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ESERCITAZIONE 3
HUMAN GENETICS
BLOOD GROUPS
RECOGNITION OF PATHERNITY
FAMILY PEDIGREE
INCOMPLETE DOMINANCE AND CO-DOMINANCE
INCOMPLETE DOMINANCE:
Is a form of intermediate inheritance in which one allele for a specific trait is
not completely expressed over its paired allele. This results in a third
phenotype in which the expressed physical trait is a combination of the
phenotype of both alleles.
(ex: color of Bella di notte petals )
CO-DOMINANCE:
In co-dominance an additional phenotype is
produced, however both alleles are expressed
completely. The eterozygous shows the traits of
both parents.
( ex: blood group)
1. The human blood groups system ABO is regulated by 3 alleles
according to the following scheme:
BLOOD
GROUP(Phenotype)
Possible Genotypes
A
IA IA
IA i
IB IB
IB i
B
AB
0
IA IB
ii
Which is the dominance relation in this series of multiple alleles?
IA is dominant over i
IB is dominant over i
IA and IB are codominant
2. Determine parental genotypes in the following crosses:
Phenotype of
crossed individuals
Phenotype in the
progeny
a)
A X A
¾ A; ¼ 0
b)
A X AB
A; AB
c)
B X 0
B; 0
d)
AB X A
B; A; 0; AB
Genotype of crossed
individuals
IA i X IA i
a) The parents have one dominant allele IA therefore the parents
could be IA IA o IA i.
In the progeny we have the phenotype 0 (recessive trait) then the
parents must be IA i
Frequencies of expected genotypes in the progeny :
IA IA(1/4), IA i (2/4), ii (1/4)
Frequencies of phenotypes: A (3/4) e O (1/4)
2. Determine parental genotypes in the following crosses :
Phenotype of
crossed individuals
Phenotype in the
progeny
a)
A X A
A; 0
b)
A X AB
½ A; ½ AB
c)
B X 0
B; 0
d)
AB X A
B; A; 0; AB
Genotype of crossed
individuals
IA IA X IA IB
b) The second parent must be IA IB and his gametes will be (½) IA
and (½) IB
The first parent could be IA IA o IA i but since we don’t have B
phenotype in the progeny then he must be homozygous IA IA
2. Determine parental genotypes in the following crosses :
Phenotype of crossed Phenotype in the
individuals
progeny
a)
A X A
A; 0
b)
A X AB
A; AB
c)
B X 0
½ B; ½ 0
d)
AB X A
B; A; 0; AB
Genotype of crossed
individuals
IB i X ii
c) The second parent is homozygous recessive ii, while the first one
could be IB IB / IB i
Since in the progeny individuals homozygous recessive appear (0
group), the first parent must be heterozygous IB i.
2. Determine parental genotypes in the following crosses :
Phenotype of crossed
individuals
Phenotype in the
progeny
a)
A X A
A; 0
b)
A X AB
A; AB
c)
B X 0
B; 0
d)
B X A
¼ B; ¼ A; ¼ 0; ¼ AB
Genotype of crossed
individuals
IB i X IA i
d) Parent Genotypes: IB IB / IB i x IA IA / IA i i
In the progeny we have individuals homozygous recessive ii
(group0) then the first parent must be heterozygous IB i and also
the second one must be heterozygous IA i
3. Rh system is determined by the gene D: the presence of antigen (Rh+
individual) is due to the dominant allele D; the absence of antigen (Rh –
individual) is homozygote for the recessive allele d.
Determine the genotypes of individuals crossed:
Phenotype of
crossed
individuals
Phenotype in the progeny
AB Rh+ X 0 Rh+
3/8 A Rh+; 3/8 B Rh+; 1/8 A Rh-; 1/8 B Rh-
B Rh- X A Rh+
1/4 AB Rh+; 1/4 A Rh+; 1/4 B Rh+; 1/4 0 Rh+
Genotype of
crossed
individuals
IA IB Dd x ii Dd
a) The first parent must be IA IB , while the second one is ii.
b) Factor Rh: both parents could be DD/Dd. Since we have
individuals with Rh- in the progeny, the parents must be
heterozygous Dd
Now use the brunch diagram to predict the expected phenotypes
classes and their frequencies in progeny.
IA IB Dd x i i Dd
Gene I
IA IB x i i
Gene D
Dd x Dd
Expected phenotypes
Rh+ (3/4)
A Rh+ (1/2 x 3/4 = 3/8)
Rh- (1/4)
Rh+ (3/4)
A Rh- (1/2 x 1/4 = 1/8)
B Rh+ (1/2 x 3/4 = 3/8)
Rh- (1/4)
B Rh- (1/2 x 1/4 = 1/8)
A (1/2)
B (1/2)
3. Rh system is determined by the gene D: the presence of antigen (Rh+
individual ) is due to the dominant allele D; the absence of antigen (Rh
– individual) is homozygote for the recessive allele d.
