Mendelian Genetics

1. Mendelian Genetics

2. LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of traits. [See SP 6.5] LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted for by Mendelian genetics. [See SP 6.3] LO 3.17 The student is able to describe representations of an appropriate example of inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits. [See SP 1.2]

3. A brief review of Mendel (and flower parts!) Pisum sativum

4. Terms you must know! Haploid Dominant Monohybrid Diploid Recessive Dihybrid Homozygous Genotype Allele Heterozygous Phenotype Hybrid

6. Check out the ratios! You have to know how to get the expected ratios to do a chi square problem that deals with a genetic data.

7. LO 3.9 The student is able to construct an explanation, using visual representations or narratives, as to how DNA in chromosomes is transmitted to the next generation via mitosis, or meiosis followed by fertilization. [See SP 6.2]

9. Law of Independent Assortment- All of the parents were yellow and smooth- but those traits didn’t always stay together in the offspring.

10. Since we know Mendelian genetics, we don’t need the Punnett square to tell us that we get a 9:3:3:1 phenotypic ratio. There is a 1/16 chance that the offspring will be short and white.

11. We also need to be able to calculate percentages- For example in a dihybrid cross between two heterozygotes, If you have 360 offspring, what are your expected values? You know that you should have a 9:3:3:1 ratio Both dominant phenotypes 9/16 = .56 = 56% = 202 One dominant; one recessive 3/16 = .19 = 19% = 68 One recessive; one dominant 3/16 = .19 = 19% = 68 Both recessive phenotypes 1/16 = .06 = 6% = 22

12. Let’s plug this into the Hardy-Weinberg equation and see if we get the same numbers. Of the 180 flowers that are tall, how many are homozygous, and how many are heterozygous? 22% are short. What is the frequency of the recessive allele? 22% is .22 .22 = q2 .47 = q What is the frequency of the dominant allele? .53 = p .53 x .53 = .28 .28 x 360 = 101 homozygous tall 2 (.53)(.47) = .5 .5 x 360 = 179 heterozygous (tall) So, we calculated that 180 of these flowers would be tall; now we know how many of the tall flowers are homozygous and how many are heterozygous

13. In a species of gecko, spots and tail size are controlled by genes that assort independently. A= spotted B= fat tail a = solid b = skinny tail A scientist collected data from several gecko breeders and got the phenotypes of 135 baby geckos born within a 6 month period. Calculate the percentage of the geckos with the 4 possible phenotypes and the number of each in this data sample if all of the parents in the trial were heterozygotes. 56% 75 19% 26 19% 26 6% 8

14. In a species of gecko, spots and tail size are controlled by genes that assort independently. A= spotted B= fat tail a = solid b = skinny tail What if the cross wasn’t between two heterozygotes? What if this was the cross? What percentages would you expect in the offspring? aabb x AaBb 25% spotted/fat tail 25% solid/fat tail 25% spotted/skinny tail 25% solid/skinny tail

15. P= smooth seeds x wrinkled seeds F1= all smooth seeds F2= 5,474 smooth seeds and 1,850 wrinkled seeds Calculate chi-square for this data and evaluate your results. . Null hypothesis- There is no significant difference between the observed and expected ratio of smooth and wrinkled seeds.

16. P= smooth seeds x wrinkled seeds F1= all smooth seeds F2= 5,474 smooth seeds and 1,850 wrinkled seeds Calculate chi-square for this data and evaluate your results. . Null hypothesis- There is no significant difference between the observed and expected ratio of smooth and wrinkled seeds. I accept my null hypothesis. My chi square value of 0.263 is less than the critical value of 3.84.

17. You don’t need Punnett squares, you can determine expected ratios using probability

18. Consider a trihybrid cross- • If a plant that is heterozygous for all three characteristics is allowed to self-fertilize, what proportion of the offspring would you expect to be as follows- AaBbDd x AaBbDd • homozygous for the 3 dominant traits? • There is a 1/4 chance for each pair of alleles to be homozygous dominant. • 1/4 x 1/4 x 1/4 = 1/64 b. homozygous for purple and tall; heterozygous for seed shape

19. The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? a. AABBCCDD b. AaBBccDd c. AaBBCCdd ¼ x ¼ x ¼ x ¼ = 1/256 ½ x ¼ x ¼ x ½ = 1/64 ½ x ¼ x ¼ x ¼ = 1/128

20. Beyond Mendel… LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of traits. [See SP 6.5] LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted for by Mendelian genetics. [See SP 6.3]

21. Beyond Mendel Incomplete Dominance F1 hybrids have an appearance that is between that of the two parents. Breeding two of the hybrids produces a 1:2:1 ratio of phenotypes in the F2 Don’t just know the definition of incomplete dominance; recognize the ratio and know how it differs from a typical Mendelian cross. You may be given the results of a genetic cross and be asked to explain how you would get these results.

