Chapter 15 - The Chromosomal
Basis of Inheritance
A. Bergeron
+AP Biology
PCHS
Do Now - Predicting Unpredictable Genotypes
As an inexperienced (albeit precocious) gardener, I am always
looking to maximize my yield of fruits and vegetables during the
summer.
Last summer, I identified two tomato plants that produced an
inordinate amount of fruit. I contemplated breeding these two
plants (either through cross-pollination or self-fertilization) but
was unsure of either plant’s genotype.
Devise a strategy for determining the unknown genotype of
both parental tomato plants using as few resources as possible.
Remember, I want to maximize the yield of fruit from all of the
future generations of tomato plants.
Do Now - Sex-linked Inheritance
The table below shows the results of a cross of two red-eyed
fruit flies. Some of the offspring have “ruby”-colored (deep
pink) eyes instead of the usual red-colored eyes.
Eye Color! !
! !
!
Red eyes! !
“Ruby” eyes! !
Numbers of Offspring
Males!!
Females
77!
!
151
75!
!
0
a. Using XR for the dominant allele and Xr for the recessive
allele, write the genotypes of the two parents.
b. How do the results of the cross indicate that ruby eye color
is sex-linked?
Testcross
Used to determine the genotype of a “parent” with the
dominant phenotype but an unknown genotype
Dominant parent “crossed/mated” with a homozygous
recessive parent
Genotype of the dominant parent can be determined based on
the offspring produced from the cross
If offspring with ONLY the dominant phenotype are produced,
the parent must have the homozygous dominant genotype
If offspring with the dominant and the recessive phenotype are
produced, the parent must have the heterozygous genotype
Sample Testcross
Reading Quiz
1. The most common phenotype in a population of organisms
is called the ____________.
2. _______ genes have loci located on the same chromosome.
3. A recombination frequency greater than 50% suggests
that two genes are (linked or unlinked)?
4. A father carries a recessive allele for hemophilia on his
X chromosome. What is the probability that he will transmit
the hemophilia allele to a son?
5. In females, one of the two X-chromosomes is inactivated
early in embryonic development. What term is used to
describe this inactivated chromosome?
Chromosomal Basis of Mendelian Inheritance
Do Now - Sex-Linked Genetics Practice Problems
1. A man with hemophilia (a recessive, sex-linked disease)
has a daughter of normal phenotype. She marries a man who
is normal for the trait. What is the probability that a daughter
of this mating will be a hemophiliac? That a son will be a
hemophiliac? If the couple has four sons, what is the
probability that all four sons will be born with hemophilia?
2. Red-green color blindness is caused by a sex-linked
recessive allele. A color-blind man marries a woman with
normal vision whose father was color blind. What is the
probability that they will have a color-blind daughter? What is
the probability that their first son will be color-blind?
Chromosomal Basis of Mendelian Inheritance
Mendel’s laws of segregation and independent assortment
assumed that the factors (i.e. alleles) that are responsible
for each of the observed traits were on different chromosomes
!
-Remember that Mendel did not know of the existence
!
of genes, chromosomes, or even DNA.
Chromosomal Theory of Inheritance
!
-Mendelian factors (genes/alleles) are located on
!
chromosomes
!
-Chromosomes segregate and assort independently
Morgan’s Discovery of Sex-linked Inheritance
Parental Cross: White-eyed male
(XwY) x red-eyed (XRXR) female fruit
flies (Drosophila)
F1 generation (males and females) had
red eyes
F2 generation had (expected)
3 (red):1(white) ratio of eye color
phenotypes but only males had white
eyes!
What could explain this peculiar
result?
Answer: Sex-linked inheritance - the
locus of some genes is located
exclusively on one of the sex c’somes
Parental Cross: White-eyed
male (XwY) x red-eyed (XRXR)
female fruit flies (Drosophila)
F1 generation (males and
females) had red eyes
F2 generation had (expected)
3 (red):1(white) ratio of eye
color phenotypes
But only males had white eyes
How Did Morgan Know that the Gene for Red
Eyes Was on a Sex Chromosome and Not an
Autosome?
1. Since Morgan hypothesized that the allele for white eyes
was recessive, a white-eyed female would need to inherit
two (2) X chromosomes with the recessive allele
!
-This was not observed in the F2 females
2. A white-eyed male would not have a wild-type (normal)
allele to “mask” the presence of the recessive allele
!
