1. RECOGNITION OF PATHERNITY
ESERCITAZIONE 3
HUMAN GENETICS
BLOOD GROUPS
RECOGNITION OF PATHERNITY
FAMILY PEDIGREE

2. INCOMPLETE DOMINANCE AND CO-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)

3. BLOOD GROUP(Phenotype)
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 B
IB IB
IB i AB
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

4. 2. Determine parental genotypes in the following crosses:
Phenotype of crossed individuals
Phenotype in the progeny
Genotype of crossed individuals 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
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)

5. 2. Determine parental genotypes in the following crosses :
Phenotype of crossed individuals
Phenotype in the progeny
Genotype of crossed individuals 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
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

6. 2. Determine parental genotypes in the following crosses :
Phenotype of crossed individuals
Phenotype in the progeny
Genotype of crossed individuals 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
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.

7. 2. Determine parental genotypes in the following crosses :
Phenotype of crossed individuals
Phenotype in the progeny
Genotype of crossed individuals 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
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

8. Determine the genotypes of individuals crossed:
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
Genotype of crossed individuals
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+
IA IB Dd x ii Dd
The first parent must be IA IB , while the second one is ii.
Factor Rh: both parents could be DD/Dd. Since we have individuals with Rh- in the progeny, the parents must be heterozygous Dd

9. 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 Gene D Expected phenotypes
IA IB x i i Dd x Dd
Rh+ (3/4) A Rh+ (1/2 x 3/4 = 3/8)
A (1/2)
Rh- (1/4) A Rh- (1/2 x 1/4 = 1/8)
Rh+ (3/4) B Rh+ (1/2 x 3/4 = 3/8)
B (1/2)
Rh- (1/4) B Rh- (1/2 x 1/4 = 1/8)

10. 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 Rh-
B 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
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.
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)

11. 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 Gene D Expected phenotypes
IB i x IA i dd x DD
A (1/4) Rh+ (1) A Rh+ (1/4)
B (1/4) Rh+ (1) B Rh+ (1/4)
AB (1/4) Rh+ (1) AB Rh+ (1/4)
O (1/4) Rh+ (1) O Rh+ (1/4)

12. 4. Mother and son have the indicated blood groups
4. Mother and son have the indicated blood groups. Could the man be the father of the child?
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.

13. Mother’s genotype: IA IA / IA i DD/Dd child’s genotype: i i DD/Dd
4. Mother and son have the indicated blood groups. Could the man be the father of the child? Rh + A -
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

14. Mother’s genotype: i i dd child’s genotype: IA IA / IA i dd
4. Mother and son have the indicated blood groups. Could the man be the father of the child? A Rh - AB +
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

15. 5. Figure out the probability of these events.
Mother’s genotype: IA IB dd
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

16. 5. Figure out the probability of these events.
Mother’s genotype: i i DD/Dd
Father’s genotype: IA IA / IA i DD/Dd
child’s genotype: 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

17. 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.

18. INHERITANCE
DOMINANT
All the generations have individuals that show the trait
RECESSIVE
The affected individuals are rare and they are not present in all the generations.
AUTOSOMAL
Male and female are usually affected with equal frequency.
SEX-LINKED
The trait emerged only in male or female

19. Mutant phenotype is present in all generations.
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 A- 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).

20. 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 aa Aa
Here we have mutant sons from normal individuals. The mutated allele is RECESSIVE.
The mutant individuals are homozygous recessive, the parents are hetorozygous.

21. 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 A a AA
1/4 Aa aa Aa AA AA
Aa (2/3) A AA
1/4 a Aa
Aa (1)
Aa (½) ?
Female III,3 is heterozygous Aa (probability=1) because she is normal and the mother is sick (homozygous recessive)
Man III,4 could be heterozygous if his father is heterozygous (II,5). The probability that II,5 is heterozygous is 2/3 (he is normal son of heterozygous 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

22. 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
III IV ?
aa (1)
Sick individuals of II generation are homozygous recessive with probability 1
AA (1)
AA (1) Aa
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
Aa (1) Aa
(1/2 )
(1/4 )
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

23. 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 Aa
(1)
III,3 e III,4 must be both heterozygous in order to produce a sick son. Aa Aa
(1)
(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

24. 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 Aa AA AA
III,1 and III,2 must be both heterozygous to have a sick son. aa
(2/3) Aa Aa Aa
(1)
(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

25. 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 (1) 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

26. Then the probability to have a second affected child is ¼.
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.
aa (1) Aa
(1) Aa
(1)
Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4