Chapter 16 ­ Genetics and Heredity Return to Contents Title: Chapter 16 (1 of 212) Section 16.1 ­ Genetics of Inheritance Return to Contents Title: 16.1 (2 of 212) Homework : Preparation Assignment Read pages 522­528, and take notes on these pages. Define what is meant by the following terms: Traits Heredity Genetics Gene Allele Variations Purebred True Breeding Title: Oct 16­9:43 AM (3 of 212) Homework Review Define what is meant by the following terms: Traits: Heredity: Title: Oct 16­9:43 AM (4 of 212) Homework Review Define what is meant by the following terms: Genetics: Gene: Allele: Title: Oct 16­9:43 AM (5 of 212) Title: Dec 9­1:15 PM (6 of 212) Homework Review Define what is meant by the following terms: Variations: Purebred: True Breeding: Title: Oct 16­9:43 AM (7 of 212) Monohybrid Cross Read Pages 529­530, and define the following terms: Timed Activity: P Generation F1 Generation F2 Generation Dominant Recessive Homozygous Heterozygous Title: Oct 16­9:43 AM (8 of 212) ��������������� Title: Dec 9­1:37 PM (9 of 212) Title: Dec 9­1:44 PM (10 of 212) Title: Dec 9­1:49 PM (11 of 212) Title: Dec 9­1:55 PM (12 of 212) Title: Dec 9­1:49 PM (13 of 212) Title: Dec 10­12:08 PM (14 of 212) Definitions Two important terms in genetics are genotype and phenotype. Genotype refers to the genetic makup of an organism. In our mice, for example, a white mouse will always have the genotype bb. A black mouse may have the genotype BB or BB. Phenotype refers to the physical appearance of a trait in an organism. For example, a mouse may be heterozygous for fur color, with the genotype Bb, but the phenotype will be that the mouse is black. Title: Oct 16­9:43 AM (15 of 212) Alleles Pea plants have several traits that are easy to identify. For example, the plants may be either tall or short. This characteristic is controlled by a single gene, which has two alleles. Tall is the dominant characteristic, so we use the notation: Title: Dec 3­1:31 PM (16 of 212) Tall allele : T Short allele: t Homozygous A true breeding tall plant will be homozygous tall. TT (both genes carry the dominant allele) A short plant will be homozygous short. tt (both genes carry the recessive allele) Title: Dec 3­1:31 PM (17 of 212) Heterozygous A heterozygous plant carries one of each allele. Tt Because the allele for tall is dominant, the tall characteristic is still expressed. Title: Dec 3­1:31 PM (18 of 212) Monohybrid Cross Describe what is meant by the Principle of Dominance and the Law of Segregation. Principle of Dominance: Law of Segregation: Title: Oct 16­9:43 AM (19 of 212) Principle of Dominance Title: Dec 3­1:42 PM (20 of 212) Law of Segregation T Title: Dec 9­12:13 PM (21 of 212) t Monohybrid Cross What are the possible results? T t T t X Title: Dec 3­1:42 PM (22 of 212) Punnett Squares A Punnett Square is a convenient way to organize the possible combinations of alleles in a crossing. T t T t X Title: Oct 16­9:43 AM (23 of 212) Monohybrid Cross Use a diagram to outline the following hypothetical crossing. One species of mouse may be either black or white, with fur color being controlled by a single gene. The allele for black fur is dominant to the allele for white fur. If a purebred black mouse is bred to a purebred white mouse, what will the physical color of the offspring be? What will their genetic makeup be? Use a diagram to explain your answer! Title: Oct 16­9:43 AM (24 of 212) All offspring of this cross will be heterozygous (genotype) and have black fur (phenotype) Title: Dec 3­1:55 PM (25 of 212) Monohybrid Cross Use a diagram to outline the following hypothetical crossing. If two F1 mice from the previous example are crossed, what ratio of physical characteristics and genetic makeup would you expect to find in the offspring? Use a Punnett square to explain your answer! Title: Oct 16­9:43 AM (26 of 212) Monohybrid Cross ­ Solution This cross will result in two phenotypes, with 75% being black and 25% being white. However, there are three possible genotypes: 25% homozygous black; 50% heterozygous (but still having black