Genetics

Genetic problems

• A monohybrid cross is a cross of individuals looking at a characteristic inherited at one gene locus

• A test cross is crossing an individual back to a homozygous recessive individual in order to determine whether it is a carrier

• A Punnett square is a tool used in genetics• Genotype refers to the alleles present in

an individual• Phenotype refers to the characteristic

shown by the individual

b

B

b

b

Bb

bb

Bb

bb

Autosomal inheritance

Both males and females have 2 alleles for the characteristic

Homozygous individuals have 2 alleles the same and produce gametes with only 1 type of allele

Heterozygous individuals have 2 different alleles and produce two types of gametes with each allele

At fertilisation gametes combine so the new individual has 2 of each allele – one from each parent

We can show the probabilities of allele combinations from different crosses by using a Punnett square

Autosomal dominant/recessive

Individuals with two dominant alleles show the dominant phenotype

Individuals with two recessive alleles show the recessive phenotype

Individuals with one of each allele show the dominant phenotype

BB

bb

Bb

Autosomal dominant recessive crosses

Crossing a homozygous dominant individual with a homozygous recessive individual leads to offspring who are all heterozygous and show the dominant trait

Crossing two heterozygous individuals leads to 1 homozygous dominant individual, showing the dominant trait : 2 heterozygous individuals, showing the dominant trait :1 homozygous recessive individual, showing the recessive trait

BB bb

Bb Bb

BB Bb

Bb bb

B b

B

b

Example – dominant recessive problem

A heterozygous black male mouse mates with a homozygous brown female mouse. Black fur is dominant over brown fur. What is the probability of having:

a) a homozygous black offspring? 0% b) a heterozygous black offspring? 50% c) a homozygous brown offspring? 50%

Bb bb

b b

B Bb Bb

b bb bb

Autosomal co-dominance

Individuals with two of the 1st allele show the first trait

Individuals with two of the 2nd allele show the second trait

Individuals with one of each allele show a mixture of both traits

SBSB

SWSW

SBSW

Autosomal co-dominant crosses

Crossing an individual homozygous for one allele with an individual homozygous with the second allele leads to offspring showing a mixture of the two traits

Crossing two heterozygous individuals leads to 1 homozygous individual showing the first trait : 2 heterozygous individuals showing the mixed trait :1 homozygous individual showing the second trait

SBSB SWSW

SBSW SBSW

SB SW

SB SBSB SBSW

SW SBSW SWSW

Example – co-dominance problem

Two heterozygous grey sheep are mated. Black wool is co-dominant to white wool. What is the probability of having:

a) a black offspring? 25% b) a grey offspring? 50% c) a white offspring? 25%

SBSW SBSW

SBSB SBSW

SBSW SWSW

SB SW

SB

SW

Sex linked inheritance

Males and females have different chromosomesMales can only show 2 phenotypes (ie males

can not be carriers)Females can show 3 phenotypes (if codominant)

or 2 phenotypes (if dominant recessive, with a carrier)

You need to show alleles on the X chromosome (Y chromosomes don’t carry an allele)

Example – sex linked recessive problemIn humans, red-green colour blindness is a relatively common condition that is

inherited as an X-linked recessive trait. a) A woman with normal vision whose father was red-green colour-blind marries

a man with normal vision.i) What proportion of her sons would you expect to be colour-blind? 0ii) What proportion of her daughters would you expect to be colour-blind? 0

b) If she married a man who was red-green colour-blind, i) what proportion of her sons would you expect to be colour-blind? ½ ii) what proportion of her daughters would you expect to be colour-blind? ½

Xh

Xh

Xh

XH

XH

XH

XH Xh XhY

Y

Y

XH Xh

XH XH

Xh Xh

XHY

XhY

XHY

Genetics problem 12. In humans, tongue rolling is dominant over not being able to tongue roll. If a

hybrid tongue rolling male mates with a non-tongue rolling female and they have four children, determine:a) the probability that the first child is a tongue roller,b) the most probable ratio of tongue rollers to non-tongue rollers in the children,c) the most probable genotypes of each child.

Tt x tt T t

t

t

Tt tt

Tt tt

a) 50%b) 1:1c) ½ Tt, ½ tt

Genetics problems 2a

6. a) In humans, the ABO blood groups are controlled by a single genes with three alleles, IA, IB and i. IA is codominant with IB, and both are dominant over i. List each phenotype, and name the genotypes possible for each.

IA IA = blood type AIA i = blood type AIA IB = blood type ABIB IB = blood type BIB i = blood type Bi i = blood type O

Genetics problems 2b

6. b) A man of blood group O marries a woman of blood group A. The woman’s father was of blood group O. What are the chances that their children will belong to blood group O?

IAi x ii = ½ IAi (blood type A), ½ ii (blood type O)

Answer = 50%

Genetics problems 2c6. c) Four babies were born in a hospital on a night in which

an electrical blackout occurred. In the confusion that followed, their identification bracelets were mixed up. Conveniently, the babies were of four different blood groups: O, A, B and AB. The four pairs of parents have the following blood groups: O and O; AB and O; A and B; B and B. Which baby belongs to which parents?

O x O parents can only have baby O, the only parents that can produce an AB baby are A x B,the B x B parents can only produce a B or an O baby – so

they have baby B,so baby A must belong to parents AB x O

Genetics problems 3a7. Haemophilia is a disease carried as a sex-linked recessive trait.a) A haemophiliac male marries a homozygous normal female. What is the

probability of this family having ai) haemophiliac daughter? 0%

ii) haemophiliac son? 0%iii) carrier daughter? 50%

iv) carrier son? 0% v) normal son? 50%

vi) normal daughter? 0%

XH XH x Xh YXH XH

Xh XH Xh XH Xh

XH Y XH YY

Genetics problems 3b7. Haemophilia is a disease carried as a sex-linked recessive trait.b) One of the daughters, from this marriage is a carrier of haemophilia.

What is the probability of her daughter being a carrier, if her husband is normal? 50% of her daughters will be carriers

c) A haemophiliac male decides to wed a non-haemophiliac female whose father had haemophilia. What are the chances that the first child will be a son with haemophilia? 25%

XH Xh x Xh YXH Xh x XH Y

b)c)

Y Y

XH

XH XHXh Xh

XhXH Xh XH Xh

XH XHXh Xh

XH Y Xh Y XH Y Xh Y

Genetics problems 48. a) In humans, Duchenne muscular dystrophy is an X-linked recessive disease which is

very rare in females. The males who have the disease usually die before they reach 18 years old. How are these two facts related?

Girls can only have a sex-linked recessive disease if they have 2 affected X chromosomes – ie they have inherited affected chromosomes from both the mother and the father. As males rarely survive to breeding age, this is not likely to occur.

b) A female has been diagnosed as a carrier of the dystrophic gene. What are the chances that her sons will be normal and healthy? 50%

c) A woman whose brother died of muscular dystrophy wants to marry a man whose family has no history of the disease.i) What is the probability that she is a carrier? 50%ii) What is the probability that her first child has the disease? 1/8 or 12.5%

b) XD Xd x XD Y c) XD Xd x XD Y or XD XD x XD Y

XD

Y Y Y

XD XDXD

XD

XD XD

XD XDXdXd

XD Xd

XD XDXD XD XD XD XD Xd

XD Y Xd Y XD Y XD YXD Y Xd Y