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Exercise #1: Monohybrid Crosses In solving a genetics problem there are five basic steps: 1. Determine the genotype of each parent. 2. Determine the possible types of gametes each parent can produce. 3. Determine all possible gene combinations that may result when these gametes combine. 4. Determine the various phenotypes possible by analysing the various genotypes. 5. Determine genotypic and phenotypic ratios as required. 1. A pure black male cat mates with a white female. Black coat colour is the product of a dominant allele. Show the genotypes and phenotypes of the parental, F1 and F2 generations. Indicate the phenotypic and genotypic ratios of the F2 generation. 2. In humans, six fingers (F) is the dominant trait and five fingers (f) is the recessive trait. Both parents are heterozygous for six-fingers. Indicate the genotypes and phenotypes of the parents and their possible offspring. What is the probability of producing a fivefingered child? 3. In cattle the polled (hornless) trait is dominant over the horned trait. A purebred polled bull is bred with a horned cow. Show the genotypes and phenotypes of the parental, F1 and F2 generations. Indicate the phenotypic and genotypic ratios of the F2 generation. 4. A single pair of genes in which the Siamese pattern is recessive to the ordinary solid coat colour pattern controls the Siamese coat pattern in cats. Predict the phenotypes, genotypes and their probable proportions in the kittens of a homozygous solid coat male with a Siamese female cat. 5. In a certain species of plant, one purebred variety has hairy leaves and another purebred variety has smooth leaves. A cross of the two varieties produces offspring that all have smooth leaves. Predict the phenotypic and genotypic ratio of the F2 generation. 6. Melanin pigments range in colour from yellow to reddish-brown to black. The amount and the colour of melanin in the skin account for differences in human skin coloration. Albinism is a genetic disorder that results in unpigmented skin and other tissues. About 1 in 20 000 humans has albinism. In humans, it can be caused by an autosomal recessive allele (a). Its dominant allele (A) results in normal pigmentation. a) If a homozygous albino marries a homozygous person of normal pigmentation, what would be the expected phenotype and genotype of their children? b) If an albino married a heterozygous person of normal pigmentation, what phenotypes would you expect in their children? c) Two parents of normal skin pigmentation had an albino child. How was this possible? 7. In Drosophila, the common fruit fly, the normal grey body is dominant to the recessive trait of black body. How can you determine the unknown genotype of a grey bodied fly? 8. Scientist believes that a mutant recessive form of an autosomal gene called BRCA1 may be associated with 5% to 10% of all cases of breast cancer. About 80% of women who inherit the gene in its defective form are likely to develop a cancerous breast tumor. Men who carry the faulty BRCA1 gene rarely develop breast cancer, but they may pass the gene to their offspring. A couple has two children, a girl and a boy. The mother has a single mutant gene for breast cancer; the father is not a carrier of the mutant BRCA1 gene. What is the probability that their daughter has inherited the mutant BRCA1 gene? 9. Black colour in guinea pigs is dominant over white. Outline a cross that would make it possible to determine if a black male guinea pig is homozygous or heterozygous. 10. Wild red foxes occasionally have silver-black pups. If two silver-black foxes mate, their offspring are all silver-black. Explain the inheritance of these coat colours in foxes. Use a Punnett square to provide evidence for your answers. 11. When 20 purebred Himalayan rabbits are mated with a solid gray male of unknown ancestry, 46 of the offspring are Himalayan and 52 are gray. A single pair of genes controls coat colour in rabbits. Himalayan rabbit 1. Is the Himalayan coat pattern controlled by a recessive or a dominant allele? 2. What are the genotypes of the female and male rabbits? 3. How many offspring of each phenotype would you expect from the cross outlined above? Use a Punnett square to provide evidence for your answer. 