Theoretical Genetics

Theoretical Genetics Stephen Taylor 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

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

1. 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? 3. Now arrange them in a karyotype (don't turn them over - leave them as they were). 4. What is the gender of your baby? Explain how gender is inherited in humans. Making Babies • Crossing-over in prophase I • Random orientation in metaphase I and II • Random fertilisation Activity from: http://www.nclark.net/Genetics 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

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

“alleles of each gene separate into different gametes when the individual produces gametes” 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. 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). Segregation Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

Crossing a singletrait. F0 Genotype: Y y Y y 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). Monohybrid Cross Y or y Y or y Gametes: Punnet Grid: F1 Genotypes: Phenotypes: Phenotype ratio: Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

Crossing a singletrait. F0 Genotype: Y y Y y 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). Ratios are written in the simplest mathematical form. Monohybrid Cross Y or y Y or y Gametes: Punnet Grid: Yy yy Yy YY F1 Genotypes: Phenotypes: 3 : 1 Phenotype ratio: Mendel from: http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

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

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

What is the expected ratio of phenotypes in this monohybrid cross? F0 Key to alleles: Y = yellow y = green Phenotype: Genotype: yy Yy Monohybrid Cross Homozygous recessive Heterozygous Punnet Grid: yy yy Yy Yy F1 Genotypes: Phenotypes: 1 : 1 Phenotype ratio: 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

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

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

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

“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 (that's 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.” Career-related Case Study From the Times Higher Education Supplement – “So Last Century” http://www.timeshighereducation.co.uk/story.asp?sectioncode=26&storycode=415941&c=2 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

“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 (that's 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.” Career-related Case Study 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 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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. I Phenylketonuria (PKU) II A B III ? • Is PKU dominant or recessive? How do you know? 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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. I Phenylketonuria (PKU) II A B III ? • 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. 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

Clinical example. A mis-sensemutation 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. Phenylketonuria (PKU) Diagnosis- blood test taken at 6-7 days after birth http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/ The Boy with PKU ideo clip from: http://www.youtube.com/watch?v=KUJVujhHxPQ 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

Clinical example. A recessivemis-sensemutation 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? Phenylketonuria (PKU) 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/ 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

Clinical example. A recessivemis-sensemutation 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 Phenylketonuria (PKU) 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/ 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

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

Key to alleles: T= Has enzyme t= no enzyme 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. Pedigree Charts Looks like 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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. Pedigree Charts $ Looks like Genotypes: D = $ = Phenotype ratio Therefore 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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. Pedigree Charts $ Looks like Genotypes: D = Tt (carrier) $ = tt (affected) Phenotype ratio 1 : 1 Normal : PKU Therefore 50% chance of a child with PKU 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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. Where the genotype is heterozygous for IA and IB, both are expressed. This is codominance. Key to alleles: i = no antigens present IA= type A anitgens present IB = type B antigens present Codominance 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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. More about blood typing Blood typing game from Nobel.org: http://nobelprize.org/educational/medicine/landsteiner/readmore.html Images and more information from: http://learn.genetics.utah.edu/content/begin/traits/blood/ 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

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

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

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

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

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

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

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

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. Sex Determination 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 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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. Sex Determination Segregation of the sex chromosomes in meiosis. 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 Therefore there is an even chance* of the offspring being male or female. http://www.hhmi.org/biointeractive/gender/lectures.html 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

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. 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 Sex Linkage 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 Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

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

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

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. Sex Linkage XNXN XNY no allele carried, none written Normal female Normal male Key to alleles: N = normal vision n = red/green colour blindness Xq28 XnY XnXn Affected female Affected male XNXn Human females can be homozygous or heterozygous with respect to sex-linked genes. Heterozygous females are carriers. Carrier female Chromosome images from Wikipedia: http://en.wikipedia.org/wiki/Y_chromosome 4.3 Theoretical Genetics http://sciencevideos.wordpress.com

Theoretical Genetics