LAB #13 GENETICS LAB #2
SEX GENETICS
The sex expressed in an individual’s body is determined by whole chromosomes, not individual genes.
There are many systems found in nature that determine sex. The most common is the XX/XY
determination system found in mammals and some insects. Although XX/XO, ZW and UV determination
systems are also well-described.
In the XX/XY determination system, X and Y are the symbols used to identify sex chromosomes in most
mammals and in some insects. In mammals, the female gender occurs when two X chromosomes are
present and the male genders results from the presence of an X and a Y. However, since the Y
chromosome possesses a gene called the SRY gene (sex-determining region of the Y chromosome), it is
the presence of the Y chromosome that determines mammal gender. If a Y chromosome with an active
SRY gene is present, you have a male. If the Y chromosome is missing, you have a female. This explains
why XO humans are considered female. This is known as Y-centered sex determination.
X-centered sex determination is also found. In fruit flies, it is the presence of 2 X chromosomes that
determines gender. Two X chromosomes define the fly as female. The Y chromosome is not important.
Therefore, XXY fruit flies are considered female, despite the presence of a Y chromosome. This is Xcentered sex determination.
Think this is all very normal? Well, here’s an interesting variation on XX/XY sex determination. In some
fish, there is another Y chromosome called Y’. XY’ fish are female. YY’ fish are males.
Finally, the number of sex chromosomes can also have an effect on gender. Fruit flies cannot survive if
more than 2 X chromosomes are present. Therefore, XXX flies cannot exist. The platypus has ten sex
chromosomes! XYXYXYXYXY patterns gets you a boy platypus. XXXXXXXXXX gets you a girl!
In the XX/XO determination system, found in many insects, females have two copies of the X
chromosome (XX) while males have one (XO). This system is also found in some mammals, such as the
spiny rat and shrews. Remember is humans, XO is considered female and is known as Turner’s
syndrome.
In other organisms, there are different sex chromosomes, such as Z and W. This is seen in birds, some
reptiles and some insects like butterflies and moths. In these organisms, the female is ZW and the male
is ZZ. In chickens, the W chromosome contains female-determining genes that work in a way similar to
the SRY gene of the mammalian Y chromosome. In moths, the presence of two Z chromosomes
produces twice the amount of specific enzymes and a male results. In the plants known as the
bryophytes, the sex chromosomes are called U and V.
I. Sex-linkage problems – X-linked disorders
In working through X-linked problems, we use pedigree charts. Pedigree charts track the progression of
genetic conditions over multiple generations and takes gender into account. However, pedigree charts
can be used to look at both Autosomal and X-linked disorders!!!
In a pedigree chart we use a circle for a female and a square to denote a male. In autosomal disorder,
we don’t write out the XX and XY genotypes for each member of the pedigree chart, but simply use the
shapes to denote gender. In X-linked disorders, we do write out the genotype and add the letter for the
condition as a superscript onto the X (e.g. Xh or XH). Affected individuals have their circle/square colored
in. Non-affected individuals have uncolored shapes. Individuals carrying an affected gene, but not
showing the phenotype for the condition are referred to as carriers.
Here is an example:
For Questions 1-9, use the pedigree chart shown below. Some of the labels may be used more than
once.
C 1. A male
A 2. A female
B 3. A “marriage”
A 4. A person who expresses the trait
C 5. A person who does not express the trait
D 6. A connection between parents and offspring
3 7. How many generations are shown on this chart?
Assuming the chart above is tracing the autosomal dominant trait of "White Forelock (F)" through the
family. F is a tuft of white hair on the forehead.
Ff 8. What is the most likely genotype of individual “A”? (FF, Ff or ff?)
ff 9. What is the most likely genotype of individual “C”? (FF, Ff or ff?)
Because some of the children are normal, homozygous recessive, then the affected parent MUST be
heterozygous
Complete the following problems in your lab notebook. If necessary, draw a pedigree chart to help you
work through these problems
10. In humans, Hemophilia is a blood clotting disorder in which one of the proteins needed to
form blood clots is missing or reduced. Individuals have difficulty forming blood clots following
injury and may suffer significant blood loss from even minor cuts and bruises. The key clotting
factor in hemophilia is called Factor VIII. The gene for Factor VIII is found on the X chromosome.
Hemophilia is caused by a lack of Factor VIII and results from a recessive allele (h) found on the X
chromosome. The condition for normal blood clotting dominates over non-clotting. A
hemophiliac male has a child with a woman who is a carrier for this condition. Using the letter H
for normal Factor VIII and h for abnormal Factor VIII, what are the chances that they have a
male child who clots normally?
From a Punnett square, there is a ¼ chance for this.
Alternatively, there is a ½ change the child will be a boy AND a ½ chance the child will clot
normally = ½ X ½ = 1/4
11. In humans, the condition for normal vision dominates color blindness. Both are linked to the X
chromosome. Using C for normal vision and c for the color-blind disorder, what are the chances that a
color-blind female and color-blind male will have a daughter with normal vision? There is no chance of
this since the color blind female carries two recessive X chromosomes and the father carries one
recessive X chromosome.
