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Theoretical Genetics Stephen Taylor http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 1 Definitions Genotype The combination of alleles of a gene carried by an organism Phenotype The expression of alleles of a gene carried by an organism Centromere This image shows a pair of homologous chromosomes. Name and annotate the labeled features. Homozygous dominant Having two copies of the same dominant allele Homozygous recessive Having two copies of the same recessive allele. Recessive alleles are only expressed when homozygous. Joins chromatids in cell division Codominant Alleles Different versions of a gene Dominant alleles = capital letter Recessive alleles = lower-case letter Carrier Heterozygous carrier of a recessive disease-causing allele http://sciencevideos.wordpress.com Pairs of alleles which are both expressed when present. Heterozygous Having two different alleles. The dominant allele is expressed. Gene loci Specific positions of genes on a chromosome 4.3 Theoretical Genetics 2 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 3 Segregation The yellow parent peas must be heterozygous. The yellow phenotype is expressed. Through meiosis and fertilisation, some offspring peas are homozygous recessive – they express a green colour. “alleles of each gene separate into different gametes when the individual produces gametes” Mendel did not know about DNA, chromosomes or meiosis. Through his experiments he did work out that ‘heritable factors’ (genes) were passed on and that these could have different versions (alleles). Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 4 Segregation “alleles of each gene separate into different gametes when the individual produces gametes” F0 F1 Genotype: Yy Yy Gametes: Y or y Y or y Punnet Grid: gametes Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 4.3 Theoretical Genetics 5 Monohybrid Cross Crossing a single trait. F0 Genotype: Yy Yy Gametes: Y or y Y or y Punnet Grid: gametes Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Fertilisation results in diploid 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). F1 Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 4.3 Theoretical Genetics 6 Monohybrid Cross Crossing a single trait. F0 Genotype: Yy Yy Gametes: Y or y Y or y Punnet Grid: F1 gametes Y y Y YY Yy y Yy yy Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. Fertilisation results in diploid 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: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 4.3 Theoretical Genetics 7 Monohybrid Cross Crossing a single trait. F0 Genotype: Yy Yy Gametes: Y or y Y or y Punnet Grid: F1 Genotypes: gametes Y YY Yy y Yy yy YY Yy Fertilisation results in diploid zygotes. y Y Yy Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes. 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). yy Ratios are written in the simplest mathematical form. Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 3:1 Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 4.3 Theoretical Genetics 8 Monohybrid Cross F0 What is the expected ratio of phenotypes in this monohybrid cross? Key to alleles: Y = yellow y = green Phenotype: Genotype: Homozygous recessive Punnet Grid: F1 Homozygous recessive gametes Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 9 Monohybrid Cross F0 Key to alleles: Y = yellow y = green Phenotype: Genotype: yy yy Homozygous recessive Punnet Grid: F1 What is the expected ratio of phenotypes in this monohybrid cross? Genotypes: Homozygous recessive gametes y y y yy yy y yy yy yy yy yy yy Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com All green 4.3 Theoretical Genetics 10 Monohybrid Cross F0 What is the expected ratio of phenotypes in this monohybrid cross? Key to alleles: Y = yellow y = green Phenotype: Genotype: Homozygous recessive Punnet Grid: F1 Heterozygous gametes Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 11 Monohybrid Cross F0 Phenotype: Genotype: yy Yy Homozygous recessive Punnet Grid: F1 What is the expected ratio of phenotypes in this monohybrid cross? Genotypes: Heterozygous gametes Y y y Yy yy y Yy yy Yy Yy Key to alleles: Y = yellow y = green yy yy Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 1:1 4.3 Theoretical Genetics 12 Monohybrid Cross F0 What is the expected ratio of phenotypes in this monohybrid cross? Key to alleles: Y = yellow y = green Phenotype: Genotype: Homozygous dominant Punnet Grid: F1 Heterozygous gametes Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 13 Monohybrid Cross F0 Phenotype: Genotype: YY Yy Homozygous dominant Punnet Grid: F1 What is the expected ratio of phenotypes in this monohybrid cross? Genotypes: Heterozygous gametes Y y Y YY Yy Y YY Yy YY YY Key to alleles: Y = yellow y = green Yy Yy Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com All yellow 4.3 Theoretical Genetics 14 Test Cross F0 Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. Key to alleles: R = Red