Download Lab 13 Genetics with answers

LABORATORY #17: Genetics Lab - Part 2
Objectives:
1. Understand how gender is determined in animals
2. Understand how X-linked traits are inherited using both Punnett squares and pedigree
charts
3. Understand how to create a pedigree chart for an X-linked trait
4. Understand the probabilities underlying X-linked traits
5. Understand multi-allelic traits such as blood type
6. Understand polygenic traits, epistasis and complementary genes
The Genetics of Sex
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 X-centered 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 get 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.
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!!!
So pedigree charts allow you to track the passage of a specific trait over several generations. In
human pedigrees, a circle indicates a female individual and a square indicates a male. A
horizontal line connecting them is a mating and the offspring are suspended below them via a
vertical line. If the circle of square is shaded in, the individual is afflicted with whatever trait is
under considered. In most genetic traits, this means the individual is homozygous recessive and
the individual shows the recessive trait. If the circle or square is unshaded, the individual does
not show the recessive trait.
A typical pedigree chart tracks the passage of a recessive trait. In this pedigree chart, the first
thing to do is fill in the genotypes for all the homozygous recessive individuals because this is the
genotype you definitely know. The next thing to do is to fill in the dominant alleles for the
unshaded, unaffected
individuals. You may
not be able to complete
the genotypes for these
unaffected individuals
because, while you know
they possess one
dominant allele (hence
the unshaded circle or
square), you may not
know the second allele.
To determine this, you
may have to look at the phenotypes and genotypes of several generations (i.e. the grandparents,
parents and kids). Remember – you only get one allele from each parent.
Some traits are passed on because they are found on the X chromosome. These traits are called
X- or sex-linked. To track the transmission of these traits, pedigree charts are also used. In these
charts, the chromosomes of X and Y, in addition to the allele being tracked, are added to the
chart. It is important to realize that a female (with two X chromosomes) only passes on an X
chromosome to ALL of her children. The male with his genotype of XY passes on an X to female
offspring and a Y to the male offspring.
The first thing you do in an X-linked pedigree chart is fill in the X and Y chromosomes (circle =
XX; square = XY). The next thing to do if fill in is the allele you are studying. This allele is
written as a superscript, attached to the X of the X chromosome (e.g. Xb). Because males have
only one X chromosome, it is easy to know their genotypes. In an X-linked recessive trait, males
showing the recessive phenotype will have an X chromosome bearing the recessive allele. If they
are not an affected make, you know their X chromosome has the dominant allele (e.g. XB). So in a
pedigree chart, you complete the males first. It is the female that can be more complicated
because of the two X chromosomes - but if she is an affected female showing the recessive
phenotype, you know she will have two recessive alleles (one on each X chromosome – i.e. XbXb).
However, if she is an unaffected female (i.e. shows the dominant phenotype), you know one of her
X-chromosomes will have the dominant allele. The other one will require you examining several
generations of the pedigree chart.
Worksheet Problems
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?
Cross: male XhY X female XHXh
From a Punnett square, there is a ¼ chance for being XhY .
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
15. 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 non-hemophiliac 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 nonhemophiliac 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
16. In humans, dark hair (B) is dominant over blondness (b), and color blindness (c) are both sexlinked recessive traits. A woman has a blond brother, a blond mother, and a dark-haired
father. Her brother and her parents have normal vision. She bears the following three
children by her blond, normal-visioned husband:
1. a dark-haired son with normal vision
2. a dark-haired daughter with normal vision, and
3. a dark-haired color-blind son
a. Make a pedigree of the entire family showing the probable genotypes of all individuals.
b. What is the probability that her next (fourth) child will be a color-blind boy? - ¼
c. If her fourth child is a boy, what is the probability that he will have dark hair? ½ of boys
d. What is the chance that her next four children will all be girls? ½ X ½ X ½ X ½ = 1/64
woman’s father = Bb and XCY – dark-haired father, normal vision
woman’s mother = bb and XCX-- blond mother, normal vision
woman’s brother = bb and XCY - blond brother, normal vision
woman = Bb and XCXc
woman’s mate = bb and XCY - blond, normal-visioned husband
1st boy = Bb and XCY - dark-haired son with normal vision
2nd boy = Bb and XcY - dark-haired color-blind son
Girl = Bband XCX-- - dark-haired daughter with normal vision
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 BO – but she has type O child (OO) which means she must be
BO
The child can only be OO
The father CANNOT be AB, AA or BB, but could be AO, BO or OO
Of course there is nothing saying that Long can’t be ll and Round be LL!
