![Possible combinations of alleles
A black mouse (BB) is crossed with a brown one (bb).
What will the offspring look like?
B
B
b
b
B
b
PURE-BREEDING organism – both alleles
are the same [BB and bb]](https://image.slidesharecdn.com/5-140824031039-phpapp01/85/Biology-Form-5-Chapter-5-5-1-Inheritance-20-320.jpg)
![Which is the dominant allele?
Parents
(true breeding
parents)
F1 generation
F2 generation
Purple flowers White flowers
All plants have
purple flowers
Fertilisation
among F1 plants
(F1 F1)
3/4 of plants
have purple flowers
1/4 of plants
have white flowers
Alleleforpurplecolour[100% purpleinF1
generation]](https://image.slidesharecdn.com/5-140824031039-phpapp01/85/Biology-Form-5-Chapter-5-5-1-Inheritance-28-320.jpg)
Uploaded byNirmala Josephine
Biology Form 5 Chapter 5 : 5.1 Inheritance
- Gregor Mendel conducted experiments with pea plants in the 1860s and is considered the founder of genetics. Through his experiments, he discovered the fundamental laws of inheritance. - Mendel determined that traits are passed from parents to offspring through "factors" that we now know as genes. His laws of inheritance include dominance, segregation, and independent assortment. - Mendel's work formed the basis for understanding how traits are inherited and laid the foundation for modern genetics.
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Biology Form 5 Chapter 5 : 5.1 Inheritance
- 1. BIOLOGY FORM 5 CHAPTER5 INHERITANCE
- 2.
- 3. is the studyof heredity is the process in which traits are passed from parents to offspring
- 4. Characters or Traitsare resemblances or differences which can be: Seen e.g. eye colour flower colour Tested for e.g. blood groups colour blindness
- 5. Gregor Mendel (1822-1884) Austrian monkwho formulated fundamental laws of heredity in early 1860s Experimented with peas Over seven years, he made crosses with 24,034 plants Called the “Father of Genetics“
- 6. 6 Mendel’s Experimental Methods Mendel hand-pollinated flowersusing a paintbrush He could snip (cut) the stamens to prevent self-pollination He traced traits through the several generations
- 7.
- 8. 8 How Mendel Began? Mendel produced pure strainsby allowing the plants to self- pollinate for several generations
- 9.
- 10.
- 11. 11 Mendel stated that physicaltraits are inherited as “particles” Mendel did not know that the “particles” were actually Chromosomes & DNA Particulate Inheritance
- 12. Let’s revise: Genes: controlthe characteristics of living organisms are carried on the chromosomes
- 13. Chromosomes are inpairs, one from each parent
- 14. Genes are inpairs Genes controlling the same characteristics occupy identical positions on homologous chromosomes The gene pairs control one characteristic gene for eye colour gene for nose shape gene for making insulin
- 16. The genes ofa corresponding pair are called alleles Homologous chromosomes have the same length and carry the same gene sequences Alleles are alternative forms of the same gene Gene
- 17. Let’s take coatcolour in mice as an example Mice can be: Black Brown
- 18. The allele forblack fur is dominant to the allele for brown fur This combination of alleles gives a BLACK mouse The dominant allele is expressed The recessive allele is masked
- 19. Alleles are representedby letters the alleles must have the same letter but the dominant allele is always in capitals Black mouse (B – dominant allele) Brown mouse (b – recessive allele)
- 20. Possible combinations ofalleles A black mouse (BB) is crossed with a brown one (bb). What will the offspring look like? B B b b B b PURE-BREEDING organism – both alleles are the same [BB and bb]
- 21. B B B B b b b b B b meiosis meiosis fertilisation All offspring will beblack (Bb) sperm mother cell ovum mother cell zygote
- 22. If two ofthe offspring (Bb) thus produced are mated, what type of offspring would result? Bb Bb x ?
