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B io Factsheet www.curriculum-press.co.uk Number 183 Variations from expected Mendelian Monohybrid Ratios Gregor Mendel demonstrated 3 key phenotypic ratios when he carried out a series of experiments investigating monohybrid crosses in garden peas. All Biologists need to know these ratios and how to use them to solve problems. However there are a few important situations where the usual outcomes do not arise. Example 2 • • • • Exam Hint: You must know the 3 key ratios and how to draw a genetic diagram to demonstrate genotypic and phenotypic outcomes. One parent is heterozygous (Tall genotype) Half the gametes will be T (dominant) and half will be t (recessive) One parent is homozygous recessive (short genotype) This parent can only produce gametes with the recessive allele (t) Male Gametes T The three key ratios From the phenotypic ratio of the progeny, you can work out the genotypes of the parents • • • • • 1:0 – Either: at least one parent is homozygous dominant; or: both parents are homozygous recessive 1:1 – One parent is heterozygous and one is homozygous recessive 3:1 – Both parents are heterozygous • tt Short Example 3 • • One parent is homozygous dominant This parent can only produce gametes with the dominant allele (T) All progeny will have the dominant phenotype – the genotype of the other parent is irrelevant Both parents are heterozygous (Tall genotype) These parents both produce gametes with either the dominant allele (T) or the recessive allele (t) Female Gametes Male Gametes T t T TT Tall t Female Gametes T t T TT Tall Tt Tall T TT Tall Tt Tall tt Short Exam Hint: You may be asked how to find the genotype of an individual with the dominant phenotype. This is the Test Cross. If, when the individual is crossed with a homozygous recessive, all the offspring show the dominant phenotype then the individual is homozygous dominant (Ratio 1:0); if the ratio is 1:1 then the parent is heterozygous. Always use the same letter for a gene – usually the first letter of the dominant allele or first letter of the characteristic Use UPPER CASE for dominant alleles and lower case for recessive alleles E.g. Mendel’s first experiment was on the height of pea plants – start by demonstrating that you understand this convention: “let T represent Tall and t represent short” Male Gametes Tt Tall Phenotypic Ratio = 1 Tall:1 Short (1/2 Tall:1/2 Short) Example 1 • • Tt Tall t Genetic Diagrams • Female Gametes t t Tt Tall Tt Tall tt Short Phenotypic Ratio = 3 Tall:1 Short (3/4 Tall:1/4 Short) When will these expected ratios not occur? These ratios will always apply to monohybrid crosses except in one of the following situations: 1. A Mutation occurs 2. There are more than 2 alleles (Multiple Alleles), in this case two or more of the alleles are usually equally dominant (CoDominance) 3. The gene has its locus on the sex chromosomes (Sex Linkage) 4. One genotype results in failure of the progeny to survive to birth (Lethal Gene) Phenotypic Ratio = All tall = 1:0 1 Bio Factsheet 183 Variations from expected Mendelian Monohybrid Ratios www.curriculum-press.co.uk Mutation Sex Linkage This is an unpredictable change to the genome of the individual. It could be that just one of the alleles is changed by as little as one base in the DNA sequence. An example of a Gene Mutation such as this is Sickle Cell Anaemia. Alternatively, there may be a change to a chromosome so that a whole gene is missing or has multiple versions. This is a Chromosome Mutation. For Example Down’s Syndrome, where the person has THREE of chromosome number 21. Sex linked genes have a locus on the sex chromosomes. The much smaller Y chromosome has many loci missing. Males only have 1 allele for the gene – on their one X chromosome. This one allele automatically expresses itself in the phenotype so recessive sex linked alleles affect males far more than females. Two major sex linked genes are those for HAEMOPHILIA and RED-GREEN COLOUR BLINDNESS. In genetic diagrams, use X and Y to represent the sex chromosomes and show the gene as superscript. Multiple alleles and Co-Dominance Each individual only receives one allele from each parent and clearly only possesses two alleles for each gene (hence diploid). If the gene has more than two alleles then clearly no individual can have them all in their genotype. The gene is said to have Multiple Alleles if there are more than two. At the same time the presence of more than two alleles affects how the gene expresses itself. Usually two or more alleles are neither dominant nor recessive in relation to each other – they are said to be Co-Dominant. They may both be dominant over the other alleles or both recessive in relation to another allele. Example 6 Consider HAEMOPHILIA with a heterozygous mother (XHXh) and a healthy father (XHY) Female Gametes Male Gametes XH Xh XH Healthy Female Y Example 4 One parent is group A and is heterozygous (dominant A, recessive O) - IA Io The