5. In humans, the ABO blood groups are controlled by three alleles (only two of which occur in
any one individual): the alleles for A and B type blood are co-dominant toward each other, and
both are dominant to the allele for O type blood.
a. If a person with type AB blood marries someone with type O blood, what are the possible
phenotypes of their offspring? Type A and type B (AB X OO cross  AO and BO)
In the following, determine the genotypes of the parents:
b. One parent has type A and the other has type B, but all four blood groups are represented in
the children. AO and BO  AO, BO, OO and AB
c. Both parents have type A, but 3/4 of the children are A and 1/4 are O. AO and AO
d. One parent has type AB and the other has type B, but of the children 1/4 have type A, 1/4 have
type AB, and 1/2 have type B. AB and BO
6. In the following cases of disputed paternity, determine the probable parent.
a. Mother is type B, child is type O. Father #1 is A; father #2 is AB.
b. Mother is type B, child is type AB. Father #1 is A; father #2 is B.
c. Mother is type O and bears non-identical twins, one type A and one type B. Father #1 is type
A; father #2 is type B.
a. Type B  OO child
Father #1 with AO genotype
b. Type B  AB child
Father #1 with A- genotype
c. Type O  A- and B- children
neither is possible – but then again, both can be possible
if she had sex with both of them with each of them
fertilizing the two ovulated eggs
7. Two babies in a maternity ward have lost their identity bands, and there is some confusion
about their footprint records. Baby #1 is type A; baby #2 is type B. If you are one of the
mothers and your blood type is O, which one of the following statements applies?
a. Neither baby could be yours.
b. The type A baby is yours.
c. The type B baby is yours.
d. Either baby could be yours.
III. Gene Interactions
Gene interactions are genetic circumstances where two or more non-allelic gene pairs influence
the expression of a trait. The common types of gene interactions that exist are:
A. Polygenetic inheritance: Polygenetic inheritance is controlled by non-allelic gene pairs,
with each allele having an additive effect. A range of phenotypes are possible for this kind
of inheritance.
B. Epistasis: Epistasis is a type of gene inheritance in which two pairs of non-allelic 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 non-allelic
gene pairs. In this case, either gene pair has veto power over each other.
1. 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, RrBB, RRBb or
RrBb. White results when both genes are homozygous and recessive – i.e. rrbb. Having one of the
two genes in recessive form results in 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?
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. Rbb 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.
2. In humans, the allele for brown eyes (B) is dominant over the allele for blue eyes (b). The
allele for normal melanin production and deposition in the iris (M) is dominant over the allele
for albinism (m). If the gene for melanin production and deposition is homozygous recessive (i.e.
mm) it results in the appearance of a red iris, as it “vetoes” the phenotype that would result
from the B and b alleles. Give the expected genotypes and phenotypes and their ratios for the
children of two individuals who are heterozygous for brown eyes and melanin production. How
many of these children will be albinos and what are their genotypes?
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 non-allelic 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. If either body color gene is homozygous for albinism, the finch is an albino, no
matter what the genotype for the other gene is. 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
b. A QQzz x qqZZ cross
a. 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
4. In cocker spaniels, the following genotypes and phenotypes are found:
AABB = white A-bb = red aabb = lemon AaB- = black aaB- = liver AABb = grey
a. A red female is mated with a liver-colored male, and one of the pups produced is
lemon-colored. What are the genotypes of the parents?
Aabb X aaBb  aaBb liver & aabb lemon pups are capable of being produced from these parents
b. What proportion of these offspring would be expected to be black?
