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Albinism is a recessive disorder where your body doesn't produce melanin. Whoever has this disorder has red eyes and their hair, etc is white. Animals can have this condition as well.
Video:
https://www.youtube.com/watch?v=i_WYW-Zrv6E

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Cystic fibrosis is a disorder where there is excessive mucus production. This disease is recessive and is called the "white man's disease". It is chromosome 7. This results in lung damage.
Video:
https://www.youtube.com/watch?v=NmyZHMaHRs0

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For every child affected by cystic fibrosis, there are around 100 people who carry the faulty gene. Everyone has two copies of the gene which, when faulty, causes cystic fibrosis. Being a carrier means that one copy of the gene is faulty, and the other is normal.

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Tay sacs is a disease that causes you to have the inability to break down lipids and it accumulates in the brain. It causes developmental problems in infants and it generally happens among the Jewish community. This is recessive.
Video:
https://www.youtube.com/watch?v=CaWoeRY-gJY

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Shortcut to Solve Pedigree Analysis Problems - Android App

Modes of Genetic Inheritance

There are six modes in which an offspring can inherit a trait from his/her parents.

Mode of Inheritance

Examples

Autosomal Dominant
Polycystic kidney disease, Huntington's disease, hereditary spherocytosis

Autosomal Recessive
Sickle Cell anemia, cystic fibrosis, Tay-Sachs disease

X-Linked Dominant
Rett Syndrome, Vitamin-D resistant rickets, Alport syndrome

X-Linked Recessive
hemophilia, color blindness

Y-Linked
Hypertrichosis pinnae, Azoospermia, Retinitis Pigmentosa

Maternal (Cytoplasmic)
All the children are affected if mother is affected.

Free Download ZoomGenetics Software (For Windows) for learning genetics.

Symbols Used in Pedigree Chart

Squares represent males. Circles are females.

Horizontal lines between square and circles represents husband and wife. Children are one rung lower.

Generation are depicted as Roman numerals. A digit stands for an individual in a generation.

A darkened shape is an affected individual - male or female. A hollow shape is a normal individual.

Identical twins are represented with slanting lines and a horizontal line between them.

Non-identical twins are slanting lines but without horizontal line between them

The figure (circle or square) of a deceased individual is represented by striking it.

Usage

Pedigree analysis is used for various purposes - criminal, legal, scientific, research, academic, etc.

Pedigree Analysis is used to determine the mode of transmission of a genetic diseases - whether is is dominant or recessive, or whether is X-linked or Y-linked or autosomal, or whether it is maternal (cytoplasmic).

Method used

Pedigree Analysis is a method which heavily based on rules of logic. Based on the facts depicted by the family tree, either a definite conclusion can be drawn regarding the nature of disease, or alternatively, if the disease is known, we can estimate the percentage chance of an offspring acquiring the disease.

Shortcut to Solve Pedigree Analysis Problems - Android App

Rules Governing Genetic Transmission of Diseases

There are certain rules governing the genetic transmission of diseases.

1. A normal individual cannot have (one or both) alleles of the dominant trait. This is logical, because by definition, even if one allele of a dominant disease trait is present, the individual would be affected.

2. A normal individual, however, can be a carrier of a recessive trait. This means that an individual may have one allele of a recessive trait and still be phenotypically normal. The recessive allele do not express and is masked by the dominant allele. This may show in subsequent generations. So the distinctive feature of recessive trait is that, it skips generations.

3. If trait is X-linked, the male would be affected even by a single recessive allele. This is because male is heterozygous regarding the sex chromosome. So a single allele on X-chromosome (even recessive) will create the expression and male would be affected.

4. When father transmits a trait to his son, it is always an Autosomal trait.

Inheritance Patterns in Pedigree Analysis

Based on the above rules, there are a few patterns of genetic inheritance which gives clue to the nature of the trait.

1. If there is skipping of generations, meaning that the trait is observed in generation-I and not in generation-II, but re-appears in subsequent generations, then the trait is "Recessive".

2. If the affected individual has both parents normal, then we can conclude that the trait is "Recessive".

3. If the normal parents have the affected "son" ('B' in figure above), then the trait is "Recessive". We cannot conclude that its Autosomal or X-Linked. If Autosomal, he got it from both parents. But if X-linked he got it from his mother only.

4. Trait is dominant, if each generation has a trait. Remember! For a trait to express only one allele is sufficient.

5. However, when an affected father transmits the disease to a daughter ('E' in figure above), it may be X-linked or autosomal; we can't say for sure. We need have to have more information to ascertain the case.

6. When a mother is affected and "all" the children - sons as well as daughters, are affected ('F' in figure above), then the trait is maternal (cytoplasmic) inheritance.