- 23. B b B b B b B b B B B b B b b b BB Bb Bb bb sperm mother cell ovummother cell meiosis Possible combinationsFertilisation sperms ova zygotes
- 24. FIRST FILIAL GENERATION (F1)the offspring produced by a parental generation xParents: SECOND FILIAL GENERATION (F2) offspring of the F1 When two F1 offspring mate, they produce the F2
- 25. Homozygous & Heterozygous HOMOZYGOUS– alleles on corresponding positions of homologous chromosomes are identical e.g. BB or bb HETEROZYGOUS – pairs of different alleles are present on corresponding positions of homologous chromosomes e.g. Bb
- 26. Genotype & Phenotype genotype:describes the genetic make-up (all of the genes) of an individual homozygous dominant heterozygous homozygous recessive phenotype: outward appearance of an individual
- 27. 27 Genotype -gene combination for a trait (e.g. RR, Rr, rr) Phenotype - the physical feature resulting from a genotype (e.g. red, white)
- 28. Which is thedominant allele? Parents (true breeding parents) F1 generation F2 generation Purple flowers White flowers All plants have purple flowers Fertilisation among F1 plants (F1 F1) 3/4 of plants have purple flowers 1/4 of plants have white flowers Alleleforpurplecolour[100% purpleinF1 generation]
- 29. Let us becomefamiliar with terms learned R represent round seed r represent wrinkled seed Round What is the: a) phenotype of a homozygous dominant plant? b) genotype of a homozygous dominant plant? c) genotype of a heterozygous plant? RR Rr
- 30. B represent yellowseed b represent green seed What is the: a) dominant allele for seed colour? b) genotype of a homozygous recessive plant? B bb c) genotype of a true breeding plant that produces green seeds? bb
- 31.
- 32. 32 Types of GeneticCrosses Monohybrid cross - cross involving a single trait e.g. flower color Dihybrid cross - cross involving two traits e.g. flower color & plant height
- 33.
- 34. A - allelefor purple flower colour a - allele for white colour. A pure breeding purple and a pure breeding white flower are crossed. What will the phenotype and genotype ratios be in the F1 generation? purple - A – AA, Aa white – a – aa Parents: Purple x White AA x aa Gametes: F1 generation: Aa Aa Aa Aa A A ax a F1 Phenotype: 100% purple F1 Genotype: 100% heterozygous
- 35. The cross doesNOT mean that FOUR offspring are produced. It shows PROBABILITY. Cross can be shown as: Parents: Purple x White AA x aa Gametes: F1 generation: Aa A x a F1 Phenotype: 100% purple F1 Genotype: 100% heterozygous IMPORTANT!!
- 36. Self-pollination occurs inone of the F1 plants. What will the phenotype and genotype ratios be in the F2 generation? purple – A – AA, Aa white – a – aa F1 generation: Purple x Purple Aa x Aa Gametes: F2 generation: xA a aA AA Aa Aa aa Phenotype- 3 purple : 1 white / 75% purple: 25% white( 3:1) Genotype- 1 AA : 2 Aa : 1 aa (1:2:1)
- 37. Remember: when both parentsare heterozygous, they produce offspring in 3:1 ratio Pp Pp x Parents Offspring
- 38. Gene diagram –Flower colour Male female RR rr parent gamete R R r r Offspring Rr RrRrRr Genotype Phenotype All red Red – R yellow – r
- 39. Gene diagram –Flower colour Male female Rr Rr parent gamete R r R r Offspring genotype RR RrRrrr Phenotype Red yellow red red 3 red : 1 yellow
- 40. Gene diagram –Flower colour Male female Rr rr parent gamete R r r r Offspring genotype Rr Rrrrrr Phenotype Red yellow yellow red Red 50% yellow 50%
- 41. x bb 50% black :50% brown x 100% black bb Bb bbBb Remember
- 42.
- 43. Reginald Punnett (1875-1967) In1902, created the Punnett Square - a chart which helped to determine the probable results of a genetic cross T t T TT Tt t Tt tt Male gametes Female gametes Tt Tt
- 44.
