Chapter 7: Extending Mendelian Genetics Notes – part one

1. Chapter 7: Extending Mendelian GeneticsNotes – part one

2. Section 1: Chromosomes & Phenotype The chromosomes on which genes are located can affect the expression of traits.

3. Autosomal Genes & Phenotype • Autosomal genes are found on autosomes • Chromosomes 1-22 in humans • Not on sex chromosomes • Phenotype is determined by two copies of an autosomal gene (one from each parent) • AA, Aa, aa • Mendel studied autosomal characteristics • Most traits in humans are determined by autosomal genes

4. Autosomal Disorders • Many genetic disorders are caused by recessive alleles on autosomes • GG: healthy • Gg: healthy but carrier for the disorder • gg: has disorder • Example : Tay Sachs disease – autosomal recessive brain disease in very young children

5. Autosomal Disorders • Very few genetic disorders are caused by dominant alleles on autosomes (some are however) • HH and Hh: has disorder • hh: healthy • Example – Achondroplastic dwarfism • DD – death of fetus; Dd – dwarf; dd – normal height

6. Autosomal Traits – sample problem • Human traits determined by one autosomal gene follow the same rules when solving genetics problems as Mendel’s monohybrid crosses • Tay Sachs is a recessive and autosomal human disease. A couple has a baby who has Tay Sachs disease. Neither parent knows of anybody in their family who had the disease. Write the genotypes of the parents and predict the percentage of their children who will have this disease using a Punnett square.

7. Parents are carriers – genotypes = Tt X Tt • Punnett square : • Odds of a child inheriting Tay Sachs disease =

8. Parents are carriers – genotypes = Tt X Tt • Punnett square : • T t T t • Odds of a child inheriting Tay Sachs disease =

9. Parents are carriers – genotypes = Tt X Tt • Punnett square : • T t T t • Odds of a child inheriting Tay Sachs disease =

10. Parents are carriers – genotypes = Tt X Tt • Punnett square : • T t T t • Odds of a child inheriting Tay Sachs disease = • ¼ or 25%

11. Sex-Linked Traits • Genes located on the sex chromosomes (X or Y) are called sex-linked genes • Genes on the Y chromosome are responsible for male characteristics only • X chromosome has many more genes that affect many traits • In males, all sex-linked traits are expressed • Males are XY • they only have one copy of each chromosome, so all alleles are expressed even if they’re recessive

12. Some disorders on the X chromosome (sex linked disorders) • Hemophilia • Muscular Dystrophy • Color blindness • SCID (bubble boy)

14. How to write genotypes for sex linked traits • To represent that the traits are on the X chromosomes, we write the dominant or recessive allele as a superscript on the X

15. Sex Linked Traits – sample problem • Color blindness is a recessive sex linked disorder. A color blind man and his wife have a son who is color blind. The wife is upset that her husband passed color blindness on to their son. Use a Punnett square to show her who she should be upset with. What percentage of their sons are predicted to be color blind?

16. – genotypes of parents (assume mom is normal) • Mom – XB XB • Dad - Xb Y • Punnett square :

17. – genotypes of parents (assume mom is normal) • Mom – XB XB • Dad - Xb Y • Punnett square : • XB XB • Xb • Y

18. – genotypes of parents (assume mom is normal) • Mom – XB XB • Dad - Xb Y • Punnett square : • XB XB • Xb • Y * A color blind dad does not pass his X on to any of his sons – just the Y - A color blind son inherited the color blindness allele from his MOM

19. – genotypes of parents – mom MUST be a carrier • Mom – XB Xb • Dad - Xb Y • Punnett square :

20. – genotypes of parents – mom MUST be a carrier • Mom – XB Xb • Dad - Xb Y • Punnett square : • XB Xb • Xb • Y

21. – genotypes of parents – mom MUST be a carrier • Mom – XB Xb • Dad - Xb Y • Punnett square : • XB Xb • Xb • Y * 50% of their sons will be color blind 50% of their daughters will be color blind (girls can be color blind but it’s very rare)

