Heredity
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- 2. Recap 2 Genes control the characteristics of living organisms Genes are carried on the chromosomes Chromosomes are in pairs, one from each parent Genes are in pairs Genes controlling the same characteristics occupy identical positions on corresponding chromosomes
- 3. Dominance 3 The gene pairs control one characteristic But they do not always control it in the same way Of the gene pair which help determine coat colour in mice, one might try to produce black fur and its partner might try to produce brown fur The gene for black fur is dominant to the gene for brown fur
- 4. Symbols 4 The genes are represented by letters The gene for black fur is given the letter B The gene for brown fur is given the letter b BB bb The genes must have the same letter but the dominant gene is always in capitals
- 5. 5 Alleles The genes of a corresponding pair are called alleles This means alternative forms of the same gene B and b are alleles of the gene for coat colour B is the dominant allele b is the recessive allele
- 6. 6 F1 A blackmale mouse (BB) is mated (crossed) with a female brown mouse (bb) In gamete production by meiosis, the alleles are separated Sperms will carry one copy of the B allele Ova will carry one copy of the b allele When the sperm fertilizes the ovum, the alleles B and b come together in the zygote
- 7. All offspring will be black (Bb) B B B meiosis fertilization sperm mother cell B B ovum mother cell b b b zygote meiosis b b
- 8. 8 The offspring fromthis cross are called the F1 (First Filial) generation They are all black because the allele for black coat colour is dominant to the allele for brown coat colour These Bb mice are called heterozygotes. Because the B and b alleles have different effects; producing either black or brown coat colour The mice are heterozygous for coat colour The BB mice are called homozygotes because the two alleles produce the same effect. Both alleles produce black coats. The bb mice are also homozygous for coat colour. Both alleles produce a brown coat colour The next slide shows what happens when the two heterozygotes are mated and produce young
- 9. F2 9 Fertilization Possible combinations sperms B B B B BB b b B sperm mother cell b Bb meiosis ovum mother cell b B B B Bb b b b b ova bb zygotes
- 10. Punnett square 10 A neater way of working out the possible combinations is to use a Punnett Square* Draw a grid Enter the alleles in the gametes Enter the possible combinations B b female gametes B BB Bb male gametes These are the b Bb bb F2 generation
- 11. 3:1 ratio 11 The offspring are in the ratio of 3 black to 1 brown Although the BB and Bb mice look identical, the Bb mice will not breed true. When mated together there is a chance that 1 in 4 of their offspring will be brown This is only a chance because sperms and ova meet at random A litter of 5, may contain no brown mice; in a litter of 12, you might expect 3 brown mice but you would not be surprised at anything between 2 and 5. The total offspring from successive matings of the heterozygotes would be expected to produce in something close to the 3:1 ratio For example, 6 successive litters might produce 35 black and 13 brown mice. This is a ratio of 2.7:1, near enough to 3:1
- 12. 12 Some terminology The offspringof the heterozgotes are the F2 generation The genetic constitution of an organism is called its genotype The visible or physiological characteristics of an organism are called its phenotype The phenotype of this mouse is BB black. Its genotype is BB The phenotype of this mouse is Bb also black, but its genotype is Bb The phenotype of this mouse is bb brown. Its genotype is bb
- 13. 13 These tobacco seedlingsare the F2 generation from a cross Between heterozygous (Cc) parents. C is the gene for chlorophyll. cc plants can make no chlorophyll. There are 75 green seedlings present. What is the ratio of green to white seedlings? What ratio would you expect?
- 14. 14 There are 21white seedlings. This is a ratio of 75:21 or 3.57:1 C c You would expect CC Cc the cross to produce C 72 green to 24 white seedlings (3:1) c Cc cc 1 CC 2 Cc and 1 cc, a ratio of 3 green to 1 white seedling Is 3.57:1 near enough to 3:1 ?*
- 15. Sex chromosomes 15 In most populations of animals there are approximately equal numbers of males and females This is the result of a pair of chromosomes; the sex chromosomes called the X and Y chromosomes The X and Y chromosomes are a homologous pair but in many animals the Y chromosome is smaller than the X Females have two X chromosomes in their cells. Males have one X and one Y in their cells At meiosis, the sex chromosomes are separated so the the gametes receive only one: either an X or a Y.