- 45. Dihybrid Cross Traits: Seedshape & Seed color Alleles: R round r wrinkled Y yellow y green Pea plants seed A Round yellow RRYY Pea plant seed B Wrinkled green rryy Parental Phenotype RY ry RrYy All round and yellow seeds Parental genotype gametes F1 genotype F1 phenotype
- 46. 46 Dihybrid Cross Traits: Seedshape & Seed color Alleles: R round r wrinkled Y yellow y green RrYy x RrYy RY Ry rY ry RY Ry rY ry All possible gamete combinations F1 x F1 gametes
- 47. 49 Dihybrid Cross RY RyrY ry RY Ry rY ry
- 48. 50 Dihybrid Cross RRYY RRYy RrYY RrYy RRYy RRyy RrYy Rryy RrYY RrYy rrYY rrYy RrYy Rryy rrYy rryy Round/Yellow: 9 Round/green:3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1 phenotypic ratio RY Ry rY ry RY Ry rY ry Alleles: R round r wrinkled Y yellow y green
- 49. 51 Dihybrid Cross Round/Yellow: 9 Round/green:3 wrinkled/Yellow: 3 wrinkled/green: 1 9:3:3:1
- 50. REMEMBER 52 MONOHYBRID CROSS eg(Rr x Rr) Phenotype Ratio : 3:1 Genotype Ratio : 1:2:1 DIHYBRID CROSS eg (RrTt x RrTt) Phenotype Ratio : 9:3:3:1
- 51. 53 Question: How many gameteswill be produced for the following allele arrangements? Remember: 2n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq
- 52. 54 Answer: 1. RrYy: 2n= 22 = 4 gametes RY Ry rY ry 2. AaBbCCDd: 2n = 23 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes
- 53. 55 Mendel’s Laws 1. Lawof Dominance 2. Law of Segregation 3. Law of Independent assortment
- 54. 56 Law of Dominance Ina cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation. All the offspring will be heterozygous and express only the dominant trait. RR x rr yields all Rr (round seeds)
- 55.
- 56. MENDEL’S LAW LAW OFSEGREGATION Each individual characteristic of a organism is determined by a pair of allele. The pairs of alleles segregate during meiosis Only one of each pair of allele can be present in a single gamete 58
- 57.
- 58. MENDEL’S LAW LAW OFINDEPENDENT ASSORTMENT Two or more pair of alleles will segregate or assort independently of one another during gamete formation 62
- 59. 63 Instead of 1 traitat a time, let’s look at how 2 traits can be passed together. Connection: Mendel’s Laws and Meiosis
- 61. Different Patterns ofInheritance As we now know, many traits do not follow Mendelian Inheritance patterns.
- 62. Co-dominance When both alleles are expressed equallyin the heterozygous individual. A and B blood type alleles are co- dominant. Black and orange color in cats are co- dominant.
- 63. BLOOD GROUPS In humans,there are four blood types: A, B, AB & O
- 64. BLOOD GROUPS sometimes acharacteristic is controlled by more than two alleles e.g. three alleles control human blood: A, B and O a person has two out of three alleles
- 65. BLOOD GROUP (phenotype) GENOTYPE A IAIA or IA IO B IB IB or IB IO AB IA IB O Io Io The alleles for groups A & B are CODOMINANT O is recessive
- 66. 70 Codominance Problem Example: homozygousmale Type B (IBIB) x heterozygous female Type A (IAi) 1/2 = IAIB 1/2 = IBi Parents: IBIB x IAIO Gametes: IB F1 generation: IB IA IO IBIOIAIB IBIO IAIB Phenotype AB B AB B
- 67. 71 Another Codominance Problem Example:male Type O (ii) x female type AB (IAIB) 1/2 = IAi 1/2 = IBi Parents: IoIO x IAIB Gametes: IO F1 generation: IO IA IB IAIOIAIO IBIO IBIO Phenotype A A B B
- 68. 2) Two parents,one with blood group A and the other with blood group B, have a child whose genotype is homozygous. a) Complete the diagram below to show how this can happen. (5) IoIo Io Io IAIo IBIo IBIA
- 69. b) What isthe chance of these parents producing a homozygous child? (1) Parents: IAIO x IBIO Gametes: IA F1 generation: xIO IB IO IAIOIAIB IOIO IBIO 25% c) What is the blood group phenotype of the homozygous child? (1) Blood group O
- 70. 74 • The differencesin human blood are due to the presence or absence of certain protein molecules called antigens and antibodies. • The antigens are on the surface of the red blood cells • the antibodies are in the blood plasma. • The blood group you belong to depends on what you have inherited from your parents. Why the different blood groups?
