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 Bb Bb b 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 BB 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

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 B b BB Bb B b Bb bb

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 SBSB 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 SWSW SBSW

Autosomal co-dominant crosses SBSB SWSW 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 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 SB SW SB SBSB SBSW SW SBSW SWSW

Sex linked inheritance Males and females have different chromosomes Males 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 problem In 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? 0 ii) What proportion of her daughters would you expect to be colour-blind? 0 XH Y Xh XhY XH Xh XHY XH XH XH 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 Y XhY Xh Xh Xh XHY XH XH Xh

Genetics problem 1 2. 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 50% 1:1 ½ Tt, ½ tt tt t Tt t tt 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 A IA i = blood type A IA IB = blood type AB IB IB = blood type B IB i = blood type B i 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 2c 6. 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 3a 7. 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 a i) 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 XH XH x Xh Y Xh XH Xh XH Xh Y XH Y XH Y

Genetics problems 3b 7. 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% c) b) XH Xh x Xh Y XH Xh x XH Y XH Xh XH Xh XH Xh Xh Xh XH Xh Xh XH XH XH Xh Y Y Y XH Y Xh Y XH Y

Genetics problems 4 b) XD Xd x XD Y c) XD Xd x XD Y or XD XD x XD Y XD 8. 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 Xd XD XD XD Xd XD XD XD XD Xd XD XD XD XD XD XD XD XD XD Xd Y XD Y Xd Y XD Y Y XD Y Y XD Y Xd Y