4. In rabbits, certain short-haired individuals crossed with long-haired ones produce only short-haired individuals. Other short-haired ones when crossed with long-haired ones produce approximately equal numbers of short-haired and long-haired offspring. When long-haired rabbits are crossed, they always produce long-haired offspring like themselves. a) Which trait is do you hypothesize is dominant? b) Give the genotypes of all the rabbits described: First cross: short-haired rabbit long-haired rabbit short-haired offspring Second cross: short-haired rabbit long-haired rabbit short-haired offspring long-haired offspring 12. In humans, brown eyes are dominant to blue. Both parents of a blue-eyed man are brown eyed. The blue-eyed man, Ed, marries a brown-eyed woman, Sharon, whose mother had brown eyes and whose father had blue eyes. The woman has a brother who has blue eyes. Ed and Sharon marry and have a brown-eyed child. Give the genotypes of all the individuals described. Ed's mother _____ Sharon's mother _____ Ed's father _____ Sharon's father _____ Ed _____ Sharon's brother _____ Sharon _____ Ed and Sharon's child _____ 13. What are the chances that the first child from a marriage of two heterozygous browneyed parents has blue eyes? If the first child has brown eyes, what are the chances that the second child will also have brown eyes? Explain your reasoning. 14. When two rough-coated guinea pigs are bred, the resulting offspring consisted of 18 rough- and 4 smooth-coated offspring. Which type of coat is dominant? What proportion of the offspring are homozygous for the dominant trait? 15. The polled or hornless condition is dominant over the horned trait in cattle. A certain polled bull is bred to three cows. Cow A, which is horned, produces a horned calf. Cow B, which is also horned, produces a polled calf. Cow C, which is polled, produces a horned calf. What are the genotypes of the animals described? What further offspring could be expected from future mating of these animals? Bull _____ Cow A _____ Her horned calf _____ Future calves _____ Cow B _____ Her polled calf _____ Future calves _____ Cow C _____ Her horned calf _____ Future calves _____ 16. In peas, inflated pods are the product of a dominant allele and constricted pods are produced by a recessive allele. Long stems are the product of a dominant allele and short stems are produced by a recessive allele. a) What symbols would be used for coding these genes? b) What are the two possible genotypes of a plant, which has the phenotype of inflated pods and short stems? c) What is the genotype of a plant, which is a hybrid for pod shape and stem length (heterozygous for both traits)? 17. In Holstein-Friesian cattle, the two colours black-and-white and red-and-white are controlled by a single pair of alleles. A black-and-white bull was bred with 20 red-and white cows. All of the resulting calves were black-and-white. a) What allele for the trait of coat colour is dominant? b) When the F1 calves grew to maturity, the ones that had the most desirable traits were mated. What would be the expected phenotypic ratio of coat colours of the F2 calves? S 18. Sickle cell anemia is caused by the sickle cell allele (Hb ) of a gene that contributes to hemoglobin (Hb) production. The abnormal hemoglobin (hemoglobin-S) produced causes red blood cells to become deformed and block capillaries. Tissue damage results. Affected individuals homozygous for the sickle cell gene rarely survive to reproductive age. Heterozygous individuals produce both normal hemoglobin and a small percentage of hemoglobin-S. These individuals are more resistant to malaria than individuals who are A homozygous for the allele for normal hemoglobin (Hb ). Their red blood cells are prone to sickling when there is a deficiency of oxygen. A S If a man and a woman who are both heterozygous for the alleles Hb and Hb have a child, what is the probability that the child would not be heterozygous? 19. In Drosophila, the common fruit fly, the normal gray body is dominant to the recessive trait of black body. A gray-bodied male fly was allowed to breed with many black-bodied females. Some of the offspring had gray bodies and some had black bodies. a) What were the genotypes of the parents and the offspring? b) Two gray-bodied flies were mated, and all the offspring had gray bodies. Can you conclusively determine the genotypes of the parents? Explain your answer. c) Two gray bodied flies were mated, and both gray and black bodies were observed in their offspring. What were the genotypes of the parents? What were the genotypic and phenotypic ratios of the offspring? 20. Marfan syndrome is an autosomal-dominant disorder of humans. Affected individuals tend to be tall and thin. They have defects in the lens of the eye and weak connective tissue around the aorta. Often, affected individuals excel in sports like volleyball or basketball, but it is not uncommon for people with this syndrome to die suddenly. 1. A man, heterozygous for Marfan syndrome and a homozygous recessive woman have a child. What is the probability that the child will be affected by Marfan syndrome? 2. If the couple’s first child has Marfan, what is the probability of their second child also having this disorder? Exercise #2: Other Patterns of Inheritance Codominance/Incomplete Dominance/Multiple Alleles 1. In the four o’clock flower, similar to a petunia, the allele for red flower colour is incompletely dominant over the allele for white flower colour. A heterozygous plant has pink flowers showing intermediate inheritance. a. Show the genotypes of the parents and F1 generation of a cross between a red flowered and a white flowered four o’clock plant. b. What would be the anticipated offspring if an F1 plant were back-crossed to the red flowered parent? c. What would be the anticipated offspring if an F1 plant were back-crossed to the white flowered parent? 