XcXc x XcY
12. A normal-visioned male marries a color-blind female. She gives birth to a color-blind daughter. The
husband claims the child is not his. Can you support his claim? How?
An XcXc x XCY cross CANNOT produce a XcXc color blind female because the father gives an XC
13. If a color-blind male has a child with a woman who carries one color-blind gene, what are the
chances that they have a male child with normal vision?
a. 0
b. ¼
c. 1/3
d. 2/3
e. ½
f. ¾
g. 1/1
14. What genotype must each parent have to produce a color-blind daughter?
The parents are
Dad = XcY
Mom = XcXc OR XCXc
14. Draw Your Own Pedigree: Hemophilia
Remember that because hemophilia is an X-linked disorder, when you identify genotypes in this
pedigree, you must use the XX/XY notation and use superscripts with each X chromosome to indicate
whether the “H” (normal) or “h” (hemophilia) allele is present. (Ex. XHY = normal male)
Hemophilia became known as the “Royal disease” after it suddenly cropped up in some of the
descendants of Great Britain’s Queen Victoria and spread through the royal families of Europe. Queen
Victoria and her husband Prince Albert had 9 children – 5 girls (Beatrice, Victoria, Alice, Helena, and
Louise – none of whom were hemophiliacs) and 4 boys (Edward, Alfred and Arthur had normal blood
clotting; their son Leopold, however was a hemophiliac). Beatrice married a man named Henry and they
had four children (sons Leopold and Maurice who were hemophiliacs, daughter Eugenie who was not a
hemophiliac, and another son who was also not a hemophiliac). Eugenie married Alfonso XIII of Spain
(non-hemophiliac) and they had 6 children (2 normal sons, 2 normal daughters and 2 hemophiliac sons).
One of those normal sons married a non-hemophiliac woman and gave birth to one son – a nonhemophiliac they named Juan Carlos (the reigning King of Spain).
Draw your own pedigree for the Royal family in your lab notebook
1. Draw the marriages and children – make the men boxes and the women circles
2. Color in all affected individuals – i.e. the ones with hemophilia
3. Fill in the X and Y chromosomes
4. Add the H and h alleles to the appropriate chromosome
HINT: Do the males first since you know if they have hemophilia they are XhY and if they don’t they
are XHY
Victoria = XHXh
Albert = XHY
Sons with hemophilia = XhY
Beatrice = XHXh
Henry = XHY
Sons with hemophilia = XhY
Eugenie = XHXh
Alfonso XIII = XHY
Sons with hemophilia = XhY
A little bit more info – for fun
Another of Victoria and Albert’s daughters was Alice (XHXh ) who married Louis (XHY). Their daughter
Alexandria (XHXh ) married Csar Nicholas of Russia (XHY). They had one hemophiliac son (Alexander= XhY)
and what we think were normal daughters. We don’t know since they along with their entire families
were murdered during the Bolshevik revolution.
One of Victoria and Albert’s sons, Edward Louis (XHY) married and had a son names George V (XHY).
George V married Mary and had a son, George VI. George the VI married Elizabeth and had the current
Queen Elizabeth II. Because George V was normal and married a non-hemophiliac women who wasn’t a
carrier, the current Royal family of Britain is now hemophilia free
Another child of Alice and Louis was the grandmother of Prince Philip, the current consort of Queen
Elizabeth II
II. Multiple Alleles
For some traits, there are more than two alleles present in the population to form genotypes. These are
known a multiple allelic traits. Although the inheritance of multi-allele traits is the same as traits with
only two alleles, the number of possible genotypes is greater.
Complete the following in your lab notebook. Use a Punnett square to help you determine your
answers.
1. In humans, the trait for type A blood and type B blood show incomplete dominance, so that a
person with both alleles as blood type AB. Both A and B are dominant over the O allele.
Therefore type A blood can have two genotypes: AA and AO. A person with type A blood has a
child with a person with type O blood. List the types of offspring they could have and the
probability for each blood type.
If the woman is AO, her children will be AO and OO – 50% probability for each
If the woman is AA, her children will all be AO
2. A woman with type O blood gives birth to a baby with type O blood. In a court case, a man with
blood type A is named as the father. Could he be the father of this child? Is he the only father or
are there other possibilities
The woman’s genotype can only be OO
With type A, the father can be AO or AA. But to have OO kids, we know his genotype to be
AO. So yes he can be the father - only if his genotype is AO.
However, someone with OO and BO genotypes could also be the father.
3. A woman with blood type B has a child with blood type O. What are the genotypes of the
mother and child? Which genotypes could the father NOT have?
The mother can be BB or NO
The child can only be OO
The father CANNOT be AB, AA or BB, but could be AO, BO or OO
4. In radishes, the shape may be long, round or oval. Crosses between long and oval gives 159 oval
and 156 long. Crosses between oval and round produce 203 oval and 199 round. Crosses
between long and round give 576 oval. Crosses between oval and oval give 121 long, 243 oval
and 119 round. Which of the shapes are defined by a homozygous dominant genotype?