Shortcut to Solve Pedigree Analysis Problems - Android App

Multiple Choice Questions

What can be the correct conclusion for the following family?

1. Mother is heterozygous.
2. Parents could not have a normal daughter for this character
3. The trait under study could not be color-blindness
4. Father is homozygous dominant

Ans: (1) Mother is heterozygous.

ZoomGenetics is a software for learning genetics for students preparing for AIPMT, AIIMS, JIPMER, PMT and other medical entrance exams. Zoom Genetics covers a major portion of about 25 to 30% of the syllabus of Biology for the students preparing for Medical Entrance Examination.

The unique features are zooming user interface, extensive graphics, bulleted notes, easy to understand description, info-graphics, memory techniques, and extensive multiple choice questions.

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Pedigree Analysis is tool with the help of which a family tree is displayed, and which shows the members of the family who are affected by a genetic trait. Males are shown as squares and females as circles. Also the affected individual is shown as black (or any colored) while unaffected is shown as unfilled square / circle.

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This disorder is where the blood doesn't clot. This is a sex-linked trait and is recessive. People that have this disorder experience excessive bleeding. Some of the symptoms are large bruises, joint swelling, and blood in urine/stool.
Video:
https://www.youtube.com/watch?v=oExbzAJP7t4

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PKU is when your body lacks the enzyme to break down phenylaline in milk. This disorder happens in infants. If the infants drink milk, then their nerves in their brain will be damaged. This disease is recessive. The treatment is that you don't feed the baby milk and instead feed them almond milk. This is recessive
Video:
https://www.youtube.com/watch?v=NFiNsq6i28s

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, IB Bio & MYP Science Teacher, HOD Science, MYP Coordinator

Published on

For the IB DP Biology course, core unit: Genetics. To get the file, please make a donation to one of my preferred charities via Biology4Good. Find out more here: http://sciencevideos.wordpress.com/about/biology4good/

1. 1. Theoretical Genetics Stephen Taylorhttp://sciencevideos.wordpress.com 4.3 Theoretical Genetics 1

2. 2. Definitions This image shows a pair of homologous chromosomes. Name and annotate the labeled features. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 2

3. 3. Definitions This image shows a pair of homologous chromosomes. Name and annotate the labeled features. Genotype The combination of alleles Homozygous dominant of a gene carried by an organism Having two copies of the same dominant allele Phenotype The expression of alleles Homozygous recessive of a gene carried by an organism Having two copies of the same recessive allele. Recessive alleles are Centromere only expressed when homozygous. Joins chromatids in cell division Codominant Alleles Pairs of alleles which are both Different versions of a gene expressed when present. Dominant alleles = capital letterRecessive alleles = lower-case letter Heterozygous Having two different alleles. The dominant allele is expressed. Carrier Gene loci Heterozygous carrier of a Specific positions of genes on a recessive disease-causing allele chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 3

4. 4. Making Babies1. Count the chromosomes in your envelope - there should be 46 in total.2. Shuffle the chromosomes, so that they are well mixed up. Which aspects of meiosis and sexual reproduction give genetic variation? * Crossing-over in prophase I * Random orientation in metaphase I and II * Random fertilisation3. Now arrange them in a karyotype (dont turn them over - leave them as they were). Activity from: http://www.nclark.net/Genetics4. What is the gender of your baby? Explain how gender is inherited in humans. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 4

5. 5. Making Babies * Crossing-over in prophase I * Random orientation in metaphase I and II * Random fertilisationList all the traits in a table. Use the key above to determine the genotypes and phenotypesof your offspring. Draw a picture of your beautiful child's face!HL identify traits which are polygenic, involve gene interactions and some which are linked. Activity from: http://www.nclark.net/Genetics http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 5

6. 6. Explain this Mendel crossed some yellow peas with some yellow peas. Most offspring were yellow but some were green! Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 6

7. 7. Segregation "alleles of each gene separate into different gametes when the individual produces gametes" The yellow parent peas must Mendel did not know about be heterozygous. The yellow DNA, chromosomes or meiosis. phenotype is expressed. Through his experiments he did Through meiosis and work out that 'heritable factors' fertilisation, some offspring (genes) were passed on and peas are homozygous that these could have different recessive - they express a versions (alleles). green colour. Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 7

8. 8. Segregation "alleles of each gene separate into different gametes when the individual produces gametes" F0 Genotype: Yy Yy Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Gametes: Y or y Y or y Punnet Grid: gametes Genotypes: F1 Phenotypes: Phenotype ratio: Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 8