- 71. 75 ABO BLOOD GROUP Bloodgroup A A antigens (on the surface of RBC) B antibodies (in blood plasma) Blood group B B antigens (on the surface of RBC) A antibodies (in blood plasma)
- 72. 76 Blood group AB A & B antigens (on the surface of RBC) No antibodies (in blood plasma) Blood group O No antigens (on the surface of RBC) A & B antibodies (in blood plasma)
- 73. BLOOD GROUP ANTIGENS& ANTIBODIES BLOOD GROUP ANTIGEN ON SURFACE OF RBC ANTIBODY IN BLOOD PLASMA A A B B B A AB A and B None O None A and B
- 75. 79 Rhesus factor bloodgrouping system Rhesus = Rh Rh + ( dominant) Rh - (recessive) Rh + = has Rhesus antigen (Cannot produce Rh antibody) Rh - = no Rhesus antigen (Able to produce antibody if he or she receives blood from a person with Rh+ blood ) A person with Rh+ blood can receive blood from a person with Rh- blood without any problems.
- 76. Rhesus Factor Rh- ( rh rh) Rh + (Rh Rh) or (Rh rh) (Rh+ - dominant)
- 77. Rh inheritance Mother Father Rh– Rh + ( rh rh) X (Rh Rh) Foetus : (Rh rh) Rh+
- 78. Rh inheritance Mother Father Rh– Rh + ( rh rh) X (Rh rh) Gamete rh Rh rh Foetus : (Rh rh) (rhrh) Phenotype Rh+ Rh-
- 79. Mother: Rh +ve,foetus : Rh –ve No problem Mother - antigen in RBC unable to diffuse through placenta Baby born alive Rh + Rh -
- 80. Rh - Rh + Mother:Rh – , Foetus: Rh + Problem During delivery, baby’s blood in placenta will mix with mother’s blood 1st baby will survive Rh antigen in RBC enter mother’s blood system Mother’s lymphocytes stimulated to produce Rh antibodies.
- 81. Rh - Rh + Mother:Rh – , Foetus: Rh + Problem Rh Antibodies remain in mother’s blood plasma Second pregnancy, if foetus also Rh+ Antibody from the mother’s blood plasma diffuse into foetus blood through placenta (leakage) Agglutination in foetus’ blood
- 82. Mom’s immune systemrecognizes that these cells are not like hers, so the baby’s blood cells are attacked. The same principle applies to rejected organ transplants and blood transfusions. Example of the immune system gone wrong…
- 83. 90 Sex Chromosomes The gender/ sex of an individual is determined genetically by the sex chromosomes. XX = female, XY = male All other chromosomes are called “autosomes” Humans have 46 chromosomes (44 autosomes+2 sex chromosomes)
- 84. 91 Sex chromosomes Every femalecell has two sex chromosomes which are alike (XX) Sex chromosome XX Human female: 44 + XX
- 85. 92 Sex chromosomes Every malecell has two chromosomes which are not alike (XY) Human male: 44 + XY
- 86. 93 Human male: 44+ XY Human female: 44 + XX
- 87. 94 cells in testisof male cell in ovary of female 44 +XY 44 +XX there are two types of sperm 22 + X there is only one type of ovum 22 + Y 22 + X 22 + X 44 +XY 44 +XY44 +XX44 +XX Genotype: 44 +XX : 44 + XY Female : Male 1 : 1 SEX DETERMINATION IN HUMANS
- 88. 96 The type ofsperm which fertilises the ovum decides the sex of the offspring X sperm – baby girl Y sperm – baby boy
- 89. 97 What is sodifferent between the X and Y chromosomes? X- over 1000 genes identified Y- 330 genes identified, many are inactive Y chromosome
- 90.
- 91. Sex-linked genes arecarried on the sex chromosomes (X chromosome) autosomes Sex chromosomes X X X Y Female carries two alleles of a gene Male carries one allele of a gene
- 92. Sex-Linkage red-green colour blindness haemophilia “3”or “8”?? examples of sex linked traits:
- 93. Males are morelikely to suffer from sex-linked diseases Normal A Females carry two alleles of a gene. If one allele is defective, female is still normal as effect is masked by the normal allele. A Normal: A Sick: a SickPhenotypically normal / carrier A A a a aa Normal Sick Female: 3 choices Male : 2 choices
- 94.