flower r = white Phenotype: R? Genotype: unknown r r Homozygous recessive Possible outcomes: F1 Phenotypes: Unknown parent = RR gametes http://sciencevideos.wordpress.com Unknown parent = Rr gametes 4.3 Theoretical Genetics 15 Test Cross F0 Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. Key to alleles: R = Red flower r = white Phenotype: R? Genotype: unknown r r Homozygous recessive Possible outcomes: F1 All red Phenotypes: Unknown parent = RR Some white, some red 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 16 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F0 Y = yellow y = green S = smooth s = rough Phenotype: Heterozygous at both loci Heterozygous at both loci Genotype: Punnet Grid: gametes F1 http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 17 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F0 Phenotype: Genotype: Punnet Grid: Smooth, yellow Smooth, yellow Heterozygous at both loci Heterozygous at both loci SsYy SsYy gametes F1 http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 18 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F0 Phenotype: Genotype: Punnet Grid: Smooth, yellow Smooth, yellow Heterozygous at both loci Heterozygous at both loci SsYy SsYy gametes SY Sy sY sy SY SSYY SSYy SsYY SsYy Sy SSYy SSyy SsYy Ssyy sY SsYY SsYy ssYY ssYy sy SsYy Ssyy ssYy ssyy F1 http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 19 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F0 Phenotype: Genotype: Punnet Grid: F1 Smooth, yellow Smooth, yellow Heterozygous at both loci Heterozygous at both loci SsYy SsYy gametes SY Sy sY sy SY SSYY SSYy SsYY SsYy Sy SSYy SSyy SsYy Ssyy sY SsYY SsYy ssYY ssYy sy SsYy Ssyy ssYy ssyy Phenotypes: 9 Smooth, yellow : 3 Smooth, green : 3 Rough, yellow : 1 Rough, green http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 20 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Calculate the predicted phenotype ratio for: F0 Key to alleles: Y = yellow y = green S = smooth s = rough Phenotype: Heterozygous at both loci Heterozygous for S, homozygous dominant for Y Genotype: Punnet Grid: F1 Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 21 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough Calculate the predicted phenotype ratio for: F0 Phenotype: Smooth, yellow Heterozygous at both loci Genotype: SsYy Smooth, yellow Heterozygous for S, homozygous dominant for Y SsYY Punnet Grid: F1 Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 22 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough Calculate the predicted phenotype ratio for: F0 Phenotype: Smooth, yellow Smooth, yellow Heterozygous for S, homozygous dominant for Y Heterozygous at both loci Genotype: Punnet Grid: SsYY SsYy gametes SY sY SY Sy sY sy F1 Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 23 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough Calculate the predicted phenotype ratio for: F0 Phenotype: Smooth, yellow Heterozygous for S, homozygous dominant for Y Heterozygous at both loci Genotype: Punnet Grid: F1 Smooth, yellow SsYY SsYy gametes SY sY SY SSYY SsYY Sy SSYy SsYy sY SsYY ssYY sy SsYy ssYy Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 24 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough Calculate the predicted phenotype ratio for: F0 Phenotype: Smooth, yellow Smooth, yellow Heterozygous for S, homozygous dominant for Y Heterozygous at both loci Genotype: Punnet Grid: SsYY SsYy gametes SY sY SY SSYY SsYY Sy SSYy SsYy sY SsYY ssYY sy SsYy ssYy 6 Smooth, yellow : 2 Rough, yellow F1 Phenotypes: 3 Smooth, yellow : 1 Rough, yellow http://sciencevideos.wordpress.com Present the ratio in the simplest mathematical form. 10.2 Dihybrid Crosses & Gene Linkage 25 Dihybrid Crosses Common expected ratios of dihybrid crosses. SsYy SsYy SsYy Heterozygous at both loci Heterozygous at both loci Heterozygous at both loci SsYy Heterozygous at one locus, homozygous dominant at the other SY sY SY SSYY SsYY Ssyy Sy SSYy SsYy ssYY ssYy sY SsYY ssYY ssYy ssyy sy SsYy ssYy SY Sy sY sy SY SSYY SSYy SsYY SsYy Sy SSYy SSyy SsYy sY SsYY SsYy sy SsYy Ssyy 3:2 9:3:3:1 Ssyy SsYy Heterozygous at both loci Heterozygous/ Homozygous recessive Sy sy SY SSYy SsYy Sy SSyy Ssyy sY SsYy ssYy sy Ssyy ssyy SSyy ssYY = All SsYy SSYY ssyy = all SyYy Ssyy ssYy =1:1:1:1 4:3:1 http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 26 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F0 Phenotype: Rough, yellow Genotype: Punnet Grid: F1 Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 27 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F0 Phenotype: Rough, yellow ssYy Genotype: Punnet Grid: F1 gametes sY sy Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 28 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F0 Phenotype: Rough, yellow ssYy or ssYY Genotype: Punnet Grid: F1 gametes sY sy sY sY Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 29 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F0 Phenotype: Rough, yellow Punnet