2. In the radish plant three shapes are observed in the root - round, long and oval. Different crosses of radishes gave the following results: a. b. c. d. long x oval produces 52 long and 48 oval long x round produces 98 oval oval x round produces 51 oval and 50 round oval x oval produces 24 long, 53 oval, and 27 round Explain the inheritance of root shape in radishes. Using symbols, demonstrate that your hypothesis is true for all crosses. 3. In a certain strain of chickens, a mating between a black chicken and a white chicken always produces offspring which have a distinctive feather appearance called blue Andalusian. A cross between two blue Andalusians produces blue, black and white offspring. a. Using Punnett squares, illustrate the two crosses described. b. What are the phenotypic and genotypic ratios of the F2 generations? c. What would be the expected phenotypic ration if a blue Andalusian hen were bred with a black rooster? 4. Blood typing: The alleles for A (IA) and B (IB) are codominant, and both are dominant to O (i). The alleles for M and N are codominant. The allele for RH+ (R) is dominant to the allele for RH- (r). The coding usually used to represent these alleles is: type A antigen IA type B antigen IB no antigen i a. Complete the table below to indicate all possible genotypes for the phenotypes indicated. Phenotypes Genotypes Type A Type B Type AB Type O b. A woman having type A blood claims that her former husband who has type B blood is the father of her baby. His mother has type B blood and his father has type A blood. The baby has type O blood. The man denies that he is the father of the child and refuses to pay child support. Show how you would determine if the man is in fact the father of the child. c. A B - If a woman with the genotype I I Rr and a man with the blood type O Rh have a child, what is the probability that the child will have blood type A Rh . d. Which children could belong to a couple in which the woman has blood type A, N, + + Rh and the man has blood type O, M, Rh . Child Blood Types + One O MN Rh Two A N Rh + Three A MN Rh - Four AB MN Rh - 5. Coat colour in mice is dependent on members of a series of multiple alleles. The hierarchy of dominance is : C+ (full colour) > Cch (chinchilla) > Cd (blonde) > c (albino) a. Complete the table below to indicate all possible genotypes for the phenotypes indicated. Phenotypes Genotypes Full colour Chinchilla Blonde Albino b. A chinchilla female which was heterozygous for albino was mated with a full colour male which was heterozygous for blonde. What phenotypes could be expected in their offspring? 1 6. Multiple alleles control the intensity of pigment in mice. The gene D designates full 2 3 colour, D designates diluted colour, and D is deadly when homozygous. The order of 1 2 3 dominance is D >D >D . When a full colour male is mated to a dilute colour female, the offspring are produced in the following ratio: two full colour to one dilute to one dead. Indicate the genotypes of the parents. Exercise #3 Dihybrid Crosses 1. Horses which are black in colour and which have a trotting gait carry autosomal dominant genes, while horses, which are chestnut in colour and have a pacing gait, carry autosomal recessive genes. A purebred black trotting stallion is bred with a chestnutpacing mare. 1. What are the parental genotypes and what are the possible sperm and eggs produced? 2. What would the genotype and phenotype of the F1 offspring be? 3. What kind of gametes could the F1 offspring produce at maturity? 4. Construct a Punnett square to show the possible F2 offspring if two F1 horses were bred maturity. 5. What is the expected phenotypic ratio of the F2 generation if many of these breeding took place? 2. In summer squash, white fruit is dominant over yellow fruit while disc is dominant over sphere shape. A plant which homozygous for white fruit and sphere shape is crosspollinated with one which is homozygous for yellow fruit colour and disc shape. 1. What will be the genotype and phenotype of the F1? 2. What is the phenotypic ratio of the F2 generation? 3. If a F1 plant were crossed with pollen from its yellow disc parent, what would be the expected phenotypes of the offspring and in what ratio? 3. In certain breeds of dogs, black colour is determined by an autosomal dominant gene and red colour is due to an autosomal recessive gene. Solid coat is dominant trait and spotted coat is recessive. A male which is heterozygous black and homozygous white spotted is bred with a female which red and white spotted. 