9. 9. Monohybrid Cross Crossing a single trait. F0 Genotype: Yy Yy Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Gametes: Y or y Y or y Fertilisation results in diploid Punnet Grid: gametes zygotes. A punnet grid can be used to deduce the potential outcomes of the cross and to calculate the expected ratio of phenotypes in the next generation (F1). Genotypes:F1 Phenotypes: Phenotype ratio: Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 9

10. 10. Monohybrid Cross Crossing a single trait. F0 Genotype: Yy Yy Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Gametes: Y or y Y or y Fertilisation results in diploid Punnet Grid: gametes Y y zygotes. A punnet grid can be used to Y YY Yy deduce the potential outcomes of the cross and to calculate the y Yy yy expected ratio of phenotypes in the next generation (F1). Genotypes:F1 Phenotypes: Phenotype ratio: Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 10

11. 11. Monohybrid Cross Crossing a single trait. F0 Genotype: Yy Yy Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Gametes: Y or y Y or y Fertilisation results in diploid Punnet Grid: gametes Y y zygotes. A punnet grid can be used to Y YY Yy deduce the potential outcomes of the cross and to calculate the y Yy yy expected ratio of phenotypes in the next generation (F1). Genotypes: YY Yy Yy yy Ratios are written in theF1 Phenotypes: simplest mathematical form. Phenotype ratio: 3:1 Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 11

12. 12. Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? F0 Phenotype: Key to alleles: Y = yellow y = green Genotype: Homozygous recessive Homozygous recessive Punnet Grid: gametes Genotypes:F1 Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 12

13. 13. Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? F0 Phenotype: Key to alleles: Y = yellow y = green Genotype: yy yy Homozygous recessive Homozygous recessive Punnet Grid: gametes y y y yy yy y yy yy Genotypes: yy yy yy yyF1 Phenotypes: Phenotype ratio: All green http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 13

14. 14. Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? F0 Phenotype: Key to alleles: Y = yellow y = green Genotype: Homozygous recessive Heterozygous Punnet Grid: gametes Genotypes:F1 Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 14

15. 15. Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? F0 Phenotype: Key to alleles: Y = yellow y = green Genotype: yy Yy Homozygous recessive Heterozygous Punnet Grid: gametes Y y y Yy yy y Yy yy Genotypes: Yy Yy yy yyF1 Phenotypes: Phenotype ratio: 1:1 http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 15

16. 16. Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? F0 Phenotype: Key to alleles: Y = yellow y = green Genotype: Homozygous dominant Heterozygous Punnet Grid: gametes Genotypes:F1 Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 16

17. 17. Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross? F0 Phenotype: Key to alleles: Y = yellow y = green Genotype: YY Yy Homozygous dominant Heterozygous Punnet Grid: gametes Y y Y YY Yy Y YY Yy Genotypes: YY YY Yy YyF1 Phenotypes: Phenotype ratio: All yellow http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 17

18. 18. Test Cross Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. F0 Phenotype: Key to alleles: R = Red flower r = white Genotype: R? r r unknown Homozygous recessive Possible outcomes: F1 Phenotypes: Unknown parent = RR Unknown parent = Rr gametes gametes http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 18

19. 19. Test Cross Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. F0 Phenotype: Key to alleles: R = Red flower r = white Genotype: R? r r unknown Homozygous recessive Possible outcomes: F1 Phenotypes: All red Some white, some red Unknown parent = RR Unknown parent = Rr gametes r r gametes r r R Rr Rr R Rr Rr R Rr Rr r rr rr http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 19

20. 20. Career-related Case Study "According to the US Bureau of Labor Statistics, the graduate of today will change career four to six times in a lifetime. By one estimate, 65 per cent of the jobs that will be available upon college graduation for students now entering high school (thats eight years from now) do not yet exist. Consider the new interdisciplinary field of genetic counselling, which combines biological science with social work and ethics - it was ranked as one of the "top 10" career choices of 2010 because it offered far more openings than could be filled by qualified applicants." From the Times Higher Education Supplement - "So Last Century" http://www.timeshighereducation.co.uk/story.asp?sectioncode=26&storycode=415941&c=2 http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 20

21. 21. Career-related Case Study "According to the US Bureau of Labor Statistics, the graduate of today will change career four to six times in a lifetime. By one estimate, 65 per cent of the jobs that will be available upon college graduation for students now entering high school (thats eight years from now) do not yet exist. Consider the new interdisciplinary field of genetic counselling, which combines biological science with social work and ethics - it was ranked as one of the "top 10" career choices of 2010 because it offered far more openings than could be filled by qualified applicants." From the Times Higher Education Supplement - "So Last Century" http://www.timeshighereducation.co.uk/story.asp?sectioncode=26&storycode=415941&c=2 You are a genetic counselor. A couple walk into your clinic and are concerned about their pregnancy. They each have one parent who is affected by phenylketonuria (PKU) and one parent who has no family history. Explain PKU and its inheritance to them. Deduce the chance of having a child with PKU and how it can be tested and treated. Use the following tools in your explanations: * Pedigree chart * Punnet grid * Diagrams http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 21