-Since the white-eye allele was proposed to be on the
!
X chromosome, a single copy of the allele would
!
guarantee the white eye phenotype in male flies
Sex-Linked Inheritance (Three Possibilities...)
X-Chromosome Inactivation in Female Mammals
Example - Female calico cats
In females, one of the two
X chromosomes is inactivated
to prevent multiple copies of a
gene from being expressed
The inactivation is random.
Either chromosome could be
inactivated.
Inactivated X c’somes are
called Barr bodies b/c the
inactivated c’some appears
as a dark mass of DNA
under the microscope
Important Genetics Vocabulary
Parental = offspring that have the same phenotype as one
or more of the parents
Recombinant = offspring whose phenotypes are different
from the parents
!
Recombinant gametes are often the products of
!
crossing over
Evidence for Linked Genes in Drosophila
Genes located on the same chromosome are considered to
be linked
Linked genes do NOT assort independently during meiosis.
Variations away from expected results of a particular cross
are indicative of a non-Mendelian inheritance pattern
What would be the expected phenotypic ratio in the offspring
if the genes for body color and wing type were unlinked
(i.e. assorted independently)?
(Assume that Morgan produced approximately
2,300 offspring in his cross)
Mendel’s Principle of Independent Assortment
Mom
Dad
A
a
A
a
B
b
b
B
A
a
A
a
B
b
b
B
A
A
a
a
B
B
b
b
Sex
cells
A
A
a
a
b
b
B
B
What would be the expected phenotypic ratio in the offspring
if the genes for body color and wing type were linked
(i.e. genes located on the same chromosome that do
not assort independently)?
???
Crossing Over
Calculating Recombination Frequency
# Recombinants
X 100%
Total # offspring
1. A wild-type fruit fly (heterozygous for gray body color and
normal wings) was mated with a black fly with vestigial
wings. The offspring had the following phenotypic
distribution:
! !
778 wild-type
! !
785 black, vestigial
! !
158 black, normal
! !
162 gray, vestigial
What is the recombination frequency between these genes
for body color and wing type?
Using Recombination Data to Calculate the
Distance Between Genes on a Chromosome
Crossover events are more likely to occur between genes that
are further apart on a chromosome
A recombination frequency that is greater than or equal to
50% indicates that genes are unlinked (i.e. on separate
chromosomes) OR are so far apart from each other on the
same chromosome that they behave as if they are on separate
chromosomes (frequent cross overs)
Creating a Genetic Map
Distance between genes on a
chromosome is measured
in centimorgans (cM) in honor
of Thomas Hunt Morgan
1 centimorgan = 1 map unit =
1% recombination frequency
One “map unit” is defined as
the distance within which a
crossover event is expected
to occur in an average of 1%
of gametes
Genetics Practice Problems
1. Determine the sequence of genes along a chromosome
based on the following recombination frequencies:
! !
A-B: 8 cM
! !
A-C: 28 cM
1 centimorgan (cM) =
! !
A-D: 25 cM
1% recombination frequency
! !
B-C: 20 cM
! !
B-D: 33 cM
2. A fruit fly of genotype BR/br is test-crossed to br/br.
(Note: Alleles grouped on either side of the slash are located
on the same chromosome) In 84% of the meioses, there
are no chiasmata between the linked genes; in 16% of the
meioses, there is one chiasmata between the genes. What
proportion of the progeny will be Br/br?
Why Shouldn’t I Include the Parental Gametes Produced
as a Result of the Crossover?
Recall…
# Recombinants
% Rec. =
X 100%
Total # offspring
The “formula” for calculating
% recombination does NOT take
the parental gametes into account
Thus, the probability of producing
recombinant gametes is
equivalent to the % recombination
Don’t forget that there is a 50%
chance that either of the two
gametes will fertilize an egg!
Genetics Practice Problems
3. In the plant Arabidopsis thaliana the loci for pod length
(L = long, l = short) and fruit hairs (H = hairy, h = smooth) are
located 20 cM apart on the same chromosome. The
following crosses were made:
!
!
L H/LH X lh/lh --> F1
!
!
Lh/Lh X lH/lH --> F1
If the F1’s from above are crossed, what proportion of the
progeny are expected to be lh/lh?