Homozygous recessive? Heterozygous?
For this problem, use the letters L and l
Long is LL
Round is ll
Oval is Ll – If a cross between long and round give 100% oval offspring, you know that long
and round MUST be homozygous and the offspring must be heterozygous because only a cross
between two homozygous genotypes will give offspring all with the same phenotype.
To check: the LL and Ll cross will give LL and Ll in a 1:1 ratio; the Ll and ll cross will give Ll and ll
in a 1:1 ratio; crosses between Ll and Ll give a 1:2:1 ratio of long:oval:round
Of course there is nothing saying that Long can’t be ll and Round be LL!
III. Gene Interactions
Gene interactions are genetic circumstances where two or more nonallelic gene pairs influence the
expression of a trait. The common types of gene interactions that exist are:
A. Polygenetic inheritance
B. Epistasis
C. Complementary genes
A. Polygenetic inheritance: Polygenetic inheritance is controlled by non-allelic gene pairs, each
having an additive effect. A range of phenotypes are possible
B. Epistasis: Epistasis is a type of gene inheritance in which two pairs of nonallelic genes determine
the expression of a trait and one gene has “veto” power over the other. Animal coat color and
human eye are the best examples of epistasis
C. Complementary genes: Complementary genes are a gene interaction between nonallelic gene
pairs. In this case, both gene pairs have veto power over each other
1. In some types of wheat, color is caused by two sets of genes. The dominant genes, R and B, are
needed for red wheat. Red wheat has the genotypes RRBB, RrBB, RRBb or RrBb. White results when
both genes are homozygous and recessive – i.e. rrbb. A strain with the genotype Rrbb is crossed with a
strain of wheat with the genotype rrBb. What is the color of each of the parent strains? Using a Punnett
square, what are the possible genotypes and colors that can result from this cross?
The important thing to realize is that if you have a dominant R and a dominant B allele in the
genotype– you get red wheat, but if you have only ONE of these alleles in dominant form (e.g. R-bb or
rrB-) you get brown wheat. If you have both in recessive forms – rrbb – you get white wheat.
Rrbb crossed with rrBb will give the following genotypes in equal numbers: RrBb, rrBb, Rrbb, rrbb
The rrbb offspring will be white, the genotype RrBb will be red. All other genotypes will give brown
wheat.
This is the version of the question from the old lab manual
In some types of wheat, color is caused by two sets of genes. The dominant gene alleles, R and B, are
needed for red wheat. For example, red wheat has the genotype RRBB or RrBb. White results when both
genes are homozygous and recessive – i.e. rrbb. Any other genotype combination produces brown
wheat. A brown wheat strain with the genotype Rrbb is crossed with another strain of brown wheat
with the genotype rrBb. Using a Punnett square, what are the possible genotypes and colors that can
result from this cross?
Ignore the statement “Any other genotype combination produces brown wheat”. To eliminate the
confusion of this question, it was re-written as question #1 above.
2. In humans, the allele for brown eyes (B) is dominant over the allele for blue eyes (b). The allele for
normal melanin production (M) is dominant over the allele for albinism (m). Give the expected
genotypes and phenotypes and their ratios for the children of two individuals who are heterozygous for
brown eyes and melanin production (i.e. albinism)
A BbMm X BbMm dihybrid cross will produce a 9:7 ratio of brown eyes with the pigment being
deposited to blue eyes (brown pigment made, no deposition or no brown pigment made)
It will result in the following:
Brown eyes since the BB makes the brown pigment and the MM allows its deposition
1 BBMM
2 BBMm
2 BbMM
4 BbMm
Blue eyes – no brown pigment made at all (M allele can be MM or Mm)
1 bbMM
2 bbMm
Albino eyes – brown pigment is made (BB or Bb), blue pigment can also result (bb) but mm means no
deposition
1 BBmm
2 Bbmm
1 bbmm
3. In zebra finches, two pairs of nonallelic genes control body color. The dominant alleles of each gene
(Q and Z) produce normal body color and the recessive alleles of each pair (q and z) produce albinos
(e.g. qqzz). If either homologous chromosome pair is homozygous for albinism (i.e. qq or zz), the finch
is an albino. Do the following crosses and determine the expected genotypic and phenotypic ratios of
the possible offspring
a. Cross between two normal colored finches who are heterozygous for both genes
This is a standard dihybrid cross between QqZz X QqZz finches
But we are looking for either qq or zz genotypes. This means that in a qqZZ finch, you might
expect a normal color because of the ZZ part of the genotype. But the qq “vetoes” this
genotype and the bird ends up an albino.
The following genotypes fulfill the condition of either homologous chromosome being
homozygous for albinism (i.e. qq or zz): QQzz, qqZZ, qqzz will be albinos and you will get 1 of
each (i.e. 1/16 chance); but 2 finches will be Qqzz and 2 finches will be qqZz. These will also be
albinos and each of their probabilities will be 1/8.
b. A QQzz x qqZZ cross between two albino finches produces QqZz offspring – all of which
will be normal colored