fur, the dominant trait); and 25% homozygous white Title: Oct 16­9:43 AM (27 of 212) Punnett Squares In the previous example using mice, the Punnet square allows us to easily see the possible results of an F 1 cross. Female gametes are generally placed across the top Male gametes are generally placed down the left side You need to know the genotype of both parents to fill in the possible gametes. B B Since the F1 mice are all heterozygous, they could each produce a B or a b b gamete (ie: a gamete carrying either the dominant or recessive allele) Title: Oct 16­9:43 AM (28 of 212) b Punnett Squares Fill in the square, and analyze the results. By filling in the possible combinations, you can see that: 1/4 offspring are BB, or homozygous dominant, and will have black fur 2/4, in other words 1/2, are Bb, or heterozygous , and will have black fur 1/4 offspring are bb, or homozygous recessive , and will have white fur Title: Oct 16­9:43 AM (29 of 212) Sample Problem Use Punnet Squares to solve the following problem: ��������������� In fruit flies, red eyes are dominant over white eyes. Show the F1 generation results of a cross between a purebreeding red­eyed fly and a purebreeding white­eyed fly. Describe the ratio of phenotypes and genotypes in this generation. Then, show the F2 generation results, and describe the ratio of phenotypes and genotypes in this generation. Give about 5 minutes to complete this question, then review the solution on the following pages. Title: Oct 16­9:43 AM (30 of 212) Sample Problem ­ Solution Because both parents are pure­breeding , they are both homozygous . As a result, each gamete from either parent will carry the same allele. Therefore, all offspring in the F 1 generation are heterozygous , and will display the dominant trait ...in this case, red eyes. Title: Oct 16­9:43 AM (31 of 212) Sample Problem ­ Solution 100% of the F1 offspring will display the red­eyed phenotype, and be heterozygous (genotype) Title: Oct 16­9:43 AM (32 of 212) Sample Problem ­ Solution When you cross two of the F 1 flies, both parents are heterozygous . In the F 2 generation, you get a distinct pattern of phenotypes, with 75% showing the dominant trait and 25% showing the recessive trait . However, when you look at the genotypes, you see that only 25% are homozygous dominant; 50% are heterozygous ; and 25% are homozygous recessive . Title: Oct 16­9:43 AM (33 of 212) Sample Problem ­ Solution Title: Oct 16­9:43 AM (34 of 212) TestCross What is the genotype of a white­eyed fruit fly? How do you know? Title: Oct 16­9:43 AM (35 of 212) TestCross What about a red­eyed fly? What are the possible genotypes? How would you determine the genotype of a red­eyed fly? Perform a test­cross. What phenotype would you use to cross the red­eyed fly with? Why? Title: Oct 16­9:43 AM (36 of 212) TestCross Outline the possible results of the testcross, and how you would interpret them. What results would indicate the red­eyed fly was homozygous? What results would indicate the red­eyed fly was homozygous? Title: Oct 16­9:43 AM (37 of 212) Homework Complete the Sample Problem activity on page 533 Title: Oct 16­9:43 AM (38 of 212) Title: Dec 10­3:07 PM (39 of 212) Title: Dec 10­3:08 PM (40 of 212) Title: Dec 10­12:34 PM (41 of 212) In Class Assignment and Homework Review Section 16.1, pages 526­534 Complete Questions #8, 10, 11, 13, and 15 on page 535 Bring two quarters to class on Thursday Title: Oct 16­9:43 AM (42 of 212) Title: Dec 10­12:46 PM (43 of 212) Title: Dec 7­1:31 PM (44 of 212) Title: Dec 7­1:33 PM (45 of 212) Title: Dec 7­1:24 PM (46 of 212) Title: Dec 7­1:25 PM (47 of 212) Title: Dec 7­1:26 PM (48 of 212) Title: Dec 7­1:17 PM (49 of 212) Title: Dec 7­1:20 PM (50 of 212) Title: Dec 7­1:15 PM (51 of 212) Title: Dec 7­12:51 PM (52 of 212) Title: Dec 7­1:00 PM (53 of 212) Title: Dec 7­1:06 PM (54 of 212) Title: Dec 7­1:12 PM (55 of 212) Determining Genotypes ­ Test Cross If you look at a fruit fly, can you determine the genotype for eye coloe by observing the phenotype? You can if the fly is white­eyed, because