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Mendelian Genetics
LO 3.15 The student is able to explain deviations from
Mendel’s model of the inheritance of traits. [See SP 6.5]
LO 3.16 The student is able to explain how the inheritance
patterns of many traits cannot be accounted for by Mendelian
genetics. [See SP 6.3]
LO 3.17 The student is able to describe representations of an
appropriate example of inheritance patterns that cannot be
explained by Mendel’s model of the inheritance of traits. [See
SP 1.2]
A brief review of Mendel
(and flower parts!)
Pisum sativum
Terms you must know!
Haploid
Dominant
Diploid
Recessive
Homozygous
Genotype
Heterozygous
Phenotype
Monohybrid
Dihybrid
Allele
Hybrid
Check out the ratios! You have to know how to get
the expected ratios to do a chi square problem that
deals with a genetic data.
LO 3.9 The student is able to construct an explanation, using visual
representations or narratives, as to how DNA in chromosomes is transmitted to
the next generation via mitosis, or meiosis followed by fertilization. [See SP 6.2]
Law of Independent
AssortmentAll of the parents were
yellow and smooth- but
those traits didn’t
always stay together
in the offspring.
Since we know Mendelian genetics, we don’t need the
Punnett square to tell us that we get a 9:3:3:1 phenotypic
ratio. There is a 1/16 chance that the offspring will be short
and white.
We also need to be able to calculate
percentagesFor example in a dihybrid cross between two heterozygotes,
If you have 360 offspring, what are your expected values?
You know that you should have a 9:3:3:1 ratio
Both dominant phenotypes 9/16 = .56 = 56% = 202
One dominant; one recessive 3/16 = .19 = 19% = 68
One recessive; one dominant 3/16 = .19 = 19% = 68
Both recessive phenotypes 1/16 = .06 = 6% = 22
Let’s plug this into the Hardy-Weinberg equation and see if we get the same
numbers. Of the 180 flowers that are tall, how many are homozygous, and how
many are heterozygous?
22% are short. What is the frequency of the recessive allele?
22% is .22
.22 = q2
.47 = q
What is the frequency of the dominant allele?
.53 = p
.53 x .53 = .28
.28 x 360 = 101 homozygous tall
2 (.53)(.47) = .5
.5 x 360 = 179 heterozygous (tall)
So, we calculated that 180 of these flowers would be tall; now we know how many of
the tall flowers are homozygous and how many are heterozygous
In a species of gecko, spots and tail size are controlled by
genes that assort independently.
A= spotted B= fat tail
a = solid
b = skinny tail
A scientist collected data from several gecko breeders and
got the phenotypes of 135 baby geckos born within a 6
month period. Calculate the percentage of the geckos
with the 4 possible phenotypes and the number of each in
this data sample if all of the parents in the trial were
heterozygotes.
56%
19%
19%
6%
75
26
26
8
In a species of gecko, spots and tail size are
controlled by genes that assort independently.
A= spotted
B= fat tail
a = solid
b = skinny tail
What if the cross wasn’t between
two heterozygotes? What if this was
the cross? What percentages would
you expect in the offspring?
aabb x AaBb
25% spotted/fat tail
25% solid/fat tail
25% spotted/skinny tail
25% solid/skinny tail
P= smooth seeds x wrinkled seeds
F1= all smooth seeds
F2= 5,474 smooth seeds and 1,850 wrinkled seeds
Calculate chi-square for this data and evaluate your results.
.
Null hypothesis- There is no
significant difference between
the observed and expected
ratio of smooth and wrinkled
seeds.
P= smooth seeds x wrinkled seeds
F1= all smooth seeds
F2= 5,474 smooth seeds and 1,850 wrinkled seeds
Calculate chi-square for this data and evaluate your results.
.
Null hypothesis- There is no significant difference between
the observed and expected ratio of smooth and wrinkled seeds.