- 16. Sex ratio 16 fertilization meiosis X X female X X Y Y male Y sperm mother cell X X X X female X X Y X male ovum mother cell X
- 17. Single gene effects 17 Very few human characteristics are controlled by a single gene Characteristics such as height or skin colour are controlled by several genes acting together Those characteristics which are controlled by a single gene are usually responsible for inherited defects (see slide 19)
- 18. ABO blood groups 18 An exception is the inheritance of the ABO blood group The IA allele produces group A The IB allele produces group B The IO allele produces group O IO is recessive to IA and IB The group A phenotype can result from genotypes IAIA or IAIO The group B phenotype can result from genotypes IBIB or IBIO The group O phenotype can result only from genotype IOIO The AB phenotype results from the genotype IAIB The alleles IA and IB are equally dominant (co-dominant)
- 19. 19 Genetic defects Cystic fibrosis(recessive) Glands of the alimentary canal produce a thick mucus which affects breathing, digestion and susceptibility to chest infection Achondroplastic dwarfism (dominant)The head and trunk grow normally but the limbs remain short Albinism (recessive) Albinos cannot to produce pigment in their skin, hair or iris Polydactyly (dominant*) an extra digit may be produced on the hands or feet Sickle cell anaemia (recessive)The red blood cells become distorted if the oxygen concentration falls. They tend to block small blood vessels in the joints
- 20. 20 Genetic counselling (Geneticdefects) If the genotypes of the parents are known, it is possible to calculate the probability of their having an affected child (i.e. one with the defect) For example if a male achondroplastic dwarf marries a normal woman, what are their chances of having an affected child? The father’s genotype must be Dd. (DD is not viable) The mother must be dd since she is not a dwarf There is a 50% probability of their having D d an affected child d Dd dd What are the probabilities if both parents d Dd dd are affected?
- 21. 21 Cystic fibrosis (recessive) Iftwo normal parents have an affected child, they must both be heterozygous (Nn) for the recessive allele n A nn parent would have cystic fibrosis N n A NN parent would produce only normal N NN Nn children n Nn nn Since the parents are now known to be heterozygous it can be predicted that their next child has a I in 4 chance of inheriting the disease This chance applies to all subsequent children*
- 22. 22 Sickle cell anaemia(recessive) Hb = haemoglobin HbA is the allele for normal haemoglobin HbS is the allele for sickle cell haemoglobin A person with the genotype HbSHbS will suffer from sickle cell anaemia A person with the genotype HbAHbA is normal The genotype HbAHbS produces sickle cell ‘trait’ because HbA is incompletely dominant to HbS The heterozygote HbAHbS has few symptoms but is a ‘carrier’ for the disease
- 23. Carriers 23 Heterozygous recessive individuals do not usually exhibit any disease symptoms but because their offspring may inherit the disease, the heterozygotes are called ‘carriers’ carriers HbA HbS HbA HbAHbA HbAHbS HbS HbAHbS HbSHbS Similarly, individuals with the genotype Nn are carriers for cystic fibrosis
- 24. Family trees 24 It is sometimes possible to work out the genotypes of parents and to track the inheritance of an allele by studying family trees = normal female = affected female = normal male = affected male Parents have normal phenotypes but produce an affected child For this to happen, both parents must have heterozygous genotypes (Nn) for the characteristic
- 25. 25 AA If one of the parents is homozygous for a dominant allele, all the children will be affected If one parent is heterozygous for a Aa aa dominant allele and the other is homozygous recessive, there is a chance that half their children will be affected Aa Aa If both parents are heterozygous for a recessive allele, there is a chance that one in four of their children will be affected
- 26. 26 grandparents marriage marriage parents children cystic fibrosis What can you deduce about the genotypes of the grandparents from this family tree?