the white­eye allele is recessive...so the only way a fly could have white eyes would be if it was homozygous. rr Title: Oct 16­9:43 AM (56 of 212) Determining Genotypes ­ Test Cross But what if the fly has red eyes? There are two possible genotypes that produce red eyes: RR Title: Oct 16­9:43 AM (57 of 212) Rr Determining Genotypes ­ Test Cross In this case, the genotype can be determined by performing a test cross. This involves crossing the fly with a fly which is homozygous recessive for the trait being observed. By observing the phenotypes of the offspring, you can then determine the genotype of the unknown parent. Use Punnett squares to predict the percentages of phenotypes if a white­ eyed (homozygous recessive) female fly is crossed with a red eyed male. You will need to create one Punnett square to show the results if the male is homozygous, and one to show the expected results if the male is heterozygous. Title: Oct 16­9:43 AM (58 of 212) Determining Genotypes ­ Test Cross Title: Oct 16­9:43 AM (59 of 212) Sample Problem In cats, long hair is recessive to short hair. 1) A true­breedingshort­haired male is mated to a true­breeding long­haired female. What will their kittens look like? 2) If you performed a test cross on a short­haired female cat, what phenotype should the male cat have? Why? 3) If the testcross resulted in eight kittens, what ratios of phenotypes would you expect if the female was homozygous? Heterozygous? 4) What if the testcross resulted in seven short­haired kittens, and one long­haired kitten? How can you explain these results? Title: Oct 16­9:43 AM (60 of 212) Title: Dec 7­1:56 PM (61 of 212) Title: Dec 7­2:00 PM (62 of 212) Title: Dec 10­2:03 PM (63 of 212) Title: Dec 10­2:16 PM (64 of 212) Title: Dec 10­2:20 PM (65 of 212) Title: Dec 10­2:22 PM (66 of 212) Title: Dec 10­2:25 PM (67 of 212) Title: Dec 10­2:30 PM (68 of 212) Title: Dec 10­2:29 PM (69 of 212) Title: Dec 10­2:33 PM (70 of 212) Determining Genotypes ­ Test Cross Title: Oct 16­9:43 AM (71 of 212) Title: Dec 11­12:05 PM (72 of 212) Homework Any questions on any of the homework problems? Title: Oct 16­9:43 AM (73 of 212) Title: Dec 11­12:48 PM (74 of 212) Title: Dec 11­12:50 PM (75 of 212) Title: Dec 11­12:53 PM (76 of 212) Section 16.2 ­ Complex Inheritance Patterns Return to Contents Title: 16.2 (77 of 212) In pairs, complete the mini­lab on page 531. Write the answers to questions #1­3 in the lab. Any questions? Title: Dec 11­12:02 PM (78 of 212) ��������������� Title: Dec 11­1:01 PM (79 of 212) Title: Dec 11­1:17 PM (80 of 212) Complex Patterns of Inheritance Multiple Traits It is possible to look at more than one trait at a time. For example, in pea plants you could look at both color and the shape of the seed. Color is controlled by a single gene with two alleles, Y (yellow) and y (green) Shape is controlled by a single gene with two alleles, R (round) and r (wrinkled) What are the results of crossing a true­breeding yellow, round plant with a true­breeding green, wrinkled plant? Title: Oct 16­9:43 AM (81 of 212) Yellow Round Plant *This is the phenotype *What is the genotype? *What alleles will all gametes contain? Title: Dec 10­2:39 PM (82 of 212) Green Wrinkled Plant *This is the phenotype *What is the genotype? *What alleles will all gametes contain? Title: Dec 10­2:39 PM (83 of 212) Give the results of this cross: This is the generation. Phenotype: Genotype: Title: Dec 10­2:42 PM (84 of 212) Sample Problem: Chickens can have brown or white feathers, with brown color dominant. They can also have long or short tail feathers, with short feathers dominant. What are the phenotypes and genotypes of chickens produced by crossing a brown, short­tailed male with a white, long­tailed female? (Assume both parents are true­breeding for both traits.) Title: Dec 10­2:46 PM (85 of 212) Title: Dec 11­1:32 PM (86 of 212) Give the results of this cross: F1 Generation Phenotype: Genotype: Title: Dec 10­2:42 PM (87 of 212) Title: Dec 10­2:55 PM (88 of 212) What if homozygous white, long­tailed male was