Do Now
4. Using the same information from #3 on the previous slide, if
the F1’s are crossed, what proportion of the progeny are
expected to be Lh/lh?
Genetics Practice Problems
5. In Drosophila, the gene for white-eyes and the gene that
produces “hairy” wings have both been mapped to the same
chromosome and have a crossover frequency of 1.5%. A
geneticist notices that in a particular stock of flies, these two
genes assorted independently; that is, they behaved as
though they were on separate chromosomes. What
explanation can you offer for this observation?
Errors and Exceptions to “Normal” Chromosomal
Inheritance
Deletion
Duplication
Inversion
Translocation
Genetics Practice Problems
6. In mice, black coat is dominant to brown and intense pigment is
dominant to diluted. One mouse with dilute black pigment is
crossed to another with intense brown pigment. The progeny are:
a. What are the genotypes of parents and progeny?
b. Draw the alleles involved on their respective chromosomes.
Genetics Practice Problems
7. The following testcross of a plant is made:
If the two loci are 15 m.u. apart, what proportion of progeny
will be (a) AB/ab? (b) ab/ab? (c) Ab/ab?
8. The A locus and the D locus are so tightly linked that no
recombination is ever observed between them. If Ad/Ad is
crossed to aD/aD, and the F1 is intercrossed, what
genotypes will be seen in the F2 and in what proportions?
Do Now - Genetics Practice Problems
A female animal with genotype A/a.B/b is crossed with a
double-recessive male a/a.b/b. Their progeny include
442 A/a.B/b, 458 a/a.b/b, 46 A/a.b/b, and 54 a/a.B/b.
Explain these proportions and draw the chromosomes of the
dihybrid parent showing the positions of the genes and alleles.
Three-point Testcross
How can we find the order and relative map distance between
genes on a chromosome?
Assume you have the following alleles at three different loci:
v (vermillion eyes) vs. v+ (red eyes)
cv (absence of crossvein on wing) vs. cv+ (presence of crossvein)
ct (snipped wing edges) vs. ct+ (smooth wing edges)
Red (left) vs. Vermillion (right)
Three-point Testcross
Available parental stocks:
v+/v+ . cv/cv . ct/ct.
v/v . cv+/cv+ . ct+/ct+
What is the F1?
Testcross F1 females to male homozygotes
Three-point Testcross
Progeny Phenotype
v
cv+
v+
cv
v
cv
v+
cv+
v
cv
v+
cv+
v
cv+
v+
cv
ct+
ct
ct+
ct
ct
ct+
ct
ct+
# Flies
580
592
45
40
89
94
3
5
1. What is the map distance between all three genetic loci?
2. Draw the gene map for these three genes.
Genetics Practice Problems
9. In corn, a triple heterozygote was obtained carrying the
mutant alleles s (shrunken), w (white aleurone), and y (waxy
endosperm), all paired with their normal wild-type alleles. This
triple heterozygote was testcrossed, and the progeny
contained 116 shrunken, white; 4 fully wild type; 2538
shrunken; 601 shrunken, waxy; 626 white; 2708 white, waxy;
2 shrunken, white, waxy; and 113 waxy.
a. Determine if any of these three loci are linked, and, if so,
show map distances.
BONUS: Show the allele arrangement on the chromosomes
of the triple heterozygote used in the testcross.
Genetics Practice Problems
Genetics Practice Problems
10. If A/A.b/b is crossed to a/a.B/B, and the F1 is testcrossed,
what percent of the testcross progeny will be a/a.b/b if the two
genes are (a) unlinked, (b) completely linked (no crossingover at all), (c) 12 map units apart, (d) 24 map units apart?
Genes of unknown linkage are shown separated by a dot, A/a.B/b
Genetics Practice Problems
Genetics Practice Problems
11. You have a Drosophila line that is homozygous for
autosomal recessive alleles a, b, and c, linked in that order.
You cross females of this line with males homozygous for the
corresponding wild-type alleles. You then cross the F1
heterozygous males with their heterozygous sisters. You obtain
the following F2 phenotypes (where letters denote recessive
phenotypes and pluses denote wild-type phenotypes):
1364 + + +, 365 a b c, 87 a b +, 84 + + c, 47 a + +, 44 + b c,
5 a + c, and 4 + b +.
(a) What is the map distance between genes a and b? b and c?
a and c?
