregation of
alleles into eggs

Rr
Segregation of
alleles into sperm
Sperm
1/
R
2
2
Eggs
4
r
2
r
R
R
1/
1/
r
2
R
R
1/
1/
1/
4
r
r
R
r
1/
4
1/
4
Solving Complex Genetics Problems with the
Rules of Probability
• each character is considered separately, and
then the individual probabilities are multiplied
© 2011 Pearson Education, Inc.
Figure 14.UN01
Probability of YYRR  1/4 (probability of YY)  1/4 (RR)  1/16
Probability of YyRR  1/2 (Yy)
 1/4 (RR)  1/8
Figure 14.UN02
ppyyRr
ppYyrr
Ppyyrr
PPyyrr
ppyyrr
1/ (yy)  1/ (Rr)
(probability
of
pp)

4
2
2
1/  1/  1/
4
2
2
1/  1/  1/
2
2
2
1/  1/  1/
4
2
2
1/  1/  1/
4
2
2
1/
Chance of at least two recessive traits
 1/16
 1/16
 2/16
 1/16
 1/16
 6/16 or 3/8
Inheritance patterns are often more complex
than predicted by simple Mendelian genetics
• basic principles of segregation and independent
assortment still apply
© 2011 Pearson Education, Inc.
Inheritance of characters by a single gene
may deviate when…
– alleles are not completely dominant or recessive
– a gene has more than two alleles
– a gene produces multiple phenotypes
© 2011 Pearson Education, Inc.
Complete Dominance
• phenotypes of the heterozygote and dominant
homozygote are identical
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Incomplete Dominance
• phenotype of F1
hybrids is
somewhere
between the
phenotypes of the
two parental
varieties
• (think “pink”omplete)
Figure 14.10-1
P Generation
White
CWCW
Red
CRCR
Gametes
CR
CW
Figure 14.10-2
P Generation
White
CWCW
Red
CRCR
Gametes
CR
CW
F1 Generation
Gametes 1/2 CR
Pink
CRCW
1/
2
CW
Figure 14.10-3
P Generation
White
CWCW
Red
CRCR
CR
Gametes
CW
F1 Generation
Pink
CRCW
1/
Gametes 1/2 CR
2
CW
Sperm
1/
F2 Generation
1/
2
CR
1/
2
CW
Eggs
2
CR
1/
2
CW
CRCR CRCW
CRCW CWCW
•
Pooh had a colony of tiggers whose stripes went
across the body. His American pen-pal, Yogi, sent
him a Tigger whose stripes ran lengthwise.
• When Pooh crossed it with one of his own animals, he
obtained plaid tiggers.
• Interbreeding among the plaid Tiggers produced
litters of a majority of plaid members, but some
crosswise- and lengthwise-striped animals were also
produced.
• Diagram the crosses made Pooh, showing the
genotypes of the Tiggers which account for the coat
patterns observed.
Codominance
• two dominant alleles affect the phenotype in
separate, distinguishable ways
Multiple Alleles
• Most genes exist in populations in more than two
allelic forms
• For example, the ABO blood group in humans
– determined by three alleles for the enzyme (I) that
attaches A or B carbohydrates to red blood cells: IA,
IB, and i.
• IA allele adds the A carbohydrate
• IB allele adds the B carbohydrate
• i allele adds neither
© 2011 Pearson Education, Inc.
Figure 14.11
(a) The three alleles for the ABO blood groups and their
carbohydrates
IA
Allele
Carbohydrate
IB
i
none
B
A
(b) Blood group genotypes and phenotypes
Genotype
IAIA or IAi
IBIB or IBi
IAIB
ii
A
B
AB
O
Red blood cell
appearance
Phenotype
(blood group)
Blood type problem
Determine the genotypes of the lettered individuals
Sex-linked gene
• gene located on either sex chromosome
• Most are X-linked
• Follow a different inheritance pattern
© 2011 Pearson Education, Inc.