I accept my null hypothesis. My
chi square value of 0.263 is less
than the critical value of 3.84.
You don’t need Punnett squares, you can determine
expected ratios using probability
Consider a trihybrid cross-
If a plant that is heterozygous for all three characteristics is
allowed to self-fertilize, what proportion of the offspring would
you expect to be as follows- AaBbDd x AaBbDd
a. homozygous for the 3 dominant traits?
There is a 1/4 chance for each pair of alleles to be homozygous
dominant.
1/4 x 1/4 x 1/4 = 1/64
b. homozygous for purple and tall; heterozygous for seed
shape
The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd.
Assuming independent assortment of these four genes,
what are the probabilities that F2 offspring will have the following
genotypes?
a. AABBCCDD
¼ x ¼ x ¼ x ¼ = 1/256
b. AaBBccDd
½ x ¼ x ¼ x ½ = 1/64
c. AaBBCCdd
½ x ¼ x ¼ x ¼ = 1/128
Beyond Mendel…
LO 3.15 The student is able to explain
deviations from Mendel’s model of the
inheritance of traits. [See SP 6.5]
LO 3.16 The student is able to explain
how the inheritance patterns of many
traits cannot be accounted for by
Mendelian genetics. [See SP 6.3]
Beyond Mendel
Incomplete Dominance
F1 hybrids have an appearance
that is between that of the
two parents. Breeding two of
the hybrids produces a 1:2:1
ratio of phenotypes in the F2
Don’t just know the definition of
incomplete dominance; recognize
the ratio and know how it differs
from a typical Mendelian cross. You
may be given the results of a genetic
cross and be asked to explain how
you would get these results.
Beyond Mendel-
Epistasis- a gene at
one locus alters the effects
of a gene at another locus
B- gray
b- tan
C- pigment
c- no pigment
If the mouse doesn’t have
a capital C, regardless of the
gene for coat color, no
pigment will be deposited in
the fur
WONKY RATIO
Two genes play a critical role in normal hearing; one is
involved in the development of a normal cochlea (A).
The other is responsible for a normal auditory nerve
(B). Either of these genes can cause deafness if the
genotype is homozygous recessive.
A man and a woman are both deaf; one as a result of a
defective cochlea and the other as a result of a
defective auditory nerve. Assume that they are
homozygous dominant for the other gene. If they have
children together (barring mutation), what is the
probability that the child will be deaf?
0 AAbb x aaBB  all AaBb
Beyond Mendel- Codominance & Multiple Alleles
More than two alleles for a trait, and two of those
alleles show dominance
Two parents, one type A and one type B,
have a type O child? Explain that!!
In the U.S. about 16% of the population is Rh negative.
The allele for Rh negative is recessive to the allele for
Rh positive. If the student population of a high school in
the U.S. is 2000, how many students would you expect for
each of the three possible genotypes?
p+q=1
p2 + 2pq + q2 = 1
16% = .16 = q2
q = .4
p = .6
Homozygous Dominant (++) 36% 720
Heterozygous (+-) 48% 960
Homozygous recessive (--) 16% 320
Beyond Mendel
Polygenic Inheritance
Two or more genes
have an additive effect
on a single character in
the phenotype
In humans,
what are some
other characteristics
that are polygenic?
Beyond MendelPleiotropy
One gene =
Multiple effects
Beyond MendelSex linked genes are located on the sex chromosomes
(x and y). They were discovered by Thomas Morgan
during his work with Drosophila. If a sex linked
gene is on the X chromosome, a man who receives
that X chromosome will have the trait. Since a woman has two
X chromosomes, it may not be expressed in her.