Phenotype of
crossed individuals
Offsprings phenotypes
Genotypes of
parents
AB Rh+ X 0 Rh+ 3/8 A Rh+; 3/8 B Rh+; 1/8 A Rh-; 1/8 B RhB Rh- X A Rh+
1/4 AB Rh+; 1/4 A Rh+; 1/4 B Rh+; 1/4 0 Rh+
IB i dd x IA i DD
a) The genotype of first parent could be IB IB o IB i (blood group) and dd for Rh
factor. The second parent could be IA IA o IAi and DD o Dd.
b) Since in the progeny the recessive trait 0 appears but Rh- does not emerge,
both parents must have one recessive allele about blood group then the
genotype will be IB i x IA i. About Rh, the second parent must be DD
4 classes of phenotypes are expected with the same frequency (1:1:1:1)
Now use the brunch diagram to predict the expected phenotypes
classes and their frequencies in the offsprings
IB i d d x IA i D D
Gene I
IB i x IA i
Gene D
dd x DD
Expected phenotypes
A (1/4)
B (1/4)
AB (1/4)
O (1/4)
Rh+ (1)
Rh+ (1)
Rh+ (1)
Rh+ (1)
A Rh+ (1/4)
B Rh+ (1/4)
AB Rh+ (1/4)
O Rh+ (1/4)
4. Mother and son have the indicated blood groups. Could the man be
the father of the child?
A Rh- B Rh+
B Rh-
-
0 Rh
Mother’s genotype: IA IA / IA i dd
child’s genotype:
IB IB / IB i DD/Dd
man’s genotype:
IB IB / IB i dd
The man is not the father of the child. He hasn’t the allele D.
0 Rh+
4. Mother and son have the indicated blood groups. Could the man be
the father of the child?
0 Rh-
0 Rh+ A Rh+
Mother’s genotype: IA IA / IA i DD/Dd
child’s genotype:
i i DD/Dd
man’s genotype:
i i dd
The man could be the father of the child if the mother’s
genotype is IA i DD/Dd
4. Mother and son have the indicated blood groups. Could the man be
the father of the child?
AB Rh+
A Rh- 0 Rh-
Mother’s genotype: i i dd
child’s genotype:
IA IA / IA i dd
man’s genotype:
IA IB DD/Dd
The man could be the father of the child if his genotype is Dd
5 – Gli individui indicati hanno avuto un figlio con il fenotipo indicato. Calcola qual è
5. Figure
outdithe
probability
of these events.
la probabilità
questo
evento?
AB Rh- 0 Rh+
0 Rh+
?
B Rh-
A Rh+
?
0 Rh-
6Mother’s
- Negli alberi
genealogici i simboli
usati sono:
genotype:
IA IBpiùdd
Father’s genotype:
i i DD/Dd
child’s genotype: IB i dd
The child is certainly heterozygous for blood group because he inherited ì
allele from the father.
Since the child is dd, the father must be heterozygous Dd
The probability that the cross IA IB dd x i i Dd produce a child IB i dd is:
Probability of phenotype B = ½ x 1 = ½
Probability of phenotype Rh- = 1 x ½ = ½
Probability of phenotipe B Rh- = ½ x ½ = 1/4
5.5 Figure
out the
probability
of un
these
– Gli individui
indicati
hanno avuto
figlio events.
con il fenotipo indicato. Calcola qual è
la probabilità di questo evento?
AB Rh-
0 Rh+
0 Rh+
A Rh+
B Rh-
?
0 Rh-
?
genotype:
i i DD/Dd
6 Mother’s
- Negli alberi
genealogici i simboli
più usati sono:
Father’s genotype:
child’s genotype:
IA IA / IA i DD/Dd
i i dd
Since the child is homozygous recessive for both genes, the father must be
heterozygous IA i Dd. The mother must be eterozygous for Rh factor: ii Dd
The probability that the cross ii Dd x IAi Dd produce a child ii dd is:
Probability of phenotype 0 = ½ x 1 = ½
Probability of phenotype Rh- = ½ x ½ =¼
Probability of phenotipe 0 Rh- = ½ x ¼ = 1/8
1 - In family trees the most used symbols are :
Male
mutant
Female
normal
I
Parents
II
children
1
2
3
A roman number is assigned to every generation and the individuals of
the same generation are progressively numbered from left to right with
Arabic numbers.
INHERITANCE
DOMINANT
All the generations have individuals that
show the trait
AUTOSOMAL
Male and female are usually affected
with equal frequency.
RECESSIVE
The affected individuals are rare and
they are not present in all the
generations.
SEX-LINKED
The trait emerged only in male or female
In the following family trees, black symbols represent a mutant phenotype. Determine if the
Nei seguenti
alberi
genealogici,
i simboli
neri indicano
untaking
fenotipo
Determina
mutant
phenotype
is due
to a dominant
or a recessive
allele,
intomutato.
consideration
that the
se il fenotipo mutato può essere dovuto a un allele dominante o recessivo,
mutant allele is rare
considerando che il carattere è raro.
a)
A-
I
aa
II
aa
Aa
Aa
aa
aa
III
Aa
aa
Aa
Mutant phenotype is present in all generations.
Then the mutant allele is DOMINANT
The mutant individuals are heterozygous or homozygous dominant
(AA o Aa) while the normal individuals are homozygous recessive
(aa).