crossed with a homozygous brown, heterozygous short­tailed female? What gametes could be produced by these parents? Genotype Possible Gametes: Title: Dec 10­2:59 PM (89 of 212) Genotype Possible Gametes: Title: Dec 11­1:43 PM (90 of 212) Title: Dec 11­1:50 PM (91 of 212) Title: Dec 12­9:26 AM (92 of 212) Multiple Traits Example from yesterday's class: In pea plants color is controlled by a single gene with two alleles, Y (yellow) and y (green) Shape is controlled by a single gene with two alleles, R (round) and r (wrinkled) What are the results of crossing a true­breeding yellow, round plant with a true­breeding green, wrinkled plant? First, you have to determine the possible parent gametes. Title: Oct 16­9:43 AM (93 of 212) Multiple Traits True­breeding yellow, round plant Title: Oct 16­9:43 AM (94 of 212) True­breeding green, wrinkled plant YYRR yyrr YR yr Multiple Traits yr What are the genotypes of the F1 plants? 100% heterozygous for both traits What are their phenotypes? 100% yellow, round (the dominant trait is displayed) Title: Oct 16­9:43 AM (95 of 212) YR YyRr Multiple Traits What percentages of phenotypes and genotypes would you expect in the F 2 generation if you crossed two plants from the F1 generation? Again, the first step is to determine the possible alleles. What are the possible gametes that the F1 plants could form? YyRr F1 Genotype: Possible Gametes: YR Title: Oct 16­9:43 AM (96 of 212) Yr yR yr Title: Dec 12­2:21 PM (97 of 212) Law of Independent Assortment Inheritance of alleles for one trait does not affect the inheritance of alleles for another trait. In other words, alleles for different characteristics are passed on to the offspring independently of one another, so that the offspring may have combinations that are different from either of the parents. What process in cell division allows for this to happen? Can you think of a situation where it would be more likely that alleles would be kept together? Title: Oct 16­9:43 AM (98 of 212) B ­ black fur b ­ brown fur Law of Segregation R t r T R ­ rough coat r ­ smooth coat T ­ long tail t ­ short tail b B Title: Dec 9­12:13 PM (99 of 212) Multiple Traits Create a Punnett square to determine the phenotypes and genotypes of the F 2 generation in the pea plant example. Remember, the F1 plants were heterozygous for both traits, and there are four possible gametes formed from each parent. Title: Oct 16­9:43 AM (100 of 212) Sample Problem: Yesterday, we looked at this sample problem: Chickens can have brown or white feathers, with brown color dominant. They can also have long or short tail feathers, with short feathers dominant. When a brown, short­tailed male is crossed with a white, long­ tailed female, what are the phenotypes and genotypes of the F 1 offspring? (Assume both parents are true­breeding for both yesterday's solution... traits.) Title: Dec 10­2:46 PM (101 of 212) Title: Dec 12­2:38 PM (102 of 212) Sample Problem: BbSs F1 Genotype: F1 Phenotype: All offspring will be brown with short tail feathers What are the possible gametes that the F1 chickens could form? Possible Gametes: BS Title: Dec 10­2:46 PM (103 of 212) Bs bS bs Create a Punnett square to determine the phenotypes and genotypes of the F2 generation resulting from crossing two of the F1 chickens. Title: Dec 11­12:56 PM (104 of 212) Title: Dec 12­2:48 PM (105 of 212) Title: Dec 12­2:54 PM (106 of 212) Sample Problem: In guinea pigs, the allele for dark fur is dominant to the allele for white fur. The allele for rough coats is dominant to the allele for smooth coats. Find the genotype and phenotype ratios of the offspring resulting from a cross of a dark, rough coated male with a white, smooth coated female.