22. X Chromosome Inactivation • Since females have two X chromosomes, one of them is randomly turned off in each cell during embryonic development – leads to “patchiness” of phenotype for X linked traits

23. X Chromosome Inactivation • Because it is random, a female will have some groups of cells with one X turned off, and others with the other X turned off X chromosomes that are turned off are visible as Barr bodies (one way to determine if cells are from a male or a female)

24. Phenotype is affected by many different factors. Section 2: Complex Patterns of Heredity

25. Incomplete dominance - heterozygous genotype shows a blending of the two homozygous phenotypes • Ex. Four o’clock plants • AA = red flowers • Aa = pink flowers • aa = white flowers

26. How to solve problems that involve traits that have alleles that show incomplete dominance • Pay close attention to genotypes and phenotypes • Otherwise they are exactly like monohybrid autosomal crosses • Sample problem : what percentage of a cross between two heterozygous four o’clocks are expected to be pink? Red? White? Color in four o’clock flowers is determined by alleles that show incomplete dominance

27. Genotypes of parents= Aa X Aa • Punnett square :

28. Genotypes of parents= Aa X Aa • Punnett square : • A a A a

29. Genotypes of parents= Aa X Aa • Punnett square : • A a A a

30. Genotypes of parents= Aa X Aa • Punnett square : • A a A a • 25% - red AA • 50% - pink Aa • 25% - white aa

31. Codominance - both alleles are dominant, so both are fully and separately expressed • Ex. Human ABO blood type - has multiple alleles (IA, IB, and i), IA and IB are codominant; i is recessive to both genotypes phenotypes • IAIA or IAi type A blood • IBIB or IBi type B blood • IAIB type AB blood • ii type O blood

32. Type AB is the universal recipient • Type O is the universal donor • A or B antigens must already be present if person is to receive blood with either antigen in it or results can be fatal

33. How to solve problems that involve codominant traits • Pay close attention to genotypes and phenotypes • Otherwise they are exactly like monohybrid autosomal crosses • Sample problem : Two parents with Type A blood are in a car accident with their baby. At the hospital they tell the doctors that their baby is definitely Type A. The doctors give the baby Type A blood. The baby quickly gets very sick and needs to go to the ICU. What happened? Explain using a Punnett square.

34. Genotypes of parents= IAi X IAi • Punnett square :

35. Genotypes of parents= IAi X IAi • Punnett square : • IA i IA i

36. Genotypes of parents= IAi X IAi • Punnett square : • IA i IA i

37. Genotypes of parents= IAi X IAi • Punnett square : • IA i IA i • The baby is Type O – each parent is a heterozygote • this is why doctors always test before starting any blood transfusion. • 25% - IAIA Type A • 50% - IAi Type A • 25% - ii Type O

38. another example of codominance : Sickle Cell Anemia symptoms : weakness, dizzy spells, blood flow issues genetic cause : codominant allele genotypes/phenotypes : HAHA= normal RBC’s HAHS= some sickle RBC’s, not enough to make person sick; resistance to malaria HSHS= sickle cell sufferer

39. Distribution of Sickle Cell Anemia mirrors that of Malaria - people who are heterozygous for Sickle Cell Anemia are more resistant to malaria (heterozygote advantage)

40. Polygenic traits - traits produced by two or more genes, leading to a range of phenotypes • Ex. human height, human eye color

41. In a polygenic trait, one gene may be epistatic, meaning it can interfere with the expression of the other genes • Ex. albinism in mammals

42. Genes are often affected by the Environment • Phenotype is usually a mixture of genes & the environment • Ex. genes determine your potential height but nutrition and diet may limit your actual height • The gender of sea turtles depends on environmental temperatures when they are developing • Females are more prevalent when temperatures are warm • More males are produced in cool temperatures