Grid: ssYy or ssYY ssyy Genotype: gametes sY sy sY sY All sy F1 Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 30 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F0 Phenotype: Rough, yellow Punnet Grid: F1 ssYy or ssYY ssyy Genotype: gametes sY sy sY sY All sy ssYy ssyy ssYy ssYy Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 31 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A rough yellow pea is test crossed to determine its genotype. F0 Phenotype: Rough, yellow Punnet Grid: F1 ssYy or ssYY ssyy Genotype: gametes sY sy sY sY All sy ssYy ssyy ssYy ssYy Phenotypes: Some green peas will be present in the offspring if the unknown parent genotype is ssYy. http://sciencevideos.wordpress.com No green peas will be present in the offspring if the unknown parent genotype is ssYY. 10.2 Dihybrid Crosses & Gene Linkage 32 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F0 Phenotype: Smooth, green Genotype: Punnet Grid: F1 Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 33 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F0 Phenotype: Smooth, green ssyy Genotype: Punnet Grid: gametes All sy F1 Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 34 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F0 Phenotype: Smooth, green Punnet Grid: F1 SSyy ssyy Genotype: gametes Sy Sy All sy Ssyy Ssyy Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 35 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F0 Phenotype: Smooth, green Punnet Grid: F1 SSyy or Ssyy ssyy Genotype: gametes Sy Sy Sy sy All sy Ssyy Ssyy Ssyy ssyy Phenotypes: http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 36 Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Key to alleles: Y = yellow y = green S = smooth s = rough A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F0 Phenotype: Smooth, green Punnet Grid: F1 SSyy or Ssyy ssyy Genotype: gametes Sy Sy Sy sy All sy Ssyy Ssyy Ssyy ssyy Phenotypes: No rough peas will be present in the offspring if the unknown parent genotype is SSyy. The presence of rough green peas in the offspring means that the unknown genotype must be Ssyy. The expected ratio in this cross is 3 smooth green : 1 rough green. This is not the same as the outcome. Remember that each reproduction event is chance and the sample size is very small. With a much larger sample size, the outcome would be closer to the expected ratio, simply due to probability. http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 37 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. Key to alleles: If they are present, they are either type A, B or both. i = no antigens present IA = type A anitgens present Where the genotype is heterozygous for IA and IB, both IB = type B antigens present are expressed. This is codominance. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 38 More about blood typing A Nobel breakthrough in medicine. Antibodies (immunoglobulins) are specific to antigens. The immune system recognises 'foreign' antigens and produces antibodies in response - so if you are given the wrong blood type your body might react fatally as the antibodies cause the blood to clot. Blood type O is known as the universal donor, as it has not antigens against which the recipient immune system can react. Type AB is the universal recipient, as it has no antibodies which will react to AB antigens. Blood typing game from Nobel.org: http://nobelprize.org/educational/medicine/landsteiner/readmore.html http://sciencevideos.wordpress.com Images and more information from: http://learn.genetics.utah.edu/content/begin/traits/blood/ 4.3 Theoretical Genetics 39 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 40 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 protection? No Yes Yes http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 41 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: F0 Phenotype: carrier Key to alleles: HbA = Normal Hb HbS = Sickle cell affected Genotype: Punnet Grid: F1 gametes Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com : Therefore 50% chance of a child with sickle cell disease. 4.3 Theoretical Genetics 42 Sickle Cell Key to alleles: HbA = Normal Hb HbS = Sickle cell Another example of codominance. Predict the phenotype ratio in this cross: F0 Phenotype: carrier Genotype: HbA Hbs Punnet Grid: F1 gametes affected HbS Hbs HbS HbS HbA HbAHbS HbAHbS HbS HbSHbS HbSHbS Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com HbAHbS & HbSHbS Carrier & Sickle cell 1:1 Therefore 50% chance of a child with sickle cell disease. 4.3 Theoretical Genetics 43 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: F0 Phenotype: carrier Key to alleles: HbA = Normal Hb HbS = Sickle cell carrier Genotype: Punnet Grid: F1 gametes Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 44 Sickle Cell Key to alleles: HbA = Normal Hb HbS = Sickle cell Another example of codominance. Predict the phenotype ratio in this cross: F0 Phenotype: carrier Genotype: HbA HbS Punnet Grid: F1 gametes carrier HbA HbS HbA HbS HbA HbAHbA HbAHbS HbS HbAHbS HbSHbS Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com HbAHb & 2 HbAHbS & HbSHbS Unaffected & Carrier & Sickle cell 1: 2 : 1 Therefore 25% chance of a child with sickle cell disease. 