- 95. 105 Features of Colourblindness Colour blindness – Inability to differentiate between red and green o hereditary disease o It is common in male but rare in female. o Caused by recessive allele located on X chromosome o Colour blindness follows criss- cross inheritance as transmitted from father to grandson through daughter. o It is never transmitted from father to son
- 96. 107 XCXC - Normalfemale XCXc - Carrier female XCY - Normal Male XcY - Affected male XcXc - Affected female
- 97. 108 Red-green colour blindness Parental PhenotypesCarrier Female x Normal Male Genotypes XBXb XBY Gametes Offspring 1 Genotypes Phenotypes Normal Female : Carrier Female : Normal Male : Colour blind Male 1 : 1 : 1 : 1 XB XbXB Y XB XBXb Y XB XB Xb Y
- 98. 109 GENETIC PEDIGREE FORCOLOUR BLINDNESS
- 99.
- 100. 111 This disease is appearedas a mutant in Queen Victoria and from her it was transmitted to her descendants. “Royal disease” XH XH - Normal female XH Xh - Carrier female XH Y - Normal Male Xh Y - Affected male Xh Xh - Affected female
- 101. 113 XhY XH XH Xh Y XHXH XHYXHXh XHY XHXh UNAFFECTED MOTHER FATHER - HAEMOPHILIA Carrier Normal Carrier Normal Female Male Female Male
- 102. 115 XHY XH Xh XH Y XHXh XhYXHXH XHY XHXh CARRIER MOTHER UNAFFECTED FATHER Normal Normal Carrier Haemophilia Female Male Female Male
- 103. 117 XhY XH Xh Xh Y XHXh XhYXHXh XHY XhXh CARRIER MOTHER FATHER - HAEMOPHILIA Carrier Normal Haemophilia Haemophilia Female Male Female Male
- 104. 119 XHY Xh Xh XH Y XhXh XhYXHXh XhY XHXh MOTHER - HEMOPHILIA Normal FATHER Carrier Haemophilia Carrier Haemophilia Female Male Female Male
- 105. 121 XhY Xh Xh Xh Y XhXh XhYXhXh XhY XhXh MOTHER - HAEMOPHILIA FATHER - HAEMOPHILIA Haemophilia Haemophilia Haemophilia Haemophilia Female Male Female Male
- 106. 122 HEREDITARY DISEASE Genetic diseasesthat offsprings inherit from their parents Eg. Haemophilia, red-green colour blindness, Duchenne muscular dystrophy are caused by recessive allele on the X chromosome. Eg. Cystic fibrosis, albinism, sickle cell anemia, thalassaemia are caused by defective genes found on the autosomes.
- 107. Sickle-cell Anaemia Causedby defective allele for synthesis of haemoglobin Autosomal gene located on chromosome number 11
- 108. This isdue to the clumping of the abnormal haemoglobin molecules in the red blood cell When blood oxygen is low the red blood cell has the shape of a sickle Sickle-cell Anaemia
- 109.
- 110. They morelikely to break, aggregate and clog the blood capillaries VIDEO SICKLE CELL DISEASE http://www.yout ube.com/watch? v=R4-c3hUhhyc
- 111. Cystic fibrosis Causedby a lack of transport protein which allows chloride ions to move across plasma membranes Normally water will pass through the plasma membranes after the chloride ions passes
- 112. Caused bycystic fibrosis gene located on chromosome number 7
- 113. Affected persons– frequent respiratory infections
- 114. Thalassaemia Anaemia Redblood cells cannot carry enough oxygen. Deficiency of iron. Caused by recessive gene Synthesis of abnormal haemoglobin in red blood cells Passed down by parents who carry thalassaemia genes in their cells
- 115. Symptoms : Appear healthy at birth After two years – become pale, listless, fussy, poor appetite Grow slowly Develop jaundice Treatment : Frequent blood transfusion Bone marrow transplant VIDEO http://www.youtube.com/watch?v=Ul7m_FNsd_c
- 116.