Figure 15.5
X
Y
Figure 15.6
44 
XY
Different
organisms
sex
determination
44 
XX
Parents
22 
22 
X or Y
22 
X
Sperm
Egg
44 
XX
or
44 
XY
(a) The X-Y system Zygotes (offspring)
22 
XX
22 
X
76 
ZW
76 
ZZ
32
(Diploid)
16
(Haploid)
(b) The X-0 system
(c) The Z-W system
(d) The haplo-diploid system
• For a recessive X-linked trait to be expressed
– A female needs two copies of the allele
(homozygous)
– A male needs only one copy of the allele
(hemizygous)
• X-linked recessive disorders are much more
common in males than in females
© 2011 Pearson Education, Inc.
Figure 15.7
XNXN
Sperm Xn
XNXn
XnY
Sperm XN
Y
XNY
XNXn
Sperm Xn
Y
XnY
Y
Eggs XN
XNXn XNY
Eggs XN
XNXN XNY
Eggs XN
XNXn XNY
XN
XNXn XNY
Xn
XNXn XnY
Xn
XnXn XnY
(a)
(b)
(c)
X-linked human disorder
– Color blindness (mostly X-linked)
– Duchenne muscular dystrophy
– Hemophilia
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Hemophilia is a recessive sex-linked trait. Specifically, the allele
that codes for this disease is found on the X chromosome
(Xh). Individuals with this disorder are unable to form blood
clots properly and often run the risk of bleeding to death from
injuries that might normally be considered minor. A couple
would like to know if the child they are expecting could be born
with hemophilia. The mother’s genotype is XHXh. The father’s
genotype is XhY and he has hemophilia.
• a) What are the possible genotypes for the couple’s children?
• b) What is the % chance that their child will have hemophilia if
it is a girl?
• c) What is the % chance that their child will have hemophilia if
it is a boy?
X Inactivation in Female Mammals
• one of the two X chromosomes in each cell is
randomly inactivated during embryonic
development
– inactive X condenses into a Barr body
• If a female is heterozygous for a particular gene
located on the X chromosome, she will be a
mosaic for that character
© 2011 Pearson Education, Inc.
Figure 15.8
X chromosomes
Allele for
orange fur
Early embryo:
Two cell
populations
in adult cat:
Allele for
black fur
Cell division and
X chromosome
inactivation
Active X
Inactive X
Active X
Black fur
Orange fur
Pleiotropy
• multiple phenotypic effects = pleiotropy
• pleiotropic alleles
– responsible for the multiple symptoms of certain
hereditary diseases, such as cystic fibrosis and
sickle-cell disease
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Epistasis
• In epistasis, a gene at one locus alters the
phenotypic expression of a gene at a second
locus
– Coat color in Labrador retrievers
• (One gene determines the pigment color (with
alleles B for black and b for brown)
• (The other gene (with alleles C for color and c for
no color) determines whether the pigment will be
deposited in the hair)
© 2011 Pearson Education, Inc.
Epistasis problem
In guinea pigs, the gene for production of melanin is epistatic to the
gene for the deposition of melanin.
• The dominant allele E causes melanin to be produced; mm
individual cannot produce the pigment.
• The dominant allele B causes the deposition of a lot of
pigment and produces a black guinea pig, whereas only a
small amount of pigment is laid down in bb animals,
producing a light-brown color.
• Without an E allele, no pigment is produced, so the allele B
has no effect, and the guinea pig is white.
• A homozygous black pig is crossed with a homozygous
recessive white : EEBB x eebb.
Give the geno- and phenotypes for the F1 and F2 generations.
Figure 14.12
BbEe
Eggs
1/
4 BE
1/
4 bE
1/
4 Be
1/
4
be
Sperm
1/ BE
4
1/
BbEe
4 bE
1/
4 Be
1/
4 be
BBEE
BbEE
BBEe
BbEe
BbEE
bbEE
BbEe
bbEe
BBEe
BbEe
BBee
Bbee
BbEe
bbEe
Bbee
bbee
9
: 3
: 4
Polygenic Inheritance
• an additive effect of two or more genes on a single
phenotype
• Skin color in humans
© 2011 Pearson Education, Inc.