Aabb X aaBb  black offspring
Aabb gametes are Ab and ab
aaBb gametes are aB and ab
Black = AaB- which is AaBB or AaBb
From the Punnett square - ¼ will be black
c. A black male (AaB-) is mated with a liver-colored female (aaB-), and they produce the
following pups:
i. 3/8 black (AaB-) & 1/8 red (A-bb)
ii. 3/8 liver-colored (aaB-) & 1/8 lemon-colored (aabb)
What are the genotypes of the two parents? AaB- and aaBBlack male = AaBb
X
liver female = aaBb
Black male gametes – AB, Ab, aB and ab
Liver female gametes – aB and ab
While there will be other pups produced, the black to red ratio is 3:1 and the liver to lemon ratio
is 3:1
Because a red puppy is produced (A-bb), the black male and liver female must have Bb genotypes
Because a lemon puppy is produced (aabb), the black male and liver female must have Bb
genotypes.
5. If two cocker spaniels of the genotypes below are mated (see below), and eight pups are born,
what is the most likely distribution of coat colors in that litter?
P1
AaBb x AABb
___1___white ___1___red ____1__lemon ___3___black ___1___liver ___1___ grey
Punnett square crosses AB Ab aB ab gametes with AB and ab gametes
Progeny:
AABB white
AABb grey
AaBB black
AaBb black
AaBb black
Aabb red
aaBb liver
aabb lemon
6. A dominant gene, A, causes yellow color in rats. The dominant allele of another independent
gene, R, produces black coat color. When the two dominant genes occur together (A-R-), they
interact to produce grey coat color. Rats of the double recessive genotype are cream-colored. If a
grey male and a yellow female are mated and produce approximately 3/8 yellow, 3/8 grey, 1/8
cream, and 1/8 black, what were the genotypes of the parents? AaRr grey and Aarr yellow
A- = yellow
R- = black
A-R- = grey
aarr = cream
A-rr = yellow
aaR- = black
if A-R- grey is crossed with A-rr yellow, the progeny will be:
3/8 A-rr = yellow
3/8 A-R- = grey
1/8 aarr = cream – BUT this phenotype tells you that both parents must have recessive ‘a’ and ‘r’
alleles
1/8 aaR- = black
So the cross is AaRr X Aarr
7. The inheritance of color blindness in humans is due to a recessive gene located on the X
chromosome (X linked).
XC (normal) > Xc (color blind)
a. If a color-blind boy is born to parents both of whom have normal vision, what are the
genotypes of the three individuals? XCY x XCXc  XcY color blink boy
b. What is the probability that the second child born to that couple will be a color-blind
daughter? 0% - a color blind daughter is not possible with a XCY x XCXc cross
c. If this couple has four children, what is the probability that the first two children will be colorblind boys and the last two children will be girls with normal vision? 1/64
chance of a color blind boy XcY = ¼
chance of a normal girl XCX- = ½
probability = ¼ x ¼ x ½ x ½ = 1/64
IV. Linkage and Crossing Over
1 . In the fruit fly, Drosophila melanogaster, crossing over is absent in the males. Suppose you
are interested in the relationship between two linked genes on chromosome 2. The genes are
for black body (b) and for curved wings (c), each of which is recessive to the normal (B) body
color and normal (C) wing shape. In your laboratory, females that are heterozygous at the
body color and wing shape loci are mated with a black-bodied male with curved wings. The
offspring of these matings were counted, with the following results:
367 normal body, curved wing = parental phenotype
131 normal body, normal wing = recombinant phenotype
139 black body, curved wing = recombinant phenotype
363 black body, normal wing = parental phenotype
Total progeny produced = 1000
a. What is the recombination frequency between these two loci?
[131+139/ 1000]x 100% = 27%
b. How many map units separate these two loci? 27 map units
2. A series of dihybrid test crosses gives the following crossover frequencies:
A-B, 3%,
A-C, 13%
B-C, 10%
a) What is the gene sequence? A, B, C
b) Another cross gives a frequency of 19% for A-D. Can you locate the position of D in
the sequence of genes you determine above? No – not enough information given
c) If the crossover for C-D is found to equal 6%, where would gene D be located? A, B, C, D
A to B distance = 3 map units
B to C distance = 10 map units
C to D distance = 6 map units
A to C distance = 13 map units
A to D distance = 19 map units