1. What is the probability of producing a pup which is red and white spotted? 2. What is the probability of these two dogs producing a solid black puppy? 4. In garden peas, the allele for tall plant height (T) is dominant over the allele for short plant height (t), and the autosomal allele for axial flower position (A) is dominant over the allele for terminal flower position (a). A plant heterozygous for both traits was crossed with plant homozygous recessive for both traits. 1. What percentage of the offspring produced would be expected to display at least one of the dominant traits? 2. Predict what fraction of the offspring will be heterozygous? 5. In pea plants, tall (T) dominant over short (t), and round seed (R) is dominant over wrinkled seed (r). A heterozygous tall-heterozygous round –seed pea plant and a shortheterozygous round-seed pea plant were crossed. 1. What is the probability of producing an offspring that is homozygous recessive for both traits? 2. What portion of the offspring are tall and round? How could you test your hypothesis of which trait is dominant? Use Punnett squares to show the results of the crossed you would use. 6. R is the allele for red flesh fruit colour in tomatoes, while r is the allele for yellow fruit. P is the allele which gives tomato stem a purplish colour, and p shows up as a greenish stem. Suppose a cross was made between two tomato plants which resulted in the following kinds of offspring: 247 red fruit, purple stem 261 red fruit, green stem, 253 yellow fruit, purple stem 256 yellow fruit, green stem What would be the phenotypes and genotypes of the parent plants? 7. A homozygous dominant snapdragon having red flowers and narrow leaves was crossed with one having white flowers and broad leaves. Indicate the genotypes and phenotypes of the parental, F1 and F2 generations. Genetics Exercise #4 Polygenic characteristics: Epistatic Genes One Gene Many Effects: Pleiotropic Genes 1. Assume that there are two gene pairs involved in determining eye colour: one codes for pigment in the front of the iris and other codes for pigment in the back of the iris. Genotype Eye Colour AABB black-brown AABb dark brown AAbb brown AaBB brown-green flecked AaBb light brown Aabb grey-blue aaBB green aaBb dark blue aabb light blue a. A man has gray-blue eyes and a woman has green eyes. Which eye colour phenotypes would be possible for children born to this man and woman? b. If one parent has light brown eyes and the other has dark brown eyes, what is the probability that they would have an offspring with gray-blue eyes? 2. In humans, normal pigmentation dominates no pigmentation (albino). Black hair dominates blonde hair. An albino person will have white hair color even though they may also have the genes for black or blonde hair colour. An albino male who is homozygous for black hair marries a woman who is heterozygous for normal pigmentation and who has blond hair. What colours of hair can their children have and what is the probability for each hair colour? 3. In mice, the gene C causes pigment to be produced, while the recessive gene c makes it impossible to produce pigment. Individuals without pigment are albino. Another gene, B, located on a different chromosome, causes a chemical reaction with the pigment and produces a black coat colour. The recessive gene, b, causes incomplete breakdown of the pigment, and a tan or light-brown colour is produced. The genes that produce black or tan coat colour rely on the gene C, which produces the pigment, but are independent of it. Indicate the phenotypes of the parents and provide the genotypic and phenotypic ratios of the F1 generations from the following crosses: 1. CCBB x Ccbb 2. ccBB x CcBb 3. CcBb x ccbb 4. CcBb x CcBb 4. The mating of a tan mouse and a black mouse produces many different offspring. The geneticist notices that one of the offspring is albino. Indicate the genotype of the tan parent. How would you determine the genotype of the black parent? 5. The gene R produces a rose comb in chickens. An independent gene, P, which is located on a different chromosome, produces a pea comb. The absence of the dominant rose comb gene and pea comb gene (rrpp) produces birds with single combs. However, when the rose and pea comb genes are present together, they interact to produce a walnut comb (R_P_). Indicate the phenotypes of the parents and give the genotypic ratios of the F1 generations from the following crosses: 1. rrPP x RRpp 2. RrPp x RRPP 3. RrPP x rrPP 4. RrPp x RrPp Chromosome Mapping The probability of a crossover between 2 loci (specific places on a chromosome where genes are located) in a chromosome is proportional to the distance between the two loci. Since a crossover can occur at any place in the chromosome, the longer the chromosome length between the loci the greater the chance that crossover will take place. By knowing recombination frequencies, the sequence of genes in a chromosome can be determined. 