Mendelian geneticscalculations

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:

Gametes

P(A)=P(a)=1/2

AA x aa

Aa

aa

AA

Aa

Aa

A a

A

A

a

a

P

F1

F2

parental generationhomozygotes (dominant and recessive)

1st filial generation

Monohybride = hybrid, that differs in a single allelic pair (heterozygote)

2nd filial generation

Genotypes of zygotes

Monohybrid cross

Genotypes of gametes

Combination square

(Punnet square)

AA Aa aa

A- aa

: :

:

AAAa Aa aa

AAAA Aa AaAa Aa aa aa

2.filial generation

Genotype ratio: 1 : 2 : 1

Phenotype ration: 3 : 1

Backcross– verification of hybrid´s heterozygosity

B1 (backcross) = F1 x parent

Testcross(F1 x recessive homozygote)

x x

Ratio 1 : 1

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 aaBBF1: AaBb double heterozygote

genotypes of gametes AB Ab aB ab

1 : 1 : 1 : 1

F2

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 gametes

Punnett (combination) square:

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-B- A-bb aaB- aabb 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 2) AABBcc x aabbCC 3)AAbbcc x aaBBCC 4)AAbbCC x aaBBcc

F1: 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

• 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)

MENDEL´S LAWS

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 brown-eyed 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 left-handed (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 Rh-

b)AB M RH+ x O N RH+

Thompson and Thompson: Clinical genetics, chapter 7 :

Mendelian inheritance, pg. 118-120, 7th edition