Some sex linked disorders
in humansHemophilia
Red/green colorblindness
Duchennes Muscular Dystrophy
In fruit flies, the mutation for bar eyes is a sex linked
dominant trait. A pure bar-eyed female is crossed
with a wild (red-eyed) male. Which of the following
statements gives the correct percentage of male
offspring that will have bar eyes and the correct
explanation for it?
a. 50%, because all the males will inherit the bar-eyed
trait from the mother
b. 50%, because all the males will inherit the bar-eyed
trait from the father
c. 100%, because all the males will inherit the bar-eyed
trait from the mother
d. 100%, because all the males will inherit the bar-eyed
trait from the father
Insert slide on how environment influences phenotype
X-inactivation
Females have two X chromosomes;
Males only have one
Autosomal Recessive
Disorders
Exactly what does this tell you about
the disorder?
If you are heterozygous, you are a carrier of the disorder. You don’t
have the disorder, but if you have children with another carrier, they have
a 25% chance of having the disorder.
Examples of autosomal recessive disorders
Cystic fibrosis- primarily Caucasians, 1 in 2500 births.
1/25 is a carrier
Abnormal protein is a membrane protein that transports chloride ions.
Results in thick sticky mucus in lungs, digestive tract and pancreas
No cure, but can be treated
Tay Sachs Disease- Primarily Jews of European descent and Cajuns
1 in 3600 births
Nonfunctional enzyme cannot breakdown lipids in brain cells
Seizures, blindness, degeneration of motor and mental function
Child dies before 5 years old
Sickle Cell Anemia
Primarily African; 1 out of 400 African Americans
Point mutation in hemoglobin
When oxygen levels are low, hemoglobin crystallizes
Pittsburgh Steelers safety
Ryan Clark will not play in
Sunday’s playoff game against
the Denver Broncos because
of a blood disorder bolstered by
high-altitude, low-oxygen
conditions. Clark, 32, has
sickle cell trait, which means
he carries an abnormal version
of the hemoglobin gene.
Although he’s spared the severe
symptoms of sickle cell disease
(in which both versions of the gene — one from each parent — are abnormal),
Clark could suffer life-threatening organ damage playing in Denver’s mile-high stadium.
A 2007 game in Denver sent Clark into sickle cell crisis, a complication that cost him his
spleen and gallbladder and ended his season.
Huntington Disease
Achondroplasia
Neurofibromatosis
Achondroplastic dwarves are
heterozygous; the homozygous
dominant condition is lethal.
What is the probability that two
achondroplastic dwarves will
have a child of normal height?
It depends on who you ask!!
If A = achondroplasia and a = no achondroplasia
AA= lethal
Aa= achondroplasia
aa= no achondroplasia
Aa x Aa
A
A
a
A A
A a
a
A a
25% lethal
50% heterozygous (achondroplasia)
25% normal height
a a
66% heterozygous (achondroplasia)
33% normal height
LO 2.28 The student is able to use representations or models to
analyze quantitatively and qualitatively the effects of
disruptions to dynamic homeostasis in biological systems. [See
SP 1.4]
Dihybrid Cross with Linkage
• In a normal case of Mendelian dihybrid
inheritance with independent assortment of
alleles, a cross between two heterozygotes
produces the expected 9:3:3:1 ration in the
offspring. In cases of dihybrid inheritance
involving linkage, the offspring of a cross
between two heterozygotes produces a greater
than predicted number of parental types and a
significantly smaller number of recombinants.
Linked Genes - genes that are on the
same chromosome tend to be inherited
together (duh)….
Unless they are separated by crossing
over (recombination)
Go back and look at the data on the previous page.
Calculate the RF- Answer is on the next page
There are 46 recombinants and a total of 427 offspring.
46/427 = .107 or 11%
Try another oneIn fruit flies, long wings (A) and gray bodies (B) are dominant to vestigial
wings and black bodies. In a cross of two heterozygotes AaBb x AaBb
you expect a 9:3:3:1 ratio. These are your results
123 long wing, gray body
21 long wing, black body
27 vestigial wing, gray body
129 vestigial wing, black body
Calculate the cross over value (recombination frequency) for the
offspring of the test cross.