* *(Both of these parent guinea pigs are true­breeding for both traits) Title: Dec 11­11:52 AM (107 of 212) First, determine the genotypes of both parents: Male Female Possible Gametes: Male Title: Dec 11­11:52 AM (108 of 212) Female Give the results of this cross: F1 Generation Phenotype: Genotype: Title: Dec 10­2:42 PM (109 of 212) Determine the results of a cross between two F1 individuals. First, determine the genotypes of both F1 parents: Male and Female Possible Gametes: Male and Female Title: Dec 11­11:52 AM (110 of 212) Create a Punnett square to determine the phenotypes and genotypes of the F2 generation resulting from crossing two of the F1 guinea pigs. Title: Dec 11­12:56 PM (111 of 212) Homework: 1. A dihybrid cross is made of pea plants. Purple flowers (P) are dominant and white flowers (p) are recessive. Tall plants (T) are dominant and and short plants (t) are recessive. Both parents are heterozygous for both traits. Prepare a Punnett square to determine the phenotypic ratios of the offspring. What Mendelian law does this ratio demonstrate? Solution: What is the genotype of both parents? PpTt What possible gamets are formed? PT Title: Dec 11­3:12 PM (112 of 212) Pt pT pt Homework 1) Solution Title: Oct 16­9:43 AM (113 of 212) Homework Complete the homework assignment sheet. Use Punnett squares where appropriate. Be sure to answer all questions. Title: Oct 16­9:43 AM (114 of 212) Title: Dec 15­12:15 PM (115 of 212) Title: Dec 16­2:14 PM (116 of 212) Test Cross with Multiple Traits A test cross when looking at multiple traits is the same as in a single trait. The only difference is, you will cross your unknown genotype with an individual that is homozygous recessive for both of the characteristics you are investigating. How can you tell an individual has a homozygous recessive genotype? Title: Oct 16­9:43 AM (117 of 212) Test Cross with Multiple Traits What are the possible genotypes of a yellow, round pea plant? (Remember, yellow peas are dominant to green peas, and round peas are dominant over wrinkled peas.) *Hint: there are four possibilities! What would the results of each testcross be? (Since there are four See results... possibilities, there will be four possible testcross results.) Title: Oct 16­9:43 AM (118 of 212) Test Cross with Multiple Traits Testcross of round, yellow pea plant. What are the results if the genotype is YYRR? Title: Oct 16­9:43 AM (119 of 212) Test Cross with Multiple Traits Testcross of round, yellow pea plant. What are the results if the genotype is YYRr? Title: Oct 16­9:43 AM (120 of 212) Test Cross with Multiple Traits Testcross of round, yellow pea plant. What are the results if the genotype is YyRR? Title: Oct 16­9:43 AM (121 of 212) Test Cross with Multiple Traits Testcross of round, yellow pea plant. What are the results if the genotype is YyRy? Title: Oct 16­9:43 AM (122 of 212) Multiple Traits Sample Problem In cats, black color is dominant to a special gene which produces cream­colored cats with dark legs, faces and tails (Siamese cats, in case you don’t recognize it). Length of fur is controlled by a single gene with two alleles; long hair is dominant to short hair. What are the F 1 expected results of a cross between a true­breeding long­haired black cat and a true­breeding short­haired Siamese colored cat? Title: Oct 16­9:43 AM (123 of 212) Multiple Traits Sample Problem If two F 1 cats are crossed, what genotype and phenotype ratios would you expect in the F 2 offspring? bl bL Bl BL BL BBLL BBLl BbLL BbLl Bl BBLl BBll BbLl Bbll bL BbLL BbLl bbLL bbLl bl BbLl Bbll bbLl bbll Black, Long: 9 Black, Short: 3 Siamese, Long: 3 Siamese, Short: 1 Title: Oct 16­9:43 AM (124 of 212) Multiple Traits Sample Problem How could you determine the phenotype of a long­haired black cat? Show all possible combinations, and the expected results? Title: Oct 16­9:43 AM (125 of 212) Title: Dec 16­2:42 PM (126 of 212) Multiple Traits Sample Problem Solution: Title: Oct 16­9:43 AM (127 of 212) Incomplete Dominance Not all traits are completely dominant or recessive. In some cases, neither allele is dominant...in these cases, a heterozygous individual will display a blending of the two traits. This is called incomplete dominance. Title: Oct 16­9:43 AM (128 of 212) Incomplete Dominance For example, some flowers have petal color controlled by a single gene with two alleles: one coding for red color, the other for white. If