Genetics Practice Problems
Genetics Quiz - Chapter 15
1. A strain of Neurospora (common bread mold) with the
genotype H.I is crossed with a strain with the genotype h.i.
Half the progeny are H.I and half are h.i. Briefly explain how
this is possible.
Genetics Practice Problems
12. The following pedigree shows a family with two rare abnormal
phenotypes: blue sclerotic (a brittle bone defect), represented by a
black-bordered symbol, and hemophilia, represented by a black
center in a symbol. Individuals represented by completely black
symbols have both disorders.
a. What pattern of inheritance is shown by each condition in
this pedigree?
Genetics Practice Problems
Blue sclerotic: Autosomal dominant
Hemophilia: Sex (X)-linked recessive
Pedigree Analysis - Important Symbols
Important Points to Consider When Performing
a Pedigree Analysis
If a trait is recessive:
!
Individuals with the trait must be homozygous
!
Unaffected parents of children with the trait will be
!
heterozygous
!
Children may exhibit the trait even if their parents do not
If a trait is dominant:
!
Any child with the trait will have at least one parent
!
with the trait
!
A trait caused by a dominant inheritance pattern
!
will show up in every generation
Inheritance Patterns & Phenotypic Characteristics
Simple (Autosomal) Recessive
-Males and females are equally
likely to have the trait
-Either parent can transmit the
trait to offspring of either gender
*The trait often skips generations
-Often, both parents of offspring
who have the trait are
heterozygous; they carry at least
one copy of the allele
-Only homozygous recessive
individuals have the trait
-The trait may appear in siblings
w/o appearing in parents
-If one parent has the trait, those
offspring who do not have it are
carriers of the trait
Pedigree Analysis
1
Autosomal
Recessive
2
3
Alkaptonuria - Is the disease caused by a dominant or
recessive allele? What are the possible genotypes of
individuals 1, 2, and 3?
Pedigree Analysis
1
Autosomal
Dominant
2
Familial Hypercholesterolemia - Is disease caused by a
dominant or recessive allele? What are the possible
genotypes of individuals 1 and 2?
Do Now - Pedigree Analysis
Determine the mode of inheritance of the trait diagrammed in
the above pedigree.
Hint: If one than one mode of inheritance is possible,
determine which is most probable.
Pedigree Analysis
Autosomal
Recessive
1
2
3
Sickle cell disease - Is disease caused by a dominant or
recessive allele? What are the possible genotypes of
individuals 1, 2, and 3?
Hint: If one than one mode of inheritance is possible,
determine which is most probable.
Pedigree Analysis
Determine the mode of inheritance of the trait diagrammed in
the above pedigree.
Hint: If one than one mode of inheritance is possible,
determine which is most probable.
Pedigree Analysis
Hint: If one than one mode
of inheritance is possible,
determine which is most
probable.
Inheritance Patterns & Phenotypic Characteristics
X-linked Recessive
-All daughters of a male who has
the trait are carriers
-There is no male-to-male
transmission
-The trait is far more common in
males than in females
-The son of a female carrier has a
50% chance of having the trait
-Mothers of males who have the
trait are either carriers or
homozygous and have the trait
-The daughters of a female carrier
have a 50% of being a carrier
Inheritance Patterns & Phenotypic Characteristics
X-linked Dominant
-All daughters of a male who has
the trait will also have the trait
-There is no male-to-male
transmission
-A female who has the trait may or
not pass the gene for that trait to
her son or daughter
Diseases Caused by Recessive Genotypes
1. Cystic fibrosis
2. Tay-Sachs disease
3. Sickle-cell disease
Diseases Caused by Dominant Genotypes
1. Achondroplasia
2. Huntington’s disease
*Be familiar with the cause of the disease (dominant or
recessive allele/s) and disease symptoms*
Fly Lab Instructions
1. I will give each group 2 petri dishes with a different fly
phenotype. Observe each phenotype under the microscope.
Be sure to observe each phenotype!
2. Practice separating male flies from female flies. You can use
any phenotype to practice.
3. Set up labeled culture vials with Drosophila medium.
!
Add 1 scoop of tap water to 1 scoop of food flakes
!
Sprinkle a few grains of yeast on top of the culture.
!
Wait 4-5 minutes before adding flies
4. Add 4-5 mating pairs of F1 flies from each cross to your
labeled culture vials. The flies should be asleep! They will
wake up later! I promise…
1) After reviewing the class data (which I will provide for you),
you should be able to write a hypothesis that explains the
mode of inheritance acting in each cross.