48 recombinants divided by total of 300 = .16 or 16%
There are 3 genes on a single chromosome: A,
B and C. They exhibit the following crossing
over frequencies:
• A-B = 35%
Start with the longest one and space it
out. I marked my “chromosome” into
• B-C = 10%
increments of 5. Answer appears on
the click. Remember it could be in
• C-D = 15%
either order.
• C-A =25%
• D-B=25%
• Determine the order of the genes on the
chromosome.
A
D
C
B
Determine the sequence of genes along a chromosome based on the
following recombination frequencies.
A-B = 8%
Do the largest first and then work from
A-C = 28%
here. You have to put a pencil to it and
A-D= 25%
draw it out. It may seem hard at first, then
B-C = 20%
it gets easier. I filled in the first one- the
B-D = 33%
answer will appear when you click again.
Map units were measured with the space bar.
C
B
A
D
IMPORTANT NOTE: If this were multiple choice, these genes could be listed
CBAD or DABC. Gene mapping only gives you the relative distance between
the genes- not their specific locus.
Another way the question may be asked-
Another way the question may be asked-
This slide combines the two skills we have just reviewed!
I don’t know about you, but this is how
I feel right now…. Wonky.
P= smooth seeds crossed with wrinkled seeds
F1= all smooth seeds
F2= 5,474 smooth seeds and 1,850 wrinkled seeds
Calculate chi-square for these results_________________
What is your null hypothesis?
Is the chi-square you have calculated within the boundary of “the possible”? Explain.
Phenotypes
Total
O
E
O-E
XXXXXXXX
(O-E)2
XXXXXXXXX
(O-E)2
E
Phenotypes
O
E
O-E
(O-E)2
(O-E)2
E
smooth
5,474
5,493
-19
361
.07
wrinkled
1,850
1,831
19
361
.20
XXXXXXXX
XXXXXXXXX
.27
Total
Where did your expected value come from? You have to know what is
the probability of each phenotype in a genetic cross. In this cross it
should be 3:1 …..75% should be smooth and 25% should be wrinkled.
There is no way you can calculate chi square if you can’t do that step.
In this case you cannot reject your null hypothesis because the chi square
value of .27 is less than the critical value of 3.84.
A wild type fly (heterozygous for gray body color and normal wings) is mated
with a black fly with vestigial wings. The offspring have the following
phenotypic distributions:
Wild type- 50
Black, vestigial- 50
Black, normal-5
Gray, vestigial- 5
(This is the cross AaBb x aabb this info was not in the original problem)
Phenotypes
Total
O
E
O-E
XXXXXXXX
(O-E)2
XXXXXXXXX
(O-E)2
E
Phenotypes
O
E
O-E
(O-E)2
(O-E)2
E
Wild (gray,
normal)
50
27.5
22.5
506.25
18.4
Black/vestigial
50
27.5
22.5
506.25
18.4
Black/normal
5
27.5
-22.5
506.25
18.4
Gray/vestigial
5
27.5
-22.5
506.25
18.4
XXXXXXXX
XXXXXXXXX
73.6
Total
In this case we will definitely reject our null hypothesis because
our chi square value is much greater than the critical value
of 7.82. Really wonky, in fact, you should be able to explain
what this is based on the data!!
You can also use chi square as a way
to analyze your data in any type of
experiment where this is an “expected”
value to compare your “observed” to.
See if this experiment looks familiar-
What is the probability that each of the following pairs of
parents will produce the indicated offspring? (Assume
independent assortment of all gene pairs.)
a. AABBCC x aabbcc------------------> AaBbCc
1x1x1=1
b. AABbCc x AaBbCc-------------------> AAbbCC ½ x ¼ x ¼ = 1/32
c. AaBbCc x AaBbCc--------------------> AaBbCc
½ x ½ x ½ = 1/8
d. aaBbCC x AABbcc--------------------> AaBbCc
1 x ½ x 1 = 1/2