Figure 14.13
AaBbCc
AaBbCc
Sperm
1/
1/
8
8
1/
1/
Eggs
8
1/
1/
8
8
1/
8
1/
1/
8
8
8
8
1/
8
1/
8
1/
1/
8
1/
8
1/
8
1/
8
Phenotypes:
Number of
dark-skin alleles:
1/
64
0
6/
64
1
15/
64
2
20/
64
3
15/
64
4
6/
64
5
1/
64
6
Linked genes
• tend to be inherited together
• are located near each other on the same
chromosome
• certain traits do not assort independently
– (b/c they’re on the same chromosome!)
© 2011 Pearson Education, Inc.
Figure 15.9-1
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body, normal wings)
Double mutant
(black body,
vestigial wings)
b b vg vg
b b vg vg
Figure 15.9-2
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body, normal wings)
Double mutant
(black body,
vestigial wings)
b b vg vg
b b vg vg
F1 dihybrid
(wild type)
b b vg vg
TESTCROSS
Double mutant
b b vg vg
Figure 15.9-3
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body, normal wings)
Double mutant
(black body,
vestigial wings)
b b vg vg
b b vg vg
F1 dihybrid
(wild type)
Double mutant
TESTCROSS
b b vg vg
b b vg vg
Testcross
offspring
Eggs b vg
b vg
Wild type
Black(gray-normal) vestigial
b vg
Grayvestigial
b vg
Blacknormal
b vg
Sperm
b b vg vg
b b vg vg
b b vg vg
b b vg vg
Figure 15.9-4
EXPERIMENT
P Generation (homozygous)
Wild type
(gray body, normal wings)
Double mutant
(black body,
vestigial wings)
b b vg vg
b b vg vg
F1 dihybrid
(wild type)
Double mutant
TESTCROSS
b b vg vg
b b vg vg
Testcross
offspring
Eggs b vg
b vg
b vg
Wild type
Black(gray-normal) vestigial
b vg
Blacknormal
Grayvestigial
b vg
Sperm
b b vg vg
b b vg vg
b b vg vg
b b vg vg
PREDICTED RATIOS
If genes are located on different chromosomes:
1
:
1
:
1
:
1
If genes are located on the same chromosome and
parental alleles are always inherited together:
1
:
1
:
0
:
0
965
:
944
:
206
:
185
RESULTS
Figure 15.UN01
F1 dihybrid female
and homozygous
recessive male
in testcross
b+ vg+
b vg
b vg
b vg
b+ vg+
b vg
Most offspring
or
b vg
b vg
• However, nonparental phenotypes were also
produced…
• What happened!?! I thought the genes were
linked!?!
© 2011 Pearson Education, Inc.
Genetic Recombination: (happened).
The production of offspring with
combinations of traits differing from
either parent
• Offspring with nonparental phenotypes are
called recombinant types, or recombinants
Recombination of Unlinked Genes:
Independent Assortment of Chromosomes
• If frequency is < 50% the genes are linked.
© 2011 Pearson Education, Inc.
Figure 15.UN02
Gametes from yellow-round
dihybrid parent (YyRr)
Gametes from greenwrinkled homozygous
recessive parent (yyrr)
YR
yr
Yr
yR
YyRr
yyrr
Yyrr
yyRr
yr
Parentaltype
offspring
Recombinant
offspring
What do you think the mechanisms
behind this recombinance is?
Recombination of Linked Genes
• That mechanism was the crossing over of
homologous chromosomes
© 2011 Pearson Education, Inc.
Animation: Crossing Over
Right-click slide / select”Play”
© 2011 Pearson Education, Inc.
Nature and Nurture:
• phenotypes are influenced by the environment
– norm of reaction is the phenotypic range of a
genotype influenced by the environment
• hydrangea flowers of the same genotype range
from blue-violet to pink, depending on soil
acidity
© 2011 Pearson Education, Inc.
Figure 14.14
Figure 14.14a
Figure 14.14b
• An organism’s phenotype reflects its overall
genotype and unique environmental history
© 2011 Pearson Education, Inc.
• Humans are not good subjects for genetic
research
• But lots of our traits follow Mendelain patterns
of inheritance
© 2011 Pearson Education, Inc.
