Chapter 15 - The Chromosomal Basis of Inheritance. A. Bergeron +AP Biology PCHS

Transcription

1 Chapter 15 - The Chromosomal Basis of Inheritance A. Bergeron +AP Biology PCHS

2 Do Now - Predicting Unpredictable Genotypes As an inexperienced (albeit precocious) gardener, I am always looking to maximize my yield of fruits and vegetables during the summer. Last summer, I identified two tomato plants that produced an inordinate amount of fruit. I contemplated breeding these two plants (either through cross-pollination or self-fertilization) but was unsure of either plant s genotype. Devise a strategy for determining the unknown genotype of both parental tomato plants using as few resources as possible. Remember, I want to maximize the yield of fruit from all of the future generations of tomato plants.

3 Do Now - Sex-linked Inheritance The table below shows the results of a cross of two red-eyed fruit flies. Some of the offspring have ruby -colored (deep pink) eyes instead of the usual red-colored eyes. Eye Color!! Numbers of Offspring!!! Males!! Females Red eyes!! 77!! 151 Ruby eyes!! 75!! 0 a. Using X R for the dominant allele and X r for the recessive allele, write the genotypes of the two parents. b. How do the results of the cross indicate that ruby eye color is sex-linked?

4 Testcross Used to determine the genotype of a parent with the dominant phenotype but an unknown genotype Dominant parent crossed/mated with a homozygous recessive parent Genotype of the dominant parent can be determined based on the offspring produced from the cross If offspring with ONLY the dominant phenotype are produced, the parent must have the homozygous dominant genotype If offspring with the dominant and the recessive phenotype are produced, the parent must have the heterozygous genotype

5 Sample Testcross

6 Reading Quiz 1. The most common phenotype in a population of organisms is called the. 2. genes have loci located on the same chromosome. 3. A recombination frequency greater than 50% suggests that two genes are (linked or unlinked)? 4. A father carries a recessive allele for hemophilia on his X chromosome. What is the probability that he will transmit the hemophilia allele to a son? 5. In females, one of the two X-chromosomes is inactivated early in embryonic development. What term is used to describe this inactivated chromosome?

7 Chromosomal Basis of Mendelian Inheritance

8 Do Now - Sex-Linked Genetics Practice Problems 1. A man with hemophilia (a recessive, sex-linked disease) has a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? That a son will be a hemophiliac? If the couple has four sons, what is the probability that all four sons will be born with hemophilia? 2. Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color blind. What is the probability that they will have a color-blind daughter? What is the probability that their first son will be color-blind?

9 Chromosomal Basis of Mendelian Inheritance Mendel s laws of segregation and independent assortment assumed that the factors (i.e. alleles) that are responsible for each of the observed traits were on different chromosomes! -Remember that Mendel did not know of the existence! of genes, chromosomes, or even DNA. Chromosomal Theory of Inheritance! -Mendelian factors (genes/alleles) are located on! chromosomes! -Chromosomes segregate and assort independently

10 Morgan s Discovery of Sex-linked Inheritance Parental Cross: White-eyed male (X w Y) x red-eyed (X R X R ) female fruit flies (Drosophila) F 1 generation (males and females) had red eyes F 2 generation had (expected) 3 (red):1(white) ratio of eye color phenotypes but only males had white eyes! What could explain this peculiar result? Answer: Sex-linked inheritance - the locus of some genes is located exclusively on one of the sex c somes

11 Parental Cross: White-eyed male (X w Y) x red-eyed (X R X R ) female fruit flies (Drosophila) F 2 generation had (expected) 3 (red):1(white) ratio of eye color phenotypes But only males had white eyes F 1 generation (males and females) had red eyes

12 How Did Morgan Know that the Gene for Red Eyes Was on a Sex Chromosome and Not an Autosome? 1. Since Morgan hypothesized that the allele for white eyes was recessive, a white-eyed female would need to inherit two (2) X chromosomes with the recessive allele! -This was not observed in the F 2 females 2. A white-eyed male would not have a wild-type (normal) allele to mask the presence of the recessive allele! -Since the white-eye allele was proposed to be on the! X chromosome, a single copy of the allele would! guarantee the white eye phenotype in male flies