22. 22. Phenylketonuria (PKU) Clinical example.Pedigree charts can be used to trace family histories and deduce genotypes and risk in the caseof inherited gene-related disorders. Here is a pedigree chart for this family history. key female maleI affected NotII Affected A B deceasedIII ? Is PKU dominant or recessive? How do you know? * * http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 22

23. 23. Phenylketonuria (PKU) Clinical example.Pedigree charts can be used to trace family histories and deduce genotypes and risk in the caseof inherited gene-related disorders. Here is a pedigree chart for this family history. key female maleI affected NotII Affected A B deceasedIII ? Is PKU dominant or recessive? How do you know? * Recessive * Unaffected mother in Gen I has produced affected II A. Mother must have been a carrier. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 23

24. 24. Phenylketonuria (PKU) Clinical example. A mis-sense mutation in the gene that produces tyrosine hydroxylase means that phenylalanine cannot be converted to tyrosine in the body - so it builds up. This results in brain developmental problems and seizures. It is progressive, so it must be diagnosed and treated early. Dairy, breastmilk, meat, nuts and aspartame must be avoided, as they are rich in phenylalanine. Diagnosis- blood test taken at 6-7 days after birth The Boy with PKU ideo clip from: http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/ http://www.youtube.com/watch?v=KUJVujhHxPQ http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 24

25. 25. Phenylketonuria (PKU) Clinical example.A recessive mis-sense mutation in the gene that produces tyrosine hydroxylase means thatphenylalanine cannot be converted to tyrosine in the body - so it builds up. Genetics review: 1. What is a missense mutation? 2. Is this disorder autosomal or sex-linked? 3. What is the locus of the tyrosine hydroxlase gene?Chromosome 12 from:http://commons.wikimedia.org/wiki/File:Chromosome_12.svg Diagnosis- blood test taken at 6-7 days after birth http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/ http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 25

26. 26. Phenylketonuria (PKU) Clinical example.A recessive mis-sense mutation in the gene that produces tyrosine hydroxylase means thatphenylalanine cannot be converted to tyrosine in the body - so it builds up. Genetics review: 1. What is a missense mutation? It is a base-substitution mutation where the change in a single base results in a different amino acid being produced in the polypeptide. 2. Is this disorder autosomal or sex-linked? Autosomal - chromosome 12 3. What is the locus of the tyrosine hydroxlase gene? 12q22 - 24Chromosome 12 from:http://commons.wikimedia.org/wiki/File:Chromosome_12.svg Diagnosis- blood test taken at 6-7 days after birth http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/ http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 26

27. 27. Phenylketonuria (PKU) Clinical example.What is the probability of two parents who are both carriers of the recessive allele producingchildren affected by PKU?F0 Phenotype: carrier carrier Key to alleles: T = Normal enzyme t = faulty enzyme Genotype: Tt Tt Punnet Grid: gametes T t T t Genotypes: F1 Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 27

28. 28. Phenylketonuria (PKU) Clinical example.What is the probability of two parents who are both carriers of the recessive allele producingchildren affected by PKU?F0 Phenotype: carrier carrier Key to alleles: T = Normal enzyme t = faulty enzyme Genotype: Tt Tt Punnet Grid: gametes T t T TT Tt t Tt tt Genotypes: TT Tt Tt tt F1 Phenotypes: Normal enzyme PKU Therefore 25% chance Phenotype ratio: 3:1 of a child with PKU http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 28

29. 29. Pedigree Charts Key to alleles: T= Has enzymePedigree charts can be used to trace family histories and deduce t = no enzymegenotypes and risk in the case of inherited gene-related disorders.Here is a pedigree chart for this family history. Key: female male affected Not Affected deceased Looks like Deduce the genotypes of these individuals: A&B C D Genotype Reason http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 29

30. 30. Pedigree Charts Key to alleles: T= Has enzymePedigree charts can be used to trace family histories and deduce t = no enzymegenotypes and risk in the case of inherited gene-related disorders.Here is a pedigree chart for this family history. Key: female male affected Not Affected deceased Looks like Deduce the genotypes of these individuals: A&B C D Genotype Both Tt tt Tt To have produced affected Trait is recessive, as both Recessive traits only child H, D must have inherited Reason are normal, yet have produced an affected child (C) expressed when homozygous. a recessive allele from either A or B http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 30