In the following family trees, black symbols represent a mutant phenotype. Determine if the
mutant phenotype is due to a dominant or a recessive allele, taking into consideration that the
mutant allele is rare
b)
Aa
I
Aa
II
aa
aa
III
7 - Considerando che le persone che entrano nella famiglia per matrimonio non siano
portatori del carattere mutato, calcola con quale probabilità può nascere un figlio
Here
wedall'incrocio
have mutant
affetto
di III, 3 sons
e III, 4 from
del 1b.normal individuals. The mutated
allele is RECESSIVE.
The mutant individuals are homozygous recessive, the parents are
hetorozygous.
Considering people that become part of the family through marriage don’t have the
mutant allele, calculate the probability that an affected (sick) child is born from cross
III,3 and III,4
b)
Aa
I
II
a
A
AA
1/4
Aa
1/
4
a
Aa
1/4
A
Aa
1/
4
A
AA
AA
aa
III
A
Aa
aa
Aa (2/3)
A AA
AA
Aa (1) ?
Aa (½)
7 - Considerando che le persone che entrano nella famiglia per matrimonio
non 1/
siano
1/4
Female
III,3 isdel
heterozygous
Aa (probability=1)
because
she is normal
and the
portatori
carattere mutato,
calcola con quale
probabilità
può nascere
un 4figlio
affetto
dall'incrocio
di III, 3recessive)
e III, 4 del 1b.
mother
is sick
(homozygous
a Aa
Aa
Man III,4 could be heterozygous if his father is heterozygous (II,5). The 1/4 1/
probability that II,5 is heterozygous is 2/3 (he is normal son of heterozygous 4
parents). The probability that III,4 inherits the recessive allele is ½ (the cross is
Aa X AA (external)).
The probability that an affected (sick) child is born from cross III,3 and III,4 is ¼.
The probability that all these events happen is the product of all the probabilities:
1 x 2/3 x ½ x ¼ = 1/12
8. in the following family trees, black symbols represent the mutant phenotype of
recessive homozygote. Considering that mutant alleles are rare, calculate the
probability of an affected child in the indicated crossed.
a)
I
II
AA (1)
AA (1)
aa (1)
aa (1)
Sick individuals of II generation are
homozygous recessive with
probability 1
III
Aa
IV
Aa (1) ?
Aa (1/2 )
(1/4 )
The individuals of IV generation could produce a sick child if they are heterozygous.
The probability that IV,1 is heterozygous is 1 because the cross is AA x aa
Probability that IV,2 is heterozygous????!!!. III,4 is heterozygous (p=1) because the cross
is aa x AA and he donates his recessive allele. The son IV,2possess the recessive allele
with the probability of ½
Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4
The probability that all these events happen is the product of all the probabilities
: 1 x ½ x ¼ = 1/8
8. in the following family trees, black symbols represent the mutant phenotype of
recessive homozygote. Considering that mutant alleles are rare, calculate the
probability of an affected child in the indicated crossed.
b)
I
II
aa
Aa
(1)
III
Aa
(1)
?
III,3 e III,4 must be both heterozygous in
order to produce a sick son.
Aa
(1/2)
(1/4)
III,3 is certainly heterozygous because his mother is homozygous recessive (p=1)
III,4 will be heterozygous if he inherited from his father the recessive allele.
The probability that the father II,3 is heterozygous is 1 (the parents are aa x Aa)
III,4 is heterozygous with probability ½.
Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4
The probability that all these events happen is the product of all the probabilities
: 1 x ½ x ¼ = 1/8
8. in the following family trees, black symbols represent the mutant phenotype of
recessive homozygote. Considering that mutant alleles are rare, calculate the
probability of an affected child in the indicated crossed.
Aa
c)
Aa
I
AA
AA
II
aa
III
Aa
(1)
(2/3) Aa
?
III,1 and III,2 must be both
heterozygous to have a sick
son.
Aa
(1/2)
III,1 is certainly heterozygous (p=1).
III,2 could be heterozygous if the father is heterozygous. The probability that II,3 is
heterozygous is 2/3 (health son of the heterozygous parents)
III,2 is heterozygous with probability of ½.
Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4
The probability that all these events happen is the product of all the probabilities:
1 x 2/3 x 1/2 x ¼ = 1/12
8. in the following family trees, black symbols represent the mutant phenotype of
recessive homozygote. Considering that mutant alleles are rare, calculate the
probability of an affected child in the indicated crossed.
d)
I
aa (1)
II
Aa
(2/3)
?
Aa
(1)
II,3 will be heterozigous with P=2/3, (health son of heterozygous parents)
II,4 is heterozygous p=1 since her mother is homozygous recessive.
Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4
The probability that all these events happen is the product of all the probabilities:
2/3 x 1 x ¼ = 2/12=1/6
9 - If individuals of cross II3-II4 have already had an affected child, which is the
probability to have a second affected child?
If these parents already have a sick child, we are sure that they
must be heterozygous Aa (1).
Then the probability to have a second affected child is ¼.
d)
I
aa (1)
II
Aa
(1)
?
Aa
(1)
Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4