4.3 Theoretical Genetics 45 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: F0 Phenotype: carrier Genotype: HbA HbS Punnet Grid: Key to alleles: HbA = Normal Hb HbS = Sickle cell unknown gametes HbA HbS F1 Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 46 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: F0 Phenotype: carrier Genotype: HbA HbS Punnet Grid: gametes Key to alleles: HbA = Normal Hb HbS = Sickle cell unknown HbA HbA or HbA HbS HbA HbA HbA HbS HbA HbS F1 Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 47 Sickle Cell Another example of codominance. Predict the phenotype ratio in this cross: F0 Phenotype: carrier Genotype: HbA HbS Punnet Grid: F1 Genotypes: Key to alleles: HbA = Normal Hb HbS = Sickle cell unknown HbA HbA or HbA HbS gametes HbA HbA HbA HbA HbAHbA HbAHbA HbAHbA HbAHbS HbS HbAHbS HbAHbS HbSHbS HbAHbS HbS 3 HbAHbA & 4 HbAHbS & 1 HbSHbS Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 3 Unaffected & 4 Carrier & 1 Sickle cell 3:4:1 Therefore 12.5% chance of a child with sickle cell disease. 4.3 Theoretical Genetics 48 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 49 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. X Y X XX XY X XX XY gametes Therefore there is an even chance* of the offspring being male or female. http://sciencevideos.wordpress.com SRY gene determines maleness. Find out more about its role and just why do men have nipples? Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://www.hhmi.org/biointeractive/gender/lectures.html 4.3 Theoretical Genetics 50 Sex Determination Non-disjunction can lead to gender disorders. XYY Syndrome: XXY: Klinefelter Syndrome: Fertile males, with increased risk of learning difficulties. Some weak connections made to violent tendencies. Males with enhanced female characteristics XO: Turner Syndrome Monosomy of X, leads to short stature, female children. XXX Syndrome: Fertile females. Some X-carrying gametes can be produced. Interactive from HHMI Biointeractive: http://www.hhmi.org/biointeractive/gender/click.html http://sciencevideos.wordpress.com Image from NCBI: http://www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mga&part=A1179 4.3 Theoretical Genetics 51 Sex Linkage X and Y chromosomes are non-homologous. The sex chromosomes are non-homologous. There are many genes on the X-chromosome which are not present on the Y-chromosome. Non-homologous region Sex-linked traits are those which are carried on the X-chromosome in the non-homologous region. They are more common in males. Non-homologous region Examples of sex-linked genetic disorders: - haemophilia - colour blindness X and Y SEM from http://www.angleseybonesetters.co.uk/bones_DNA.html http://sciencevideos.wordpress.com Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 4.3 Theoretical Genetics 52 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 53 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 54 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 n X n X N X n X Affected female Carrier female XN Y Normal male n X Y Affected male no allele carried, none written Key to alleles: N = normal vision n = red/green colour Xq28 blindness Human females can be homozygous or heterozygous with respect to sex-linked genes. Chromosome images from Wikipedia: Heterozygous females are carriers. http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 55 Sex Linkage X and Y chromosomes are non-homologous. What chance of a colour-blind child in the cross between a normal male and a carrier mother? F0 Genotype: Phenotype: N X n X Carrier female X N X Y Key to alleles: N = normal vision n = red/green colour blindness Normal male Punnet Grid: F1 Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 56 Sex Linkage X and Y chromosomes are non-homologous. What chance of a colour-blind child in the cross between a normal male and a carrier mother? F0 Genotype: Phenotype: Punnet Grid: F1 N X n X Carrier female X N X Y Key to alleles: N = normal vision n = red/green colour blindness Normal male XN Y N XN N X X XN Y n X N X n X Xn Y Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 57 Sex Linkage X and Y chromosomes are non-homologous. What chance of a colour-blind child in the cross between a normal male and a carrier mother? F0 Genotype: Phenotype: Punnet Grid: F1 N X n X Carrier female X N X Y Key to alleles: N = normal vision n = red/green colour blindness Normal male XN Y N XN N X X XN Y Normal female n X N X n X Carrier female Normal male Xn Y Affected male What ratios would we expect in a cross between: a. a colour-blind male and a homozygous normal female? b. a normal male and a colour-blind female? http://sciencevideos.wordpress.com There is a 1 in 4 (25%) chance of an affected child. Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 4.3 Theoretical Genetics 58 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 to the Y chromosome, red-green colour blindness is known as a sexlinked disorder. The male has no allele on the Y chromosome to combat a recessive faulty allele on the X chromosome. Xq28 Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 59 http://illinoisreview.typepad.com/illinoisrevie w/2015/02/myths-and-facts-about-parcc-inillinois.html Colour Blind cartoon from: http://www.almeidacartoons.com/Med_toons1.html http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 60 Phenylketonuria (PKU) Clinical example. Pedigree charts can be used to trace family histories and deduce genotypes and risk in the case of inherited gene-related disorders. Here is a pedigree chart for this family history. key I female male affected Not Affected II A III B deceased ? Is PKU dominant or recessive? How do you know? • • http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 61 Phenylketonuria (PKU) Clinical example. Pedigree charts can be used to trace family histories and deduce genotypes and risk in the case of inherited gene-related disorders. Here is a pedigree chart for this family history. key I female male affected Not Affected II A III B deceased ? 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 62 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. The Boy with PKU ideo clip from: http://www.youtube.com/watch?v=KUJVujhHxPQ http://sciencevideos.wordpress.com Diagnosis- blood test taken at 6-7 days after birth http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/ 4.3 Theoretical Genetics 63 Phenylketonuria (PKU) Clinical example. A recessive 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. 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 64 Phenylketonuria (PKU) Clinical example. A recessive 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. 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 - 24 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 65 Phenylketonuria (PKU) Clinical example. What is the probability of two parents who are both carriers of the recessive allele producing children affected by PKU? F0 Phenotype: carrier Genotype: Tt Punnet Grid: gametes carrier Tt T Key to alleles: T = Normal enzyme t = faulty enzyme t T t F1 Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 66 Phenylketonuria (PKU) Clinical example. What is the probability of two parents who are both carriers of the recessive allele producing children affected by PKU? F0 F1 Phenotype: carrier Genotype: Tt Punnet Grid: gametes T t T TT Tt t Tt tt Genotypes: Phenotypes: Phenotype ratio: http://sciencevideos.wordpress.com TT carrier Tt Tt Tt Normal enzyme Key to alleles: T = Normal enzyme t = faulty enzyme tt PKU 3:1 Therefore 25% chance of a child with PKU 4.3 Theoretical Genetics 67 Pedigree Charts Pedigree charts can be used to trace family histories and deduce genotypes and risk in the case of inherited gene-related disorders. Here is a pedigree chart for this family history. Key to alleles: T= Has enzyme t = no enzyme 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 68 Pedigree Charts Pedigree charts can be used to trace family histories and deduce genotypes and risk in the case of inherited gene-related disorders. Here is a pedigree chart for this family history. Key to alleles: T= Has enzyme t = no enzyme Key: female male affected Not Affected deceased Looks like Deduce the genotypes of these individuals: A&B C D Genotype Both Tt tt Tt Reason Trait is recessive, as both are normal, yet have produced an affected child (C) Recessive traits only expressed when homozygous. To have produced affected child H, D must have inherited a recessive allele from either A or B http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 69 Pedigree Charts Key to alleles: T= Has enzyme t = no enzyme Individuals D and $ are planning to have another child. Calculate 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 70 Pedigree Charts Key to alleles: T= Has enzyme t = no enzyme Individuals D and $ are planning to have another child. Calculate the chances of the child having PKU. Key: female male affected $ Not Affected deceased Looks like Genotypes: D = Tt (carrier) $ = tt (affected) T t t Tt tt t Tt tt Gametes http://sciencevideos.wordpress.com Phenotype ratio 1 : 1 Normal : PKU Therefore 50% chance of a child with PKU 4.3 Theoretical Genetics 71 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 these factors not being produced. Therefore, the clotting response to injury does not work and the patient can bleed to death. XH XH Normal female Xh Xh Affected female XH Xh Carrier female XH Y Normal male Xh Y no allele carried, none written Key to alleles: XH = healthy clotting factors Xh = no clotting factor Affected male Human females can be homozygous or 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 72 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 73 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 74 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 75 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. 