31. 31. Pedigree Charts Key to alleles: T= Has enzymeIndividuals D and $ are planning to have another child. t = no enzymeCalculate the chances of the child having PKU. Key: female male affected Not $ Affected deceased Looks like Genotypes: D= Gametes Phenotype ratio $= Therefore http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 31

32. 32. Pedigree Charts Key to alleles: T= Has enzymeIndividuals D and $ are planning to have another child. t = no enzymeCalculate the chances of the child having PKU. Key: female male affected Not $ Affected deceased Looks like Genotypes: D = Tt (carrier) Gametes T t Phenotype ratio t Tt tt 1 : 1 Normal : PKU $ = tt (affected) Therefore 50% chance of a t Tt tt child with PKU http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 32

33. 33. Codominance Some genes have more than two alleles. Where alleles are codominant, they are both expressed.Human ABO blood typing is an example of multiple alleles and codominance.The gene is for cell-surface antigens (immunoglobulin receptors).These are either absent (type O) or present.If they are present, they are either type A, B or both. Key to alleles: i = no antigens presentWhere the genotype is heterozygous for IA and IB, both IA = type A anitgens presentare expressed. This is codominance. IB = type B antigens present http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 33

34. 34. More about blood typing A Nobel breakthrough in medicine.Antibodies (immunoglobulins) are specific to antigens.The immune system recognises foreign antigens andproduces antibodies in response - so if you are given thewrong blood type your body might react fatally as theantibodies cause the blood to clot.Blood type O is known as the universal donor, as it hasnot antigens against which the recipient immune systemcan react. Type AB is the universal recipient, as it has noantibodies which will react to AB antigens. Blood typing game from Nobel.org: Images and more information from: http://nobelprize.org/educational/medicine/landsteiner/readmore.html http://learn.genetics.utah.edu/content/begin/traits/blood/ http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 34

35. 35. Sickle Cell Another example of codominance. Remember the notation used: superscripts represent codominant alleles. In codominance, heterozygous individuals have a mixed phenotype. The mixed phenotype gives protection against malaria, but does not exhibit full-blown sickle cell anemia. Complete the table for these individuals: Genotype Description Homozygous HbA Heterozygous Homozygous HbS Phenotype Malaria protection? http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 35

36. 36. Sickle Cell Another example of codominance. Remember the notation used: superscripts represent codominant alleles. In codominance, heterozygous individuals have a mixed phenotype. The mixed phenotype gives protection against malaria, but does not exhibit full-blown sickle cell anemia. Complete the table for these individuals: Genotype HbA HbA HbA HbS HbS HbS Description Homozygous HbA Heterozygous Homozygous HbS Phenotype normal carrier Sickle cell disease Malaria No Yes Yes protection? http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 36

37. 37. Sickle Cell Another example of codominance. Key to alleles: HbA = Normal Hb HbS = Sickle cell Predict the phenotype ratio in this cross: F0 Phenotype: carrier affected Genotype: Punnet Grid: gametes Genotypes: F1 Phenotypes: Phenotype ratio: : Therefore 50% chance of a child with sickle cell disease. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 37

38. 38. Sickle Cell Another example of codominance. Key to alleles: HbA = Normal Hb HbS = Sickle cell Predict the phenotype ratio in this cross: F0 Phenotype: carrier affected Genotype: HbA Hbs HbS Hbs Punnet Grid: gametes HbS HbS HbA HbAHbS HbAHbS HbS HbSHbS HbSHbS Genotypes: HbAHbS & HbSHbS F1 Phenotypes: Carrier & Sickle cell Phenotype ratio: 1:1 Therefore 50% chance of a child with sickle cell disease. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 38

39. 39. Sickle Cell Another example of codominance. Key to alleles: HbA = Normal Hb HbS = Sickle cell Predict the phenotype ratio in this cross: F0 Phenotype: carrier carrier Genotype: Punnet Grid: gametes Genotypes: F1 Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 39

40. 40. Sickle Cell Another example of codominance. Key to alleles: HbA = Normal Hb HbS = Sickle cell Predict the phenotype ratio in this cross: F0 Phenotype: carrier carrier Genotype: HbA HbS HbA HbS Punnet Grid: gametes HbA HbS HbA HbAHbA HbAHbS HbS HbAHbS HbSHbS Genotypes: HbAHb & 2 HbAHbS & HbSHbS F1 Phenotypes: Unaffected & Carrier & Sickle cell Phenotype ratio: 1: 2 : 1 Therefore 25% chance of a child with sickle cell disease. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 40

41. 41. Sickle Cell Another example of codominance. Key to alleles: HbA = Normal Hb HbS = Sickle cell Predict the phenotype ratio in this cross: F0 Phenotype: carrier unknown Genotype: HbA HbS Punnet Grid: gametes HbA HbS Genotypes: F1 Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 41