other parent is group B and is heterozygous - IB Io Male Gametes IA Female Gametes IB Io IA IB IA Io AB Io IB Io B Io Io O IA Female Gametes IB Io IA Io IA Io A IB IB Io B Haemophiliac Male Either the homozygous dominant or the homozygous recessive condition results in failure of the offspring to develop properly so they do not appear in the progeny, affecting the expected ratio. An example of this is in Manx cats. These cats have a dominant allele that causes them to have a much-shortened tail. However, if the dominant allele is inherited from both parents the condition is lethal. Hence all surviving short-tailed Manx cats are heterozygous, but the phenotypic ratio of “Manx” to “normal-tailed” cats is not the expected 3:1, but 2:1 (See example 3 above). One parent is group AB and is co-dominant (dominant A, dominant B) - IA IB The other parent is group O and is homozygous recessive - Io Io Male Gametes Xh Y Lethal Gene Example 5 • Healthy Female (carrier) Exam Hint: You must know how to use Superscript in genetic diagrams to demonstrate how multiple alleles and sex linkage affect inheritance. A Offspring could be any blood group, with an equal chance (1/4) of each occurring • XH Xh All daughters are healthy – though half are carriers. Half of the sons suffer from Haemophilia. Unfortunately, the allele controls production of “Factor VIII”, one of the factors in the blood clotting process. This means that the blood of Haemophiliacs will not clot, they suffer frequent bouts of serious internal bleeding (haemorrhages) and until recently did not survive beyond childhood. The inheritance of haemophilia can be observed in the British Royal Family. Queen Victoria was a carrier for the condition and passed the recessive allele on to several of her 9 children (probability of half of them!) and it continued through European Royalty from there, though not in the current British Royal Family. Try to find a pedigree diagram of Queen Victoria’s descendants and predict the genotype of each individual. When constructing genetic diagrams a few new “rules” apply: • There are 3 alleles (“multiple alleles”) so the usual method of denoting a gene using UPPER or lower case cannot apply. • Use a letter for the gene and superscript for the alleles e.g. IA • XH Y Healthy Male An example of this is the ABO blood-grouping gene in humans, where the alleles for antigen A and antigen B are co-dominant over the allele for no antigen (O). • XH XH Example 7 A Male Gametes M IB Io B Female Gametes M m MM Mm Manx Lethal m Offspring all have different blood groups from parents – ½ A and ½ B. A person with phenotype AB cannot pass this on to their children, they can only pass on one of the alleles. Mm mm Manx Phenotypic Ratio = 2 Manx:1 Normal length tail (2/3 Manx:1/3 Normal) 2 Normal Bio Factsheet 183 Variations from expected Mendelian Monohybrid Ratios www.curriculum-press.co.uk Practice Questions 1. A homozygous tall pea plant is crossed with a heterozygous tall plant. All the offspring are tall. Use a genetic diagram to explain this, with the letter T representing the gene. 2. Explain how you would find which of the offspring are homozygous and which are heterozygous. 3. Use a genetic diagram to explain the phenotypic ratio when a person with blood group AB has children with a group A person if a) the second parent is homozygous; b) is heterozygous 4. Use a genetic diagram to explain the phenotypic ratio when a colour-blind man has children with a woman who is a carrier for the condition. Answers 1. Male Gametes T T Female Gametes T t TT Tall Tt TT Tall Tt Tall Tall Phenotypic Ratio = All tall = 1:0; all progeny must receive a dominant allele from their homozygous parent. 2. Use a test cross. Breed the plants with a short (homozygous recessive) plant. Progeny from the homozygous tall plants will all be tall (ratio 1:0); half of those from the heterozygous parent will be tall and half will be short (ratio 1:1) 3. (a) Male Gametes Female Gametes IA IB IA IA IA A IA IA A IB IB IA AB IB IA B Phenotypic ratio = 1 A: 1 AB – half the children have group A and half have group AB (b) Male Gametes Female Gametes I0 IA IA IA IA A IA I0 A IB IB IA AB I B I0 B The heterozygous parent must have genotype IA Io, the phenotypic ratio = 2 A: 1 AB: 1 B (½ AB, ¼ A, ¼ B) 4. Let XH represent the healthy allele, Xh the colour-blind allele, and Y the male sex chromosome without the gene locus. (XC and Xc are not recommended as the letter C is the same in upper and lower case) Female Gametes Male Gametes XH Xh Xh XH Xh Healthy Female (carrier) Y XH Y Healthy Male Xh Xh Colour blind Female Xh Y Colour blind Male Phenotypic ratio = 1 healthy male: 1 colour blind male: 1 healthy female: 1 colour blind female (¼ healthy male: ¼ colour blind male: ¼ healthy female: ¼ colour blind female) Acknowledgements: This Factsheet was researched and written by AndyPeters. Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU. Bio Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136 3