the individual is homozygous for either allele, this is the petal color that will be displayed. Phenotype: Phenotype: Genotype: Genotype: Title: Oct 16­9:43 AM (129 of 212) Incomplete Dominance However, if the individual is heterozygous, neither of these traits is displayed...instead, the flower shows a blending of red and white, resulting in pink petals. Phenotype: Genotype: Title: Oct 16­9:43 AM (130 of 212) Codominance Codominance is similar to incomplete dominance, except both traits are dominant. In this case, homozygous individuals display one trait, while heterozygous individuals display both traits, not as a blending of the two, but showing both characteristics partially. Title: Oct 16­9:43 AM (131 of 212) Codominance For example, color in one type of cow is controlled by a single gene with two alleles, one coding for brown fur and one for white fur. Homozygous cows are either brown or white. Phenotype: Phenotype: Genotype: Genotype: Title: Oct 16­9:43 AM (132 of 212) Codominance However, heterozygous cows will display brown and white patches, rather than being light brown all over. Phenotype: Genotype: Title: Oct 16­9:43 AM (133 of 212) Multiple Alleles Not all traits are controlled by genes with only two alleles; some have multiple alleles. In humans, for example, the A, B and O blood types are determined by three alleles, two of which are co­dominant and one of which is recessive. The IA allele results in blood type A; the IB allele results in blood type B; and the i allele, being recessive, results in blood type O. Remember that the blood types are determined by antigens on the surface of blood cells. These antigens are proteins, so the gene codes for the production of these antigens...in type O, the gene does not code for production of an antigen. Title: Oct 16­9:43 AM (134 of 212) Multiple Alleles List all possible genotypes which could result in the following phenotypes (blood types). Blood Type Title: Oct 16­9:43 AM (135 of 212) Genotype Homework Review Section 16.2, and complete Questions #2­4 on page 544 Title: Oct 16­9:43 AM (136 of 212) Homework Page 544 #1­2 1) A dihybrid cross is made of pea plants. Purple flowers (P) are dominant and white flowers are recessive. Tall plants (T) are dominant and short plants are recessive. Both parents are heterozygous for both traits. Prepare a Punnett square to determine the phenotype ratios of the offspring. What Mendelian law does this ratio demonstrate? Title: Oct 16­9:43 AM (137 of 212) Homework 1) Solution Title: Oct 16­9:43 AM (138 of 212) Homework 2) A rose­combed rooster is mated with two rose­combed hens. Hen A produces 14 chicks, all rose­combed. Hen B produces 9 chicks, 7 of which are rose­combed and 2 single­combed. What are the likely genotypes of the parent birds? Rose comb Title: Oct 16­9:43 AM (139 of 212) Single comb Homework 2) Solution Title: Oct 16­9:43 AM (140 of 212) Homework 2) Solution Title: Oct 16­9:43 AM (141 of 212) Homework 2) Solution Title: Oct 16­9:43 AM (142 of 212) Sample Problem You know that one type of flower has petal color determined by a single gene with two alleles, red or white, and they display incomplete domininance. If a red flower is crossed with a white flower, what are the genotypes and phenotypes of the F 1 generation offspring? What are the expected genotypes and phenotypes of the F 2 offspring if two pink flowers are cross pollinated? Title: Oct 16­9:43 AM (143 of 212) Sample Problem Solution Title: Oct 16­9:43 AM (144 of 212) Sample Problem Solution Title: Oct 16­9:43 AM (145 of 212) Assignment: Inheritance of Coat Color in Rabbits Complete theThinking Lab on page 543. Be sure to answer all questions completely, including the Punnett squares. white chinchilla Darkgrey Himalayan Title: Oct 16­9:43 AM (146 of 212) Assignment: Inheritance of Coat Color in Rabbits 1) List all the possible genotypes for a: (a) Dark grey rabbit (b) Chinchilla rabbit (c) Himalayan rabbit (d) White rabbit Title: Oct 16­9:43 AM (147 of 212) Assignment: Inheritance of Coat Color in Rabbits 2) Predict the phenotype