42. 42. Sickle Cell Another example of codominance. Key to alleles: HbA = Normal Hb HbS = Sickle cell Predict the phenotype ratio in this cross: F0 Phenotype: carrier unknown Genotype: HbA HbS HbA HbA or HbA HbS Punnet Grid: gametes HbA HbA HbA HbS HbA HbS Genotypes: F1 Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 42

43. 43. Sickle Cell Another example of codominance. Key to alleles: HbA = Normal Hb HbS = Sickle cell Predict the phenotype ratio in this cross: F0 Phenotype: carrier unknown Genotype: HbA HbS HbA HbA or HbA HbS Punnet Grid: gametes HbA HbA HbA HbS HbA HbAHbA HbAHbA HbAHbA HbAHbS HbS HbAHbS HbAHbS HbAHbS HbSHbS Genotypes: 3 HbAHbA & 4 HbAHbS & 1 HbSHbS F1 Phenotypes: 3 Unaffected & 4 Carrier & 1 Sickle cell Phenotype ratio: 3:4:1 Therefore 12.5% chance of a child with sickle cell disease. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 43

44. 44. Sex Determination It's all about X and Y... Humans have 23 pairs of chromosomes in diploid somatic cells (n=2). 22 pairs of these are autosomes, which are homologous pairs. One pair is the sex chromosomes. XX gives the female gender, XY gives male.Karyotype of a human male, showing X and Y chromosomes:http://en.wikipedia.org/wiki/Karyotype SRY The X chromosome is much larger than the Y. X carries many genes in the non-homologous region which are not present on Y. The presence and expression of the SRY gene on Y leads to male development. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 44

45. 45. Sex Determination It's all about X and Y... Chromosome pairs segregate in meiosis. Females (XX) produce only eggs containing the X chromosome. Males (XY) produce sperm which can contain either X or Y chromosomes.Segregation of the sex chromosomes in meiosis. SRY gene determines maleness. gametes X Y Find out more about its role and X XX XY just why do men have nipples? X XX XY Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome Therefore there is an even chance* of the offspring being male or female. http://www.hhmi.org/biointeractive/gender/lectures.html http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 45

46. 46. Sex Determination Non-disjunction can lead to gender disorders.XYY Syndrome: XXY: Klinefelter Syndrome:Fertile males, with increased risk of learning difficulties. Males with enhanced female characteristicsSome weak connections made to violent tendencies.XO: Turner SyndromeMonosomy of X, leads to short stature, female children.XXX Syndrome:Fertile females. Some X-carrying gametes can be produced. Interactive from HHMI Biointeractive: Image from NCBI: http://www.hhmi.org/biointeractive/gender/click.html http://www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mga&part=A1179 http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 46

47. 47. Sex Linkage X and Y chromosomes are non-homologous.The sex chromosomes are non-homologous.There are many genes on the X-chromosome Non-homologouswhich are not present on the Y-chromosome. regionSex-linked traits are those which are carried onthe X-chromosome in the non-homologousregion. They are more common in males. Non-homologous region Examples of sex-linked genetic disorders: - haemophilia - colour blindnessX and Y SEM from Chromosome images from Wikipedia:http://www.angleseybonesetters.co.uk/bones_DNA.html http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 47

48. 48. Sex Linkage X and Y chromosomes are non-homologous.What number do you see? Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 48

49. 49. Sex Linkage X and Y chromosomes are non-homologous.What number do you see? 5 = normal vision 2 = red/green colour blindness Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 49

50. 50. Sex Linkage X and Y chromosomes are non-homologous.How is colour-blindness inherited? The red-green gene is carried at locus Xq28. This locus is in the non-homologous region, so there is no corresponding gene (or allele) on the Y chromosome. Normal vision is dominant over colour-blindness. XN XN Normal female XN Y Normal male no allele carried, none written Key to alleles: n n n N = normal vision Xq28 X X Affected female X Y Affected male n = red/green colour blindness N n Human females can be homozygous or X X Carrier female heterozygous with respect to sex-linked genes. Heterozygous females are carriers. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 50

51. 51. Sex Linkage X and Y chromosomes are non-homologous.What chance of a colour-blind child in the cross between a Key to alleles:normal male and a carrier mother? N = normal vision n = red/green colourF0 Genotype: XN Xn XN Y blindness Phenotype: Carrier female X Normal male Punnet Grid: F1 Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 51

52. 52. Sex Linkage X and Y chromosomes are non-homologous.What chance of a colour-blind child in the cross between a Key to alleles:normal male and a carrier mother? N = normal vision n = red/green colourF0 Genotype: XN Xn XN Y blindness Phenotype: Carrier female X Normal male Punnet Grid: XN Y XN XN XN XN Y F1 Xn XN Xn Xn Y Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 52