Example - Monohybrid/dihybrid (autosomal) cross
!
Monohybrid/dihybrid (sex-linked) cross
2) You should assume that chance has been responsible for
the independent assortment of alleles that produced the F1
and F2 generations of flies.
ANY variation of your observed results when compared to
the expected results is due to CHANCE
3) In this experiment, variation due to chance alone is called
the “null hypothesis”
4) Can the class data be used to support the null hypothesis?
Chi-Square (Χ2) Statistical Analysis
Χ2 is used to determine how well observed ratios fit expected
ratios
Chi-square formula
Χ2 = Σ (o-e)2
!
-------!
!
e
You will have observed values from the class data set
The expected values will come from your expected phenotype
ratios
!
For example, if you predict that the offspring should exist in a 3:1
!
!
ratio and there are 400 offspring, 300 offspring should have one
phenotype and 100 offspring should have the other phenotype
The experimental Χ2 value is compared to actual Χ2 values from
a statistical table to determine significance
Critical Values of the Chi-Square Distribution
Unless told otherwise, use Χ2 values for the 5% probability value
Degrees of Freedom
Prob. due to 1
chance
0.05 (5%)
3.84
2
3
4
5
5.99
7.82
9.49
11.1
0.01 (1%)
6.64
9.21
11.3
13.2
15.1
0.001 (0.1%) 10.8
13.8
16.3
18.5
20.5
Degrees of freedom = (# phenotypes - 1)
Practice Problem
A true-breeding corn plant with purple corn kernels is crossed with a corn
plant with yellow corn kernels. An F1 offspring is “selfed” to produce the
F2 generation. The results of the F1 cross are shown in the table below:
Are differences between the observed and expected values due to
chance alone or is there an underlying cause of the variation? Perform a
Χ2 analysis to support your answer.
Phenotype # Observed (o)
Yellow
149
kernels
Purple
388
kernels
Total
537
# Expected (e)
(o-e)
(o-e)2 (o-e)2/2
Chi-Square Analysis
Phenotype # Observed (o)
Yellow
# Expected (e)
(o-e)
(o-e)2 (o-e)2/2
149
kernels
Purple
388
kernels
Total
537
Chi-square formula:
Χ2 = Σ (o-e)2
!
-------!
!
e
Null hypothesis:
There is no significant difference
between the expected and observed
results. The difference/s are due
to chance.
Chi-Square Analysis
Phenotype # Observed (o)
# Expected (e)
(o-e)
(o-e)2 (o-e)2/e
Yellow
149
134
15
225
1.68
388
403
-15
225
.56
537
537
kernels
Purple
kernels
Total
Chi-square formula:
Χ2 = Σ (o-e)2
!
-------!
!
e
2.24
Null hypothesis:
There is no significant difference
between the expected and observed
results. The difference/s are due
to chance.
Critical Values of the Chi-Square Distribution
Degrees of freedom:
(# phenotypes - 1)
Degrees of Freedom
Probability
due to
chance
1
2
3
4
5
0.05 (5%)
3.84
5.99
7.82
9.49
11.1
0.01 (1%)
6.64
9.21
11.3
13.2
15.1
0.001 (0.1%) 10.8
13.8
16.3
18.5
20.5
Making Conclusions Using Chi-Square
If the calculated X2 value is less than the critical value for a
given probability of variation due to chance, you must accept
your null hypothesis.
!
-There is no significant difference between the
!
expected and observed values.
!
-Any difference between the expected and observed
!
values probably occurred due to chance.
If the calculated X2 value is greater than the critical value
for a given probability of variation due to chance, you
must reject your null hypothesis
!
-There is a significant difference between the expected
!
and observed values
!
-There is an underlying reason why a difference exists
!
between the expected and observed values
Practice Problem - Chi Square Analysis
Suppose that a scientist crossed pink-flowered snapdragons
and obtained 236 offspring. 66 of the offspring grew into plants
with red flowers, 115 with pink flowers, and 55 with white
flowers.
a. What are the expected phenotypic ratios of the offspring in
the F1 generation?
b. What are the expected number of offspring with each type of
flower color?
c. Are the differences between the observed and expected
results significant enough to conclude that the results did
NOT occur by chance alone?