of a rabbit with the following genotypes. Explain your answers. (a) c h cch Title: Oct 16­9:43 AM (148 of 212) (b) C c h Assignment: Inheritance of Coat Color in Rabbits 3) Would it be possible to obtain white rabbits if one parent is white and the other is chinchilla? Explain. Title: Oct 16­9:43 AM (149 of 212) Title: Dec 13­1:01 PM (150 of 212) Assignment: Inheritance of Coat Color in Rabbits 4) Would it be possible to obtain chinchilla rabbits if one parent is Himalayan and the other is white? Explain. Title: Oct 16­9:43 AM (151 of 212) Title: Dec 16­3:13 PM (152 of 212) Assignment: Inheritance of Coat Color in Rabbits 5) A chinchilla rabbit is mated with a Himalayan. Some of the offspring are white. What are the parents' genotypes? Title: Oct 16­9:43 AM (153 of 212) Title: Dec 16­3:16 PM (154 of 212) Assignment: Inheritance of Coat Color in Rabbits 5) A chinchilla rabbit is mated with a Himalayan. Some of the offspring are white. What are the parents' genotypes? Title: Oct 16­9:43 AM (155 of 212) Assignment: Inheritance of Coat Color in Rabbits Extension Question: A male chinchilla rabbit and a female dark grey rabbit are crossed. The resulting litter contains 11 kits; 6 are dark grey, and 2 are chinchilla. However, 3 of the kits are Himalayan! Show how this is possible. Is there any way for this cross to produce white rabbits? Title: Oct 16­9:43 AM (156 of 212) Assignment: Inheritance of Coat Color in Rabbits Title: Oct 16­9:43 AM (157 of 212) Title: Dec 13­1:19 PM (158 of 212) Title: Dec 13­1:22 PM (159 of 212) Title: Dec 13­1:20 PM (160 of 212) Assignment: Inheritance of Coat Color in Rabbits Title: Oct 16­9:43 AM (161 of 212) Title: Dec 17­12:11 PM (162 of 212) Title: Dec 17­12:57 PM (163 of 212) Title: Dec 17­1:01 PM (164 of 212) Title: Dec 17­1:02 PM (165 of 212) Title: Dec 17­1:03 PM (166 of 212) Testcross Question If you had a brown cow and wanted to know its genotype, how would you perform a testcross? A testcross is not necessary if traits show incomplete or co­ dominance...because the phenotype tells you the genotype! Title: Oct 16­9:43 AM (167 of 212) Rabbit Question You crossed a dark grey male rabbit and a Himalayan female rabbit, with 6 kits produced: 2 dark gray 3 white 1 Himalayan What are the genotypes of the parents? Which parent genotype is it impossible to know for sure? Title: Oct 16­9:43 AM (168 of 212) Title: Dec 17­1:18 PM (169 of 212) Rabbit Question You crossed a dark grey male rabbit and a Himalayan female rabbit, with 6 kits produced: 2 dark gray 3 white 1 Himalayan What are the genotypes of the parents? Which parent genotype is it impossible to know for sure? Title: Oct 16­9:43 AM (170 of 212) Assignment: Human Blood Types Complete the Sample Problem on page 542. Check yourself against the solution on page 811. Title: Oct 16­9:43 AM (171 of 212) Title: Dec 17­1:28 PM (172 of 212) Class Assignment: Oompa Loompa Genetics Complete the assignment sheet on Oompa Loompa genetics. Be sure to show all work, including Punnett squares, for all problems! Title: Oct 16­9:43 AM (173 of 212) Homework Read Section 16.3, pages 545­553, taking notes on this section. Define the following terms in your own words! Sex­linked inheritance Barr body Polygenic inheritance Modifier genes Deletion error Inversion error Duplication error Translocation error Nondisjunction Monosomy Down Syndrome Fragile X Syndrome What is the result of the following genotype: XO Title: Oct 16­9:43 AM (174 of 212) XXY Title: Dec 3­12:12 PM (175 of 212) Title: Dec 7­10:02 AM (176 of 212) Section 16.3 ­ Chromosomes and Heredity Return to Contents Title: 16.3 (177 of 212) Chromosomes and Heredity Mendel did his work by esperimentation, before scientists knew what chromosomes or DNA were. Still, he was able to conclude that characteristics were determined by pairs of "factors", and that one factor was donated to an offspring by each parent. He also stated that inheritance of one factor did not affect the inheritance of other factors; this became the Law of Independent Assortment. Title: Oct 16­9:43 AM (178 of 