53. 53. Sex Linkage X and Y chromosomes are non-homologous. What chance of a colour-blind child in the cross between a Key to alleles: normal male and a carrier mother? N = normal vision n = red/green colour F0 Genotype: XN Xn XN Y blindness Phenotype: Carrier female X Normal male Punnet Grid: XN Y XN XN XN XN Y Normal female Normal male F1 Xn XN Xn Carrier female Xn Y Affected male There is a 1 in 4 (25%) chance of an affected child.What ratios would we expect in a cross between:a. a colour-blind male and a homozygous normal female? Chromosome images from Wikipedia:b. a normal male and a colour-blind female? http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 53

54. 54. Red-Green Colour Blindness How does it work? The OPN1MW and OPN1LW genes are found at locus Xq28. They are responsible for producing photoreceptive pigments in the cone cells in the eye. If one of these genes is a mutant, the pigments are not produced properly and the eye cannot distinguish between green (medium) wavelengths and red (long) wavelengths in the visible spectrum.Because the Xq28 gene is in a non-homologous region when compared Xq28to the Y chromosome, red-green colour blindness is known as a sex-linked disorder. The male has no allele on the Y chromosome tocombat a recessive faulty allele on the X chromosome. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 54

55. 55. Colour Blind cartoon from:http://www.almeidacartoons.com/Med_toons1.htmlhttp://sciencevideos.wordpress.com 4.3 Theoretical Genetics 55

56. 56. Hemophilia Another sex-linked disorder.Blood clotting is an example of a metabolic pathway -a series of enzyme-controlled biochemical reactions.It requires globular proteins called clotting factors.A recessive X-linked mutation in hemophiliacs results in one of thesefactors not being produced. Therefore, the clotting response toinjury does not work and the patient can bleed to death.XH XHNormal female XH Y Normal male no allele carried, none written Key to alleles: XH = healthy clotting factors Xh Xh Affected female Xh Y Affected male Xh = no clotting factor Human females can be homozygous or XH Xh Carrier female heterozygous with respect to sex-linked genes. Heterozygous females are carriers. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 56

57. 57. Hemophilia results from a lack of clotting factors. These are globular proteins, which act as enzymes in the clotting pathway. Read/ research/ review: How can gene transfer be used to treat hemophiliacs? What is the relevance of "the genetic code is universal" in this process? Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 57

58. 58. Hemophilia results from a lack of clotting factors. These are globular proteins, which act as enzymes in the clotting pathway. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 58

59. 59. Hemophilia This pedigree chart of the English Royal Family gives us a picture of the inheritance of this X-linked disorder.Royal Family Pedigree Chart from:http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 59

60. 60. Hemophilia Pedigree chart practice State the genotypes of the following family members: 1. Leopold 2. Alice 3. Bob was killed in a tragic croquet accident before his phenotype was determined. Key: female male 4. Britney affectedKey to alleles: NotH = healthy clotting factors Affectedh = no clotting factor deceased Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 60

61. 61. Hemophilia Pedigree chart practice State the genotypes of the following family members: 1. Leopold Xh Y 2. Alice 3. Bob was killed in a tragic croquet accident before his phenotype was determined. Key: female male 4. Britney affectedKey to alleles: NotH = healthy clotting factors Affectedh = no clotting factor deceased Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 61

62. 62. Hemophilia Pedigree chart practice State the genotypes of the following family members: 1. Leopold Xh Y 2. Alice XH Xh 3. Bob was killed in a tragic croquet accident before his phenotype was determined. Key: female male 4. Britney affectedKey to alleles: NotH = healthy clotting factors Affectedh = no clotting factor deceased Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 62

63. 63. Hemophilia Pedigree chart practice State the genotypes of the following family members: 1. Leopold Xh Y 2. Alice XH Xh 3. Bob was killed in a tragic croquet accident before his phenotype was determined. XH Y or Xh Y Key: female male 4. Britney affectedKey to alleles: NotH = healthy clotting factors Affectedh = no clotting factor deceased Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 63

64. 64. Hemophilia Pedigree chart practice State the genotypes of the following family members: 1. Leopold Xh Y 2. Alice XH Xh 3. Bob was killed in a tragic croquet accident before his phenotype was determined. XH Y or Xh Y Key: female male 4. Britney XH XH or XH Xh affectedKey to alleles: NotH = healthy clotting factors Affectedh = no clotting factor deceased Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 64

65. 65. Pedigree Chart Practice Key: female male affected Not Affected deceasedDominant or Recessive? Autosomal or Sex-linked? http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 65

