Mendelian genetics
calculations
Seminar
No 404 Heredity
Key words:
heredity, variability, gene, genotype, trait, phenotype,
genome, allele, standard and mutant alleles, allele
polymorphism, allelic heterogeneity, locus, locus
(nonallelic) heterogeneity, homozygote, heterozygote,
hemizygote, dominance, recessivity, codominance,
incomplete dominance (intermediary heredity), hybrid,
parental generation, filial generation, genotype and
phenotype ratio, breeding news, backcross, testcross,
monohybrid, dihybrid and trihybrid crosses,
polyhybridism, hybridologic analysis, independent
combination, Mendelian laws, genocopy, phenocopy,
pleiotropy
Calculation of genotype and phenotype
ratios:
Monohybrid cross
AA
P
x
parental generation
homozygotes (dominant and recessive)
aa
Aa
1st filial generation
F1
Gametes
A
Genotypes
of gametes
F2
Combination
square
(Punnet square)
A
A
AA
a
P(A)=P(a)=1/2
Monohybride = hybrid, that
differs in a single allelic pair
(heterozygote)
a
Aa
2nd filial generation
a
Aa
aa
Genotypes of zygotes
2.filial generation
Genotype ratio: 1 : 2 : 1
AA : Aa : aa
Phenotype ration: 3 : 1
A- : aa
Backcross– verification of hybrid´s heterozygosity
B1 (backcross) = F1 x parent
Aa x AA
AA AA
Aa Aa
Testcross
(F1 x recessive
homozygote)
Aa x aa
Aa Aa
Ratio 1
aa
:
1
aa
Dihybrid cross
2 allelic pairs – carried by different pairs of
homologous chromosomes – segregate
independently
two types of parental crosses
P: AABB x aabb
AAbb x aaBB
F1:
AaBb double heterozygote
genotypes of gametes AB Ab aB ab
1 : 1 : 1 : 1
Punnett (combination) square:
F2
Genotypes
of gametes
AB
Ab
aB
ab
AB
AABB
AABb
AaBB
AaBb
Ab
AABb
AAbb
AaBb
Aabb
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
Genotypes of zygotes:
AABB AABb AAbb AaBB AaBb Aabb aaBB aaBb aabb
1
: 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1
= (1 : 2 : 1)(1 : 2 : 1)
Phenotype:
A-BA-bb
aaBaabb
9
:
3
:
3
:
1
= (3 : 1)(3 : 1)
Breeding news = homozygotic recombined forms = new
combinations of parental traits
For parental cross: AABB x aabb → AAbb, aaBB
for
AAbb x aaBB → AABB, aabb
Backcross - testcross
AaBb x aabb heterozygote x recessive homozygote
B1
AB
ab
Ab
aB
ab
AaBb Aabb aaBb aabb
Ratio of phenotypes in B1
1
:
1
:
1
:
1
Trihybrid cross– 3 allelic pairs
4 type of homozygote crosses
P: 1) AABBCC x aabbcc
3)AAbbcc x aaBBCC
F1:
2) AABBcc x aabbCC
4)AAbbCC x aaBBcc
AaBbCc
8 types of gamets:
ABC ABc AbC Abc aBC aBc abC abc
1 : 1 : 1 : 1 : 1 : 1 : 1 : 1
F2
64 of zygotic combinations
Calculation of ratios:
- by combinatory square (Punnet square)
- by combination of monohybrid ratios
(1AA:2Aa:1aa)(1BB:2Bb:1bb)(1CC:2Cc:1cc)
- according to probability count
Polyhybrid cross
n - number of followed traits …….
