Extending Mendelian Genetics 23 October, 2002 Text Chapter 14

1. Extending Mendelian Genetics23 October, 2002 Text Chapter 14

2. Incomplete Dominance Alleles of some genes show incomplete dominance. In these cases, the heterozygote has an intermediate phenotype - a mixture of the phenotypes conferred by its alleles. Is this an example of blending? Why or why not? No. Two pink individuals can mate to produce red, white offspring.

4. Codominance In a set of codominant alleles, the heterozygote does not show an intermediate phenotype. The phenotypes of both alleles are individually expressed.

6. What is a Dominant Allele? • Usually, dominant alleles are recipes for functional proteins. • Recessive alleles are altered recipes for non-functional proteins. Think about flower color in pea plants. Substrate (colorless) Product (purple) The P allele is a recipe for a functional enzyme. The p allele is a recipe for a non-functional enzyme. Purple is dominant because one copy of a functional recipe is enough. Enzyme P In the analogous situation in snapdragons, one copy is not enough, And an intermediate phenotype is seen. At the molecular level, both functional and non-functional proteins are present. This is more like codominance.

7. Epistasis In epistasis, the alleles of one gene alter the expression of alleles of another gene.

8. Polyfactorial Characters Most interesting characters are influenced not by one or two genes, but by dozens or hundreds. In this example, alleles at three loci control skin color. The distribution shown here is characteristic of quantitative characters. These characters vary continuously, rather than in a few discrete states. Environmental factors can also influence these polyfactorial characters

10. Pleiotropy Pleiotropy is the converse of the concept of polyfactorial characters. In this case, one allele causes many different phenotypes.

11. Pedigree Analysis Pedigree analysis is a useful tool for studying inheritance in families.

12. Mendelian Inheritance in Humans In humans, many disorders follow Mendelian patterns of inheritance. Cystic fibrosis, sickle cell disease, Tay-Sachs syndrome, and many others are inherited as recessive alleles. The recessive allele codes for a non-functional protein. Usually, one functional allele is enough, so heterozygotes are asymptomatic. Why are cystic fibrosis and sickle cell alleles so common? In both cases, heterozygotes are protected from other diseases - recessive CFTR alleles protect against typhoid infection, while HbS alleles protect against malaria. Lethal dominant alleles are (for obvious reasons) less common. Huntingtons disease is caused by a dominant allele whose effect on phenotype is not obvious until after age 40.

13. Problem 1 What is the mode of inheritance for this disorder? Fill in as many genotypes as possible. Use “A-” when you are not sure of the genotype of a dominant-phenotype individual.

14. A man with type O blood has a sister with type AB blood. What are the genotypes and phenotypes of their parents? Both parents must have i allele. Sister must have gotten IA from one parent. IB from the other. So parents must be IAi and IBi, type A and type B blood.

15. Four babies, of blood types A, B, AB, and O, were born in a hospital one night. Two of these babies were twins (non-identical twins, produced from two different fertilized eggs). The three sets of parents had the following blood types: P1) A and B P2) O and B <-- parents of the twins P3) O and AB Assuming that the twins were born to the parents with blood types O and B, which baby belonged to which parent? What were the genotypes of the parents and babies in each case? AB must get A from one parent, B from the other, so AB baby goes home with P1. O baby has to get I from both, so she is one of the twins. A baby has to get A from somewhere, so congratulations P3. This leaves B baby as the twin brother. Wasn’t that fun?!

16. A form of blindness in humans, retinitis pigmentosa, is caused by either a dominant allele R, or a recessive allele a. Hence, only rrA- persons are normal. A man with the abnormality whose parents were normal marries a woman of the genotype RrAa. Assume independent assortment. a. What proportion of their children would be expected to have the abnormality? Since the man’s parents were normal, his genotype must be rraa. All his gametes will be ra. The woman’s gamtetes will be 1/4 RA, 1/4 Ra, 1/4 rA. 1/4 ra Three ways to be blind: rraa, RrAa, Rraa, each 1/4 probability. Add to give 3/4 children blind.

17. b. What are the genotypes of the man's parents? Since the man’s parents are normal, they must be rrA-. But, since the man is blind, he must be rraa. This means that both of his parents must be rrAa. c. If this couple has 3 children, what is the probability that 2 are blind and 1 has normal vision? For any child, 3/4 chance of being blind. There are three ways to get 2 blind, 1 normal: BNB, BBN, NBB. The probability of each way is (3/4)(3/4)(1/4) = 9/64. Add to 27/64. d. If the parents have 3 children, what is the probability that the first 2 are blind and the youngest child has normal vision? This is just the middle way from part c. The answer is 9/64.

18. Summer squash can be found in three shapes: disk, spherical, and elongate. In one experiment, two squash plants with disk-shaped fruits were crossed. The first 160 seeds planted from this cross produced plants with fruit shapes as follows: 89 disk, 61 sphere, and 10 elongate. What is the mode of inheritance of fruit shape in summer squash? This is very close to a 9:6:1 ratio. This ratio could result from two genes in an epistatic relationship. In this case, the genotype of the parents would be AaBb. Then A-B- would be disks, any plant with either aa or bb would be spherical, and aabb plants would be elongate.