66. 66. Pedigree Chart Practice Key: female male affected Not Affected deceasedDominant or Recessive? Autosomal or Sex-linked?Dominant.A and B are both affected but have producedunaffected (D & F). Therefore A and B must havebeen carrying recessive healthy alleles.If it were recessive, it would need to behomozygous to be expressed in A & B - and thenall offspring would be homozygous recessive. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 66

67. 67. Pedigree Chart Practice Key: female male affected Not Affected deceasedDominant or Recessive? Autosomal or Sex-linked?Dominant. Autosomal.A and B are both affected but have produced Male C can only pass on one X chromosome. If itunaffected (D & F). Therefore A and B must have were carried on X, daughter H would be affectedbeen carrying recessive healthy alleles. by the dominant allele.If it were recessive, it would need to be Tip: Don't get hung up on the number ofhomozygous to be expressed in A & B - and then individuals with each phenotype - eachall offspring would be homozygous recessive. reproductive event is a matter of chance. Instead focus on possible and impossible genotypes. Draw out the punnet grids if needed. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 67

68. 68. Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% B. 12.5% C. 25% D. 50% Key: female male affected Not Affected deceased http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 68

69. 69. Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% Key to alleles: B. 12.5% XH = healthy clotting factors Xh = no clotting factor C. 25% D. 50% Key: female male affected Not Affected deceased http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 69

70. 70. Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% Key to alleles: B. 12.5% XH = healthy clotting factors Xh = no clotting factor C. 25% D. 50% What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y Key: female male affected Not Affected XH deceased Y http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 70

71. 71. Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% Key to alleles: B. 12.5% XH = healthy clotting factors Xh = no clotting factor C. 25% D. 50% What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y Key: female male There is an equal chance of F being XH XH or XH Xh So: affected Not XH XH XH Xh Affected XH deceased Y http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 71

72. 72. Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% Key to alleles: B. 12.5% XH = healthy clotting factors Xh = no clotting factor C. 25% D. 50% What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y Key: female male There is an equal chance of F being XH XH or XH Xh So: affected Not XH XH XH Xh Affected XH XH XH XH XH XH XH XH Xh deceased Y XH Y XH Y XH Y Xh Y http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 72

73. 73. Super Evil Past Paper Question In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected? A. 0% Key to alleles: B. 12.5% XH = healthy clotting factors Xh = no clotting factor C. 25% D. 50% What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y Key: female male There is an equal chance of F being XH XH or XH Xh So: affected Not XH XH XH Xh Affected XH XH XH XH XH XH XH XH Xh deceased Y XH Y XH Y XH Y Xh Y So there is a 1 in 8 (12.5%) chance of the offspring being affected! http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 73

74. 74. Whirling Gene activity from the awesome Learn.Genetics site:http://learn.genetics.utah.edu/archive/pedigree/mapgene.html http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 74

75. 75. For more IB Biology resources: http://sciencevideos.wordpress.com This presentation is free to view. Please make a donation to one of my chosen charities at Gifts4Good and I will send you the editable pptx file. Click here for more information about Biology4Good charity donations.This is a Creative Commons presentation. It may be linked and embedded but not sold or re-hosted. 4.3 Theoretical Genetics 75

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IB Bio & MYP Science Teacher, HOD Science, MYP Coordinator Published on For the IB DP Biology course, core unit: Genetics. To get the file, please make a donation to one of my preferred charities via Biology4Good. Find out more here: http://sciencevideos.wordpress.com/about/biology4good/ 1. Theoretical Genetics Stephen Taylorhttp://sciencevideos.wordpress.com 4.3 Theoretical Genetics 1 2.

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Sickle Cell Anemia is recessive and codominant and is a point mutation. People with this have abnormal RBC. It generally affects the black community. Sickle cell anemia results in heart damage.
Video:
https://www.youtube.com/watch?v=R4-c3hUhhyc

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genetic pedigrees genetic pedigrees are like family trees

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Genetic Pedigrees Genetic pedigrees are like family trees - ppt video online download

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Creating a Pedigree H h H h hh H ? H ? Key H = Healthy h = PKU H h H h hh H ? H ? 1) Manny and Rosa are married and both are healthy. 2) They have 3 children. 3) Their oldest child, Jacob, has a recessive disorder called PKU.

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Huntington's disease is dominant. Only one person needs a copy of the gene to develop the disorder. This disorder deteriorates the brain and it is fatal. People usually get Huntington's when they are about 30 to 50 and eventually gets pneumonia, has heart failure, or another horrible result.
Video: https://www.youtube.com/watch?v=-s6h-JuYHEE

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Late onset Huntington's disease is an autosomal dominant genetic disorder that affects the central nervous system of human beings. This disease typically shows up when a person reaches their mid...