Pedigree Analysis
• A pedigree is a family tree
• Used to make predictions about offspring
© 2011 Pearson Education, Inc.
Figure 14.15
Key
Male
1st
generation
Affected
male
Female
Affected
female
Mating
1st
generation
Ww
ww
Ww
ww
2nd
generation
Ww
ww
3rd
generation
WW
or
Ww
Widow’s
peak
ff
ff
(a) Is a widow’s peak a dominant or
recessive trait?
Ff
Ff
Ff
ff
ff
FF
or
Ff
3rd
generation
ww
No widow’s
peak
ff
Ff
2nd
generation
FF or Ff
Ww ww ww Ww
Ff
Offspring
Attached
earlobe
Free
earlobe
b) Is an attached earlobe a dominant
or recessive trait?
Figure 14.15a
Widow’s
peak
Figure 14.15b
No widow’s
peak
Figure 14.15c
Attached
earlobe
Figure 14.15d
Free
earlobe
© 2011 Pearson Education, Inc.
Recessively Inherited Disorders
• Many genetic disorders are inherited in a
recessive manner
• Show up only in homozygous individuals
• Carriers – heterozygous for condition (normal
phenotype)
• Albinism
© 2011 Pearson Education, Inc.
Figure 14.16
Parents
Normal
Aa
Normal
Aa
Sperm
A
a
A
AA
Normal
Aa
Normal
(carrier)
a
Aa
Normal
(carrier)
aa
Albino
Eggs
• Consanguineous matings (i.e., matings
between close relatives/inbreeding) increase
the chance of mating between two carriers of
the same rare allele
© 2011 Pearson Education, Inc.
Fetal Testing
• In amniocentesis,
• In chorionic villus sampling (CVS), a sample
of the placenta is removed and tested
• ultrasound and fetoscopy,
© 2011 Pearson Education, Inc.
Figure 14.19
(a) Amniocentesis
1
(b) Chorionic villus sampling (CVS)
Ultrasound monitor
Amniotic
fluid
withdrawn
Ultrasound
monitor
Fetus
1
Placenta
Chorionic villi
Fetus
Placenta
Uterus
Cervix
Cervix
Uterus
Suction
tube
inserted
through
cervix
Centrifugation
Fluid
Fetal
cells
Several hours
2
Several
weeks
Biochemical
and genetic
tests
Several
hours
Fetal cells
2
Several hours
Several weeks
3
Karyotyping
Figure 15.10
Black body, vestigial wings
(double mutant)
Gray body, normal wings
(F1 dihybrid)
Testcross
parents
b vg
b vg
b vg
b vg
Replication
of chromosomes
Meiosis I
Replication
of chromosomes
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
Meiosis I and II
b vg
b vg
b vg
Meiosis II
Recombinant
chromosomes
bvg
b vg
b vg
b vg
944
Blackvestigial
206
Grayvestigial
185
Blacknormal
Eggs
Testcross
offspring
965
Wild type
(gray-normal)
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
Parental-type offspring
Recombinant offspring
391 recombinants
Recombination

 100  17%
frequency
2,300 total offspring
b vg
Sperm
Figure 15.10a
Gray body, normal wings
(F1 dihybrid)
Testcross
parents
Black body, vestigial wings
(double mutant)
b vg
b vg
b vg
b vg
Replication
of chromosomes
Replication
of chromosomes
Meiosis I
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
Meiosis I and II
b vg
b vg
b vg
Meiosis II
bvg
Eggs
Recombinant
chromosomes
b vg
b vg
b vg
b vg
Sperm
Figure 15.10b
Recombinant
chromosomes
Eggs
Testcross
offspring
bvg
965
Wild type
(gray-normal)
b vg
b vg
b vg
944
Blackvestigial
206
Grayvestigial
185
Blacknormal
b vg
b vg
b vg
b vg
b vg
b vg
b vg
b vg
Parental-type offspring
Recombinant offspring
Recombination
391 recombinants  100  17%

frequency
2,300 total offspring
b vg
Sperm