• number of gametes 2n
• number of zygotes 4n
• genotype ratio:
( 1 : 2 : 1 )n
• phenotype ratio:
( 3 : 1 )n
MENDEL´S LAWS
• 1. uniformity of F1 generation – identity
of reciprocal crosses
• 2. principle of segregation
alleles segregate individually to gametes
• 3. principle of combination (independent
assortment)
alleles of two or more allelic pairs assort
independently of one another– there are as many
types of gamets as possible combinations among
them (genes on different chromosomes behave
independently in gamete production)
Probability count:
• 2 independent events P (A+B) = P (A) . P (B)
• 2 mutual exclusive events: P(A or B) = P (A) + P (B)
•
•
•
•
•
•
Monohybrid cross
P (A) = P (a) = 1/2
P (AA) = P (A ) . P (A) = 1/2 . 1/2 = 1/4
P ( aa ) = P (a ) . P (a ) =1/2 . 1/2 = 1/4
P (Aa ) = P (A ) . P (a ) =1/2 . 1/2 = 1/4
P (aA ) = P (a ) . P (A ) =1/2 . 1/2 = 1/4
P ( Aa or aA ) = 1/4 +1/4 = 1/2
1/4 : 1/2 :1/4 = 1 : 2 : 1
Dihybrid cross
•
•
•
•
P (AB) = P (Ab) = P (aB) = P (ab) = 1/4
P (AA BB ) = P (AB ) . P (AB ) = 1/4 . 1/4 = 1/16
P ( AA bb ) = P ( Ab ) . P (Ab ) = 1/4 . 1/4 = 1/16
P (AA Bb) = P (AB) . P (Ab) + P (Ab) . P (AB) =
1/16 +1/16 = 1/8
• P (Aa Bb) = P (Ab) . P (aB) + P (aB) . P (Ab) +
P (AB).P(ab) + P (ab) . P(AB) =
1/16+1/16+1/16+1/16 = 1/4
• etc.
1. How many gametes is produced by individual who is
a)heterozygous in 1 locus?
b)heterozygous in 2 loci?
c)heterozygous in 3 loci?
d)heterozygous in n loci?
2. What is the probability that a boy or a girl would
be born?
3. What is the probability of three boys born in three
pregnancies?
4. What is the probability of three babies of the
same sex born in three pregnancies ?
5. What is the probability of two boys and one girl born
from three pregnancies?
6. Determine the probability of obtaining three times
number 6 in three cube tosses.
7. Determine the probability of obtaining three times the
same number in three cube tosses.
8. What is theprobability that couple of Aa x aa
genotypes will have a child with aa genotype?
9. What is the probability that couple of Aa x aa
genotypes will have a heterozygous child?
13. The lack of pigmentation (albinism) is determined by
recessive allele a and normal pigmentation is the result of
dominant allele A. Phenotypically normal parents have an
albinotic child.
a) What are genotypes of all family members?
b) What is the probability that their next child will be
albinotic?
c) What is the probability that their next two children will be
albinotic?
d) What is the probability that their next two children will be
normal?
e) What is the probability that their next two children will
be one normal and one albinotic?
f) What is the probability that their next two children will be
both heterozygous?
g) What is the probability that their two healthy children will
be heterozygous?
15. There is hypotese that a dominant allele B determines brown
eyes (allele B) and recessive allele b determines blue eyes.
a) Blue-eyed female and brown-eyed male had all offsprings
brown-eyed. What are family members genotypes?
b) Brown-eyed man (with brown-eyed parents) married browneyed woman. Her father had brown eyes too and her
mother was blue- eyed. The couple had one blue-eyed
baby. What are genotypes of all family members?
c) Parents are heterozygous brown-eyed. What is their chance of
having first child blue-eyed?
d) What is their chance of having second child blue-eyed when
their first child had brown eyes?
e) What is their chance of having one blue-eyed child and one
child with brown eyes?
f) What is their chance of having four children, all with brown
eyes ?
16. What are genotypes of individuals of blood groups:
A, B, AB, 0?
17. What blood groups could be expected in children of
couples with blood groups: AB x AB, AB x 0, A x 0, B
x 0.
19. Rh positivity is determined by dominant allele D and
Rh negativity by recessive genotype dd. Explain what
types of crossing (genotypes) would produce clinically
significant maternal-fetal incompatibility?
20. Both parents have blood group AB. What is the
probability that their dizygotic twins will have identical
blood group? What is the probability that their
monozygotic twins will have identical blood group?
21. What types of gametes are produced by individuals
with genotypes: AaBb, aaBb, AABB, aaBB?
What is the ratio of individual genotypes?
24. Let‘s presume that allele B determines brown eyes
and recessive allele b (genotype bb) determines blue
eyes.
Right-handed ability (allele R) is dominant to lefthanded (recessive allele r).
a) Blue-eyed and right-handed man (whose father was
left-handed) married a brown-eyed left-handed
woman. She was from family with brown-eyed
members in several generations. What will be
phenotype of their children?
b) Brown- eyed man married blue-eyed woman. Both were
right handed. Their first child had blue eyes and was left
handed. What will be phenotype of their other children?
27. Which of two couples are parents of child with O MN Rh+
blood groups?
a)A MN Rh+ x B M Rhb)AB M RH+ x O N RH+
Thompson and Thompson: Clinical genetics, chapter 7 :
Mendelian inheritance, pg. 118-120, 7th edition