13 Sex-Linked Inheritance (Three Possibilities...)

14 X-Chromosome Inactivation in Female Mammals Example - Female calico cats In females, one of the two X chromosomes is inactivated to prevent multiple copies of a gene from being expressed The inactivation is random. Either chromosome could be inactivated. Inactivated X c somes are called Barr bodies b/c the inactivated c some appears as a dark mass of DNA under the microscope

15 Important Genetics Vocabulary Parental = offspring that have the same phenotype as one or more of the parents Recombinant = offspring whose phenotypes are different from the parents! Recombinant gametes are often the products of! crossing over Evidence for Linked Genes in Drosophila Genes located on the same chromosome are considered to be linked Linked genes do NOT assort independently during meiosis. Variations away from expected results of a particular cross are indicative of a non-mendelian inheritance pattern

16 What would be the expected phenotypic ratio in the offspring if the genes for body color and wing type were unlinked (i.e. assorted independently)? (Assume that Morgan produced approximately 2,300 offspring in his cross)

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18 Mendel s Principle of Independent Assortment Mom Dad A B a b A b a B A a A a B b b B A B A B a b a b Sex cells A b A b a B a B

19 What would be the expected phenotypic ratio in the offspring if the genes for body color and wing type were linked (i.e. genes located on the same chromosome that do not assort independently)?

20 ???

21 Crossing Over

22 Calculating Recombination Frequency # Recombinants Total # offspring X 100% 1. A wild-type fruit fly (heterozygous for gray body color and normal wings) was mated with a black fly with vestigial wings. The offspring had the following phenotypic distribution:!! 778 wild-type!! 785 black, vestigial!! 158 black, normal!! 162 gray, vestigial What is the recombination frequency between these genes for body color and wing type?

23 Using Recombination Data to Calculate the Distance Between Genes on a Chromosome Crossover events are more likely to occur between genes that are further apart on a chromosome A recombination frequency that is greater than or equal to 50% indicates that genes are unlinked (i.e. on separate chromosomes) OR are so far apart from each other on the same chromosome that they behave as if they are on separate chromosomes (frequent cross overs)

24 Creating a Genetic Map Distance between genes on a chromosome is measured in centimorgans (cm) in honor of Thomas Hunt Morgan 1 centimorgan = 1 map unit = 1% recombination frequency One map unit is defined as the distance within which a crossover event is expected to occur in an average of 1% of gametes

25 Genetics Practice Problems 1. Determine the sequence of genes along a chromosome based on the following recombination frequencies:!! A-B: 8 cm!! A-C: 28 cm!! A-D: 25 cm!! B-C: 20 cm!! B-D: 33 cm 1 centimorgan (cm) = 1% recombination frequency 2. A fruit fly of genotype BR/br is test-crossed to br/br. (Note: Alleles grouped on either side of the slash are located on the same chromosome) In 84% of the meioses, there are no chiasmata between the linked genes; in 16% of the meioses, there is one chiasmata between the genes. What proportion of the progeny will be Br/br?

26 Why Shouldn t I Include the Parental Gametes Produced as a Result of the Crossover? Recall % Rec. = # Recombinants Total # offspring The formula for calculating % recombination does NOT take the parental gametes into account Thus, the probability of producing recombinant gametes is equivalent to the % recombination Don t forget that there is a 50% chance that either of the two gametes will fertilize an egg! X 100%

27 Genetics Practice Problems 3. In the plant Arabidopsis thaliana the loci for pod length (L = long, l = short) and fruit hairs (H = hairy, h = smooth) are located 20 cm apart on the same chromosome. The following crosses were made:!! L H/LH X lh/lh --> F 1!! Lh/Lh X lh/lh --> F 1 If the F1 s from above are crossed, what proportion of the progeny are expected to be lh/lh? Do Now 4. Using the same information from #3 on the previous slide, if the F1 s are crossed, what proportion of the progeny are expected to be Lh/lh?