22. Beyond Mendel- Epistasis- a gene at one locus alters the effects of a gene at another locus B- gray b- tan C- pigment c- no pigment If the mouse doesn’t have a capital C, regardless of the gene for coat color, no pigment will be deposited in the fur WONKY RATIO

23. Two genes play a critical role in normal hearing; one is involved in the development of a normal cochlea (A). The other is responsible for a normal auditory nerve (B). Either of these genes can cause deafness if the genotype is homozygous recessive. A man and a woman are both deaf; one as a result of a defective cochlea and the other as a result of a defective auditory nerve. Assume that they are homozygous dominant for the other gene. If they have children together (barring mutation), what is the probability that the child will be deaf? 0 AAbb x aaBB  all AaBb

24. Beyond Mendel- Codominance & Multiple Alleles More than two alleles for a trait, and two of those alleles show dominance

25. Two parents, one type A and one type B, have a type O child? Explain that!!

26. In the U.S. about 16% of the population is Rh negative. The allele for Rh negative is recessive to the allele for Rh positive. If the student population of a high school in the U.S. is 2000, how many students would you expect for each of the three possible genotypes? p + q = 1 p2 + 2pq + q2 = 1 16% = .16 = q2 q = .4 p = .6 Homozygous Dominant (++) 36% 720 Heterozygous (+-) 48% 960 Homozygous recessive (--) 16% 320

27. Beyond Mendel Polygenic Inheritance Two or more genes have an additive effect on a single character in the phenotype In humans, what are some other characteristics that are polygenic?

28. Beyond Mendel- Pleiotropy One gene = Multiple effects

29. Beyond Mendel- Sex linked genes are located on the sex chromosomes (x and y). They were discovered by Thomas Morgan during his work with Drosophila. If a sex linked gene is on the X chromosome, a man who receives that X chromosome will have the trait. Since a woman has two X chromosomes, it may not be expressed in her. Some sex linked disorders in humans- Hemophilia Red/green colorblindness Duchennes Muscular Dystrophy

30. In fruit flies, the mutation for bar eyes is a sex linked dominant trait. A pure bar-eyed female is crossed with a wild (red-eyed) male. Which of the following statements gives the correct percentage of male offspring that will have bar eyes and the correct explanation for it? • 50%, because all the males will inherit the bar-eyed trait from the mother • 50%, because all the males will inherit the bar-eyed trait from the father • 100%, because all the males will inherit the bar-eyed trait from the mother • 100%, because all the males will inherit the bar-eyed trait from the father

31. Insert slide on how environment influences phenotype

32. X-inactivation Females have two X chromosomes; Males only have one

35. Autosomal Recessive Disorders Exactly what does this tell you about the disorder? If you are heterozygous, you are a carrier of the disorder. You don’t have the disorder, but if you have children with another carrier, they have a 25% chance of having the disorder.

36. Examples of autosomal recessive disorders Cystic fibrosis- primarily Caucasians, 1 in 2500 births. 1/25 is a carrier Abnormal protein is a membrane protein that transports chloride ions. Results in thick sticky mucus in lungs, digestive tract and pancreas No cure, but can be treated Tay Sachs Disease- Primarily Jews of European descent and Cajuns 1 in 3600 births Nonfunctional enzyme cannot breakdown lipids in brain cells Seizures, blindness, degeneration of motor and mental function Child dies before 5 years old

37. Sickle Cell Anemia Primarily African; 1 out of 400 African Americans Point mutation in hemoglobin When oxygen levels are low, hemoglobin crystallizes

38. Heterozygote Advantage

39. Pittsburgh Steelers safety  Ryan Clark will not play in Sunday’s playoff game against the Denver Broncos because of a blood disorder bolstered by high-altitude, low-oxygen conditions. Clark, 32, has sickle cell trait, which means he carries an abnormal version of the hemoglobin gene. Although he’s spared the severe symptoms of sickle cell disease (in which both versions of the gene — one from each parent — are abnormal), Clark could suffer life-threatening organ damage playing in Denver’s mile-high stadium. A 2007 game in Denver sent Clark into sickle cell crisis, a complication that cost him his spleen and gallbladder and ended his season.

40. Autosomal Dominant Disorders Huntington Disease Achondroplasia Neurofibromatosis Achondroplastic dwarves are heterozygous; the homozygous dominant condition is lethal. What is the probability that two achondroplastic dwarves will have a child of normal height?

41. It depends on who you ask!! If A = achondroplasia and a = no achondroplasia AA= lethal Aa= achondroplasia aa= no achondroplasia Aa x Aa a A 25% lethal 50% heterozygous (achondroplasia) 25% normal height A a A A A a a A a a 66% heterozygous (achondroplasia) 33% normal height

47. LO 2.28 The student is able to use representations or models to analyze quantitatively and qualitatively the effects of disruptions to dynamic homeostasis in biological systems. [See SP 1.4]

49. Dihybrid Cross with Linkage • In a normal case of Mendelian dihybrid inheritance with independent assortment of alleles, a cross between two heterozygotes produces the expected 9:3:3:1 ration in the offspring. In cases of dihybrid inheritance involving linkage, the offspring of a cross between two heterozygotes produces a greater than predicted number of parental types and a significantly smaller number of recombinants.

50. Linked Genes - genes that are on the same chromosome tend to be inherited together (duh)…. Unless they are separated by crossing over (recombination)