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Theoretical Genetics

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Information about Theoretical Genetics
Education

Published on May 10, 2008

Author: gurustip

Source: slideshare.net

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 af

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