28 Genetics Practice Problems 5. In Drosophila, the gene for white-eyes and the gene that produces hairy wings have both been mapped to the same chromosome and have a crossover frequency of 1.5%. A geneticist notices that in a particular stock of flies, these two genes assorted independently; that is, they behaved as though they were on separate chromosomes. What explanation can you offer for this observation?

29 Errors and Exceptions to Normal Chromosomal Inheritance Deletion Duplication Inversion Translocation

30 Genetics Practice Problems 6. In mice, black coat is dominant to brown and intense pigment is dominant to diluted. One mouse with dilute black pigment is crossed to another with intense brown pigment. The progeny are: a. What are the genotypes of parents and progeny? b. Draw the alleles involved on their respective chromosomes.

31 Genetics Practice Problems 7. The following testcross of a plant is made: If the two loci are 15 m.u. apart, what proportion of progeny will be (a) AB/ab? (b) ab/ab? (c) Ab/ab? 8. The A locus and the D locus are so tightly linked that no recombination is ever observed between them. If Ad/Ad is crossed to ad/ad, and the F 1 is intercrossed, what genotypes will be seen in the F 2 and in what proportions?

32 Do Now - Genetics Practice Problems A female animal with genotype A/a.B/b is crossed with a double-recessive male a/a.b/b. Their progeny include 442 A/a.B/b, 458 a/a.b/b, 46 A/a.b/b, and 54 a/a.b/b. Explain these proportions and draw the chromosomes of the dihybrid parent showing the positions of the genes and alleles.

33 Three-point Testcross How can we find the order and relative map distance between genes on a chromosome? Assume you have the following alleles at three different loci: v (vermillion eyes) vs. v + (red eyes) cv (absence of crossvein on wing) vs. cv + (presence of crossvein) ct (snipped wing edges) vs. ct + (smooth wing edges) Red (left) vs. Vermillion (right)

34 Three-point Testcross Available parental stocks: v + /v +. cv/cv. ct/ct. v/v. cv + /cv +. ct + /ct + What is the F1? Testcross F1 females to male homozygotes

35 Three-point Testcross Progeny Phenotype # Flies v cv + ct v+ cv ct 592 v cv ct+ 45 v+ cv+ ct 40 v cv ct 89 v+ cv+ ct+ 94 v cv+ ct 3 v+ cv ct What is the map distance between all three genetic loci? 2. Draw the gene map for these three genes.

36 Genetics Practice Problems 9. In corn, a triple heterozygote was obtained carrying the mutant alleles s (shrunken), w (white aleurone), and y (waxy endosperm), all paired with their normal wild-type alleles. This triple heterozygote was testcrossed, and the progeny contained 116 shrunken, white; 4 fully wild type; 2538 shrunken; 601 shrunken, waxy; 626 white; 2708 white, waxy; 2 shrunken, white, waxy; and 113 waxy. a. Determine if any of these three loci are linked, and, if so, show map distances. BONUS: Show the allele arrangement on the chromosomes of the triple heterozygote used in the testcross.

37 Genetics Practice Problems

38 Genetics Practice Problems 10. If A/A.b/b is crossed to a/a.b/b, and the F 1 is testcrossed, what percent of the testcross progeny will be a/a.b/b if the two genes are (a) unlinked, (b) completely linked (no crossingover at all), (c) 12 map units apart, (d) 24 map units apart? Genes of unknown linkage are shown separated by a dot, A/a.B/b

39 Genetics Practice Problems

40 Genetics Practice Problems 11. You have a Drosophila line that is homozygous for autosomal recessive alleles a, b, and c, linked in that order. You cross females of this line with males homozygous for the corresponding wild-type alleles. You then cross the F 1 heterozygous males with their heterozygous sisters. You obtain the following F 2 phenotypes (where letters denote recessive phenotypes and pluses denote wild-type phenotypes): , 365 a b c, 87 a b +, c, 47 a + +, 44 + b c, 5 a + c, and 4 + b +. (a) What is the map distance between genes a and b? b and c? a and c?