212) Chromosomes and Heredity It wasn't until the 1900s that scientists were able to determine that Mendel's "factors" were located on chromosomes. Walter Sutton and Theodor Boveri made the connection while studying the behavior of chromosomes during meiosis. They noticed that chromosomes occur in pairs, and that during meiosis, the chromosome pairs align at the cell's equator, and are then segregated (separated) independently. This meant that each gamete receives one chromosome from each pair, and this one does not influence which other chromosomes end up in the gamete. This allowed them to come up with the Chromosome Theory of Inheritance. Title: Oct 16­9:43 AM (179 of 212) Chromosomes Theory of Inheritance The theory contains two main points: 1) Mendel's factors of inheritance, or genes, are carried on the chromosomes 2) The segregation and independent assortment of chromosomes during meiosis accounts for the patterns of inheritance. Title: Oct 16­9:43 AM (180 of 212) Chromosomes Theory: Independent Assortment Title: Oct 16­9:43 AM (181 of 212) Linked Traits Not all characteristics follow the law of independent assortment completely. Title: Oct 16­9:43 AM (182 of 212) Chromosomes Theory: Linked Traits Title: Oct 16­9:43 AM (183 of 212) Sex­Linked Traits Characteristics which occur on the X or Y chromosomes are called sex­linked traits. For example, the gene for eye color on fruit flies is located on the X chromosome. The "wild type" or red eyes are dominant; white eyes are recessive. Title: Oct 16­9:43 AM (184 of 212) Chromosomes Theory: Sex Linked Traits Title: Oct 16­9:43 AM (185 of 212) Problem If you cross a homozygous red­eyed female fruit fly with a white­eyed male, what will the phenotype ratios of the offspring be? Remember, the gene for eye color on fruit flies is located on the X chromosome, and the white­eyes allele is recessive. First, determine the parent (P) genotypes: Male Genotype XrY Title: Oct 16­9:43 AM (186 of 212) Female Genotype XRXR Problem Fill in the Punnett square. XR Xr Y XRXr XRY Title: Oct 16­9:43 AM (187 of 212) XR XRXr XRY In the F1 Generation: 50% of the young are female, and all of the females have red eyes. (However, all carry the recessive white­eyed allele) 50% of the young are male, and all of the males have red eyes. (However, they only carry the red­ eyed allele on the X chromosome.) Problem What about the F 2 Generation? If you crossed one of these males with one of the females, what would the F 2 results be? XR XR XRXR Y XRY Xr In the F2 Generation: XRXr 50% of the young are female, and all of the females have red eyes. (However, half carry the recessive white­eyed allele) XrY 50% of the youndg are male, and half of the males have red eyes, while the other half have white eyes. Because eye color is carried on the X chromosome, which parent do the males inherit eye color from? Title: Oct 16­9:43 AM (188 of 212) Homework Read Section 16.3, pages 545­549. Complete Questions #1, 5­9 on page 554. Title: Oct 16­9:43 AM (189 of 212) Title: Jan 7­11:26 AM (190 of 212) Title: Jan 7­2:32 PM (191 of 212) Title: Jan 7­2:35 PM (192 of 212) Title: Jan 7­2:39 PM (193 of 212) Title: Jan 7­2:43 PM (194 of 212) Title: Jan 7­2:44 PM (195 of 212) Title: Jan 7­2:46 PM (196 of 212) Review of Tests Title: Oct 16­9:43 AM (197 of 212) Virtual Fruit Fly Lab http://www.sciencecourseware.org/vcise/drosophila/ Title: Oct 16­9:43 AM (198 of 212) Title: Dec 17­9:34 AM (199 of 212) Title: Jan 7­2:59 PM (200 of 212) Title: Jan 7­3:09 PM (201 of 212) Title: Jan 7­3:13 PM (202 of 212) Virtual Fruit Fly Lab 2 http://www.biologylabsonline.com/protected/FlyLab/ Title: Oct 16­9:43 AM (203 of 212) Title: Dec 14­9:03 AM (204 of 212) Section 16.4 ­ Human Genetics Return to Contents Title: 16.4 (205 of 212) Title: Jan 5­2:06 PM (206 of 212) Title: Jan 5­2:28 PM (207 of 212) Title: Jan 5­2:41 PM (208 of 212) Title: Jan 5­2:47 PM (209 of 212) Title: Jan 5­2:51 PM (210 of 212) Title: Jan 5­2:59 PM (211 of 212) Title: Jan 5­3:04 PM (212 of 212) Attachments Endrocrine Gland Function.wpd codontable2[1]