- 27. 27 Cystic fibrosis iscaused by a recessive gene An affected person must therefore have the genotype nn Since neither of the grandparents is affected, they must be either NN or Nn genotypes If they were both NN, none of their children or grandchildren could be affected If one was Nn and the other NN, then there is a chance that 50% of their children could be carriers Nn If one of the carriers marries another carrier, there is a 1 in 4 chance of their having an affected child The genotypes of the grand parents must be either both Nn or one NN and the other Nn
- 28. 28 D d D DD Dd d Dd dd If both parents have the Dd genotype there is a 75% chance of their having affected children, but the DD individual is unlikely to survive
- 29. 29 Question 1 Which of the following are heterozygous genotypes? (a) Aa (b) bb (c) nn (d) Bb
- 30. 30 Question 2 Which of these genes are alleles? A B C chromosomes A b c (a) A and A (b) A and B (c) B and C (d) B and b
- 31. 31 Question 3 Which of the following processes separates homologous chromosomes ? (a) mitosis (b) cell division (c) meiosis (d) fertilization
- 32. 32 Question 4 Which of the following terms correctly describes the genotype bb ? (a) homozygous dominant (b) heterozygous dominant (c) homozygous recessive (d) heterozygous recessive
- 33. 33 Question 5 What is the likely ratio of affected children born to parents both of whom are heterozygous for cystic fibrosis ? (a) 1 affected: 3 normal (b) 3 affected: 1 normal (c) 2 affected: 2 normal (d) all affected
- 34. 34 Question 6 Which of the following phenotypes corresponds to the Genotype IAIO ? (a) Blood group A (b) Blood group B (c) Blood group O (d) Blood group AB
- 35. 35 Question 7 What is the expected ratio of offspring from a black rabbit Bb and a white rabbit bb ? (a) 3 black: 1 white (b) 1 black: 3 white (c) 50% white; 50% black (d) all black
- 36. 36 Question 8 Which of these Punnett squares correctly represents a cross between two heterozygous individuals ? (a) A a (b) A a A AA aa A AA Aa a AA aa a Aa aa (c) (d) A a a a A AA Aa A Aa Aa a Aa Aa a aa aa
- 37. 37 Question 9 A married couple has a family of 6 boys. What are the chances that the next child will be a girl ? (a) 6:1 (b) 1:6 (c) 3:1 (d) 1:1
- 38. 38 Question 10 Which of the following is a ‘carrier’ genotype for a disease caused by a recessive gene ? (a) nn (b) NN (c) Nn
- 39. 39 Question 11 If normal parents have a child with cystic fibrosis (a) one of them must be heterozygous (b) both of them must be heterozygous (c) one of them must be homozygous (d) both of them must be homozygous
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Editor's Notes
- #2 The slide shows tobacco seedlings some of which have no chlorophyll, See slide No.13
- #11 * After Dr. R.C. Punnett, Professor of Genetics in Cambridge in the first half of the 20 th century.
- #12 The 3:1 ratio is sometimes called a Mendelian ratio after Gregor Mendel (1822-84), an Austrian Monk with an interest in plant breeding and mathematics. He made experimental crosses, particularly of pea plants, to see how characteristics were inherited. He realised that although certain characteristics seemed to disappear in the first generation, they could reappear in the second generation, and so he developed the idea of dominant and recessive characteristics with the recessive characteristics being suppressed in the first generation.. His second generation crosses, involving single characteristics, produced ratios approximating to 3;1. From these results he drew accurate conclusions about the mechanism of inheritance although he knew nothing about chromosomes and genes at this time.
- #15 *There are 75 green seedlings + 21 white seedlings. This is the ‘observed result’. If the 3:1 ratio is operating, the ‘expected result’ would be 72 green and 24 white seedlings. A statistical test on these figures tells us that the difference between the observed and expected results is sufficiently small to be ignored and the 75:21 ratio is near enough to 3:1 to be confident that this is the outcome of a genetic process.
- #17 This shows that there is a 50% chance of a boy or a girl baby. Because this ratio depends on the chance meeting of X and Y gametes and human families are small, the 1:1 ratio is rarely seen, even in large families. A family with 6 boys may hope for a girl next time but there is still only a 50% chance of getting a girl. In the population as a whole, the 1:1 ratio holds good.
- #20 *Even though the allele is dominant, it may not always be fully expressed. X-rays will reveal an extra metacarpal but this may not form a separate digit so the hand looks normal, In some cases only one of the hands shows the extra digit.
- #21 The incidence of achondroplasia is about 1 in 30000. These cases arise as a result of a mutation; the affected children are born to normal parents. A mutation is a spontaneous change in a gene or a chromosome. Mutations are mostly harmful.
- #22 Even if the parents have five normal children, there is still a 1 in 4 chance of the next child being affected. If the heterozygotes could be detected before they had children, they could be counselled about the likelihood of having affected children. Families at risk (I.e. cystic fibrosis in a relative) may be offered DNA testing to look for one or more of the genes which cause the disease.