41 Genetics Practice Problems

42 Genetics Quiz - Chapter A strain of Neurospora (common bread mold) with the genotype H.I is crossed with a strain with the genotype h.i. Half the progeny are H.I and half are h.i. Briefly explain how this is possible.

43 Genetics Practice Problems 12. The following pedigree shows a family with two rare abnormal phenotypes: blue sclerotic (a brittle bone defect), represented by a black-bordered symbol, and hemophilia, represented by a black center in a symbol. Individuals represented by completely black symbols have both disorders. a. What pattern of inheritance is shown by each condition in this pedigree?

44 Genetics Practice Problems Blue sclerotic: Autosomal dominant Hemophilia: Sex (X)-linked recessive

45 Pedigree Analysis - Important Symbols

46 Important Points to Consider When Performing a Pedigree Analysis If a trait is recessive:! Individuals with the trait must be homozygous! Unaffected parents of children with the trait will be! heterozygous! Children may exhibit the trait even if their parents do not If a trait is dominant:! Any child with the trait will have at least one parent! with the trait! A trait caused by a dominant inheritance pattern! will show up in every generation

47 Inheritance Patterns & Phenotypic Characteristics Simple (Autosomal) Recessive -Males and females are equally likely to have the trait -Either parent can transmit the trait to offspring of either gender *The trait often skips generations -Often, both parents of offspring who have the trait are heterozygous; they carry at least one copy of the allele -Only homozygous recessive individuals have the trait -The trait may appear in siblings w/o appearing in parents -If one parent has the trait, those offspring who do not have it are carriers of the trait

48 Pedigree Analysis 1 Autosomal Recessive 2 3 Alkaptonuria - Is the disease caused by a dominant or recessive allele? What are the possible genotypes of individuals 1, 2, and 3?

49 Pedigree Analysis 1 Autosomal Dominant 2 Familial Hypercholesterolemia - Is disease caused by a dominant or recessive allele? What are the possible genotypes of individuals 1 and 2?

50 Do Now - Pedigree Analysis Determine the mode of inheritance of the trait diagrammed in the above pedigree. Hint: If one than one mode of inheritance is possible, determine which is most probable.

51 Pedigree Analysis Autosomal Recessive Sickle cell disease - Is disease caused by a dominant or recessive allele? What are the possible genotypes of individuals 1, 2, and 3? Hint: If one than one mode of inheritance is possible, determine which is most probable.

52 Pedigree Analysis Determine the mode of inheritance of the trait diagrammed in the above pedigree. Hint: If one than one mode of inheritance is possible, determine which is most probable.

53 Pedigree Analysis Hint: If one than one mode of inheritance is possible, determine which is most probable.

54 Inheritance Patterns & Phenotypic Characteristics X-linked Recessive -All daughters of a male who has the trait are carriers -There is no male-to-male transmission -The trait is far more common in males than in females -The son of a female carrier has a 50% chance of having the trait -Mothers of males who have the trait are either carriers or homozygous and have the trait -The daughters of a female carrier have a 50% of being a carrier

55 Inheritance Patterns & Phenotypic Characteristics X-linked Dominant -All daughters of a male who has the trait will also have the trait -There is no male-to-male transmission -A female who has the trait may or not pass the gene for that trait to her son or daughter

56 Diseases Caused by Recessive Genotypes 1. Cystic fibrosis 2. Tay-Sachs disease 3. Sickle-cell disease Diseases Caused by Dominant Genotypes 1. Achondroplasia 2. Huntington s disease *Be familiar with the cause of the disease (dominant or recessive allele/s) and disease symptoms*