4. Britney Key: female male affected Key to alleles: H = healthy clotting factors h = no clotting factor Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com Not Affected deceased 4.3 Theoretical Genetics 76 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. 4. Britney Key: female male affected Key to alleles: H = healthy clotting factors h = no clotting factor Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com Not Affected deceased 4.3 Theoretical Genetics 77 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. 4. Britney Key: female male affected Key to alleles: H = healthy clotting factors h = no clotting factor Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com Not Affected deceased 4.3 Theoretical Genetics 78 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 4. Britney Key: female male affected Key to alleles: H = healthy clotting factors h = no clotting factor Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com Not Affected deceased 4.3 Theoretical Genetics 79 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 4. Britney Key to alleles: H = healthy clotting factors h = no clotting factor Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp http://sciencevideos.wordpress.com XH XH or XH Xh Key: female male affected Not Affected deceased 4.3 Theoretical Genetics 80 Pedigree Chart Practice Key: female male affected Not Affected deceased Dominant or Recessive? http://sciencevideos.wordpress.com Autosomal or Sex-linked? 4.3 Theoretical Genetics 81 Pedigree Chart Practice Key: female male affected Not Affected deceased Dominant or Recessive? Dominant. Autosomal or Sex-linked? A and B are both affected but have produced unaffected (D & F). Therefore A and B must have been carrying recessive healthy alleles. If it were recessive, it would need to be homozygous to be expressed in A & B – and then all offspring would be homozygous recessive. http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 82 Pedigree Chart Practice Key: female male affected Not Affected deceased Dominant or Recessive? Dominant. Autosomal or Sex-linked? Autosomal. A and B are both affected but have produced unaffected (D & F). Therefore A and B must have been carrying recessive healthy alleles. Male C can only pass on one X chromosome. If it were carried on X, daughter H would be affected by the dominant allele. If it were recessive, it would need to be homozygous to be expressed in A & B – and then all offspring would be homozygous recessive. Tip: Don’t get hung up on the number of http://sciencevideos.wordpress.com individuals with each phenotype – each reproductive event is a matter of chance. Instead focus on possible and impossible genotypes. Draw out the punnet grids if needed. 4.3 Theoretical Genetics 83 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% female Key: male affected Not Affected deceased http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 84 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% Key to alleles: XH = healthy clotting factors Xh = no clotting factor D. 50% female Key: male affected Not Affected deceased http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 85 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% Key to alleles: XH = healthy clotting factors Xh = no clotting factor D. 50% female Key: male What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y affected Not Affected XH deceased Y http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 86 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% Key to alleles: XH = healthy clotting factors Xh = no clotting factor D. 50% female Key: male affected Not Affected What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y There is an equal chance of F being XH XH or XH Xh So: XH XH XH Xh XH deceased Y http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 87 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% Key to alleles: XH = healthy clotting factors Xh = no clotting factor D. 50% female Key: male affected Not Affected What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y There is an equal chance of F being XH XH or XH Xh So: XH XH XH Xh XH XH XH XH XH XH XH X H Xh Y XH Y XH Y XH Y Xh Y deceased http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 88 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% Key to alleles: XH = healthy clotting factors Xh = no clotting factor D. 50% female Key: male affected Not Affected What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y There is an equal chance of F being XH XH or XH Xh So: XH XH XH Xh XH XH XH XH XH XH XH X H Xh Y XH Y XH Y XH Y Xh Y deceased So there is a 1 in 8 (12.5%) chance of the offspring being affected! http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 89 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 90 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 91