57 Fly Lab Instructions 1. I will give each group 2 petri dishes with a different fly phenotype. Observe each phenotype under the microscope. Be sure to observe each phenotype! 2. Practice separating male flies from female flies. You can use any phenotype to practice. 3. Set up labeled culture vials with Drosophila medium.! Add 1 scoop of tap water to 1 scoop of food flakes! Sprinkle a few grains of yeast on top of the culture.! Wait 4-5 minutes before adding flies 4. Add 4-5 mating pairs of F 1 flies from each cross to your labeled culture vials. The flies should be asleep! They will wake up later! I promise

58 1) After reviewing the class data (which I will provide for you), you should be able to write a hypothesis that explains the mode of inheritance acting in each cross. Example - Monohybrid/dihybrid (autosomal) cross! Monohybrid/dihybrid (sex-linked) cross 2) You should assume that chance has been responsible for the independent assortment of alleles that produced the F1 and F2 generations of flies. ANY variation of your observed results when compared to the expected results is due to CHANCE 3) In this experiment, variation due to chance alone is called the null hypothesis 4) Can the class data be used to support the null hypothesis?

59 Chi-Square (Χ 2 ) Statistical Analysis Χ 2 is used to determine how well observed ratios fit expected ratios Chi-square formula Χ 2 = Σ (o-e) 2! !! e You will have observed values from the class data set The expected values will come from your expected phenotype ratios! For example, if you predict that the offspring should exist in a 3:1! ratio and there are 400 offspring, 300 offspring should have one! phenotype and 100 offspring should have the other phenotype The experimental Χ 2 value is compared to actual Χ 2 values from a statistical table to determine significance

60 Critical Values of the Chi-Square Distribution Unless told otherwise, use Χ 2 values for the 5% probability value Degrees of Freedom Prob. due to chance 0.05 (5%) (1%) (0.1%) Degrees of freedom = (# phenotypes - 1)

61 Practice Problem A true-breeding corn plant with purple corn kernels is crossed with a corn plant with yellow corn kernels. An F1 offspring is selfed to produce the F2 generation. The results of the F1 cross are shown in the table below: Are differences between the observed and expected values due to chance alone or is there an underlying cause of the variation? Perform a Χ 2 analysis to support your answer. Phenotype # Observed (o) # Expected (e) (o-e) (o-e) 2 (o-e) 2 /2 Yellow 149 kernels Purple kernels 388 Total 537

62 Chi-Square Analysis Phenotype # Observed (o) # Expected (e) (o-e) (o-e) 2 (o-e) 2 /2 Yellow kernels Purple kernels Total 537 Chi-square formula: Χ 2 = Σ (o-e) 2! !! e Null hypothesis: There is no significant difference between the expected and observed results. The difference/s are due to chance.

63 Chi-Square Analysis Phenotype # Observed (o) # Expected (e) (o-e) (o-e) 2 (o-e) 2 /e Yellow kernels Purple kernels Total Chi-square formula: Χ 2 = Σ (o-e) 2! !! e Null hypothesis: There is no significant difference between the expected and observed results. The difference/s are due to chance.

64 Critical Values of the Chi-Square Distribution Degrees of freedom: (# phenotypes - 1) Degrees of Freedom Probability due to chance (5%) (1%) (0.1%)

65 Making Conclusions Using Chi-Square If the calculated X 2 value is less than the critical value for a given probability of variation due to chance, you must accept your null hypothesis.! -There is no significant difference between the! expected and observed values.! -Any difference between the expected and observed! values probably occurred due to chance. If the calculated X 2 value is greater than the critical value for a given probability of variation due to chance, you must reject your null hypothesis! -There is a significant difference between the expected! and observed values! -There is an underlying reason why a difference exists! between the expected and observed values

66 Practice Problem - Chi Square Analysis Suppose that a scientist crossed pink-flowered snapdragons and obtained 236 offspring. 66 of the offspring grew into plants with red flowers, 115 with pink flowers, and 55 with white flowers. a. What are the expected phenotypic ratios of the offspring in the F 1 generation? b. What are the expected number of offspring with each type of flower color? c. Are the differences between the observed and expected results significant enough to conclude that the results did NOT occur by chance alone?