1. The recombination frequency between gene A and gene B is 9%; between A and C is 17%; and between B and C is 26%. What is the sequence of genes in the chromosome? Sketch a map of the chromosome, showing the map unit distances between the genes. 2. The crossing-over frequency between genes A and B is 35%; between B and C 10%; between C and D 15%; between A and C 25%; between A and D is 10%, and between B and D 25%. What is the sequence of genes in the chromosome? Sketch a map of the chromosome, showing the map unit distances between the genes. 3. The crossing-over frequency between genes A and B is 5%; between B and C 17%; between C and D 18%; between A and C 12%; between B and D 1%; and between A and D 6%. What is the sequence of genes in the chromosome? Sketch a map of the chromosome, showing the map unit distances between the genes. 4. In a series of breeding experiments, a group of genes located on the same chromosome was found to show the recombination frequencies in the chart below. Using this data, map the chromosome. GENE A B A B C D E 8 12 4 1 8 - 4 C 12 4 - 12 9 Recombinations 16 13 per 100 fertilized eggs D 4 12 16 - 3 E 1 9 13 3 - 5. For a breeding experiment, a linkage group composed of genes W, X, Y, and Z was found to show the following gene combinations. Use this data to construct a gene map. Gene WXYZ W - 5 78 X 5 - 23 Y 7 2 -1 Z 8 3 1- 6. The use of marker genes and the analysis of crossover frequencies of genes have enabled geneticists to map the location of many genes on human chromosomes. Blue colour vision and blue colour blindness (tritanopia) are both controlled by a gene on chromosome 7. The gene for the production of trypsin (a digestive enzyme) and the gene responsible for cystic fibrosis are also found on chromosome 7. Some crossover frequencies of these genes are shown below. Pair of Genes Crossover Frequency Marker gene cystic fibrosis 18% Marker gene tritanopia 13% Cystic fibrosis trypsin 6% Trypsin tritanopia 1% Which of the following gene maps shows the correct sequence of these genes on chromosome 7? Crossover Frequencies for Some Genes on an Autosome of Organism Z Genes Crossover Frequency P and Q 5% P and R 8% P and S 12% Q and R 13% Q and S 17% 7. Which chromosome map best represents the sequence of genes on the autosome from organism Z? Note: For all genetics problems involving sex-linked traits, state the results of the two sexes separately. Exercise #6: Sex linkage 1. Colour blindness in humans is a sex-linked recessive trait. A woman of normal colour vision whose father was colour blind marries a man of normal vision whose father was also colour blind. What type of colour vision will be expected in their children? 2. Mr. and Mrs. White have normal colour vision. They have three children: Bob, who is colour blind and who has a daughter of normal colour vision; Joan, who has normal colour vision and who had one son who is colour blind and one son who is normal; and Susan, who also has normal colour vision and who has five sons of normal colour vision. What are the genotypes of the individuals described? 3. Peter's maternal grandmother had normal colour vision and his maternal grandfather was colour blind. Peter's mother is colour blind, and his father has normal colour vision. A. What are the genotypes of the individuals described? B. What type of colour vision does Peter have? C. What type of colour vision do his sisters have? D. If Peter were to marry a woman who is genotypically the same as his sister Joan, what type of colour vision would be expected in their children? 4. Eye colour in Drosophila is sex-linked. It is controlled by a pair of genes in which red is dominant over white. A. If a red-eyed male fruit fly is crossed with a white-eyed female, what will be the phenotypes of the offspring? B. If a male F1 is mated with a white-eyed female, what will be the appearance of the offspring produced? C. If a female F1 is mated with a red-eyed male, what will be the appearance of the offspring produced? D. If a white-eyed female is crossed with a red-eyed male and the F1 are allowed to freely interbreed, what will be the eye colour of the F2 offspring? 5. "Alligator men" or "fish women" were exhibited for their physical abnormalities in fairs or circuses earlier this century. These people probably suffered from X-linked ichthyosis, which produces symmetric dark scales on the body. The disease occurs in 1 in 6 000 males and is much more rarely in females. Ichthysosis is likely a recessive disorder. A. If an "alligator man" were to marry a woman homozygous for the normal condition, what is the percentage probability that their children would have ichthyosis? B N 6. Baldness is a sex-influenced trait. Two autosomal alleles (H and H ) exist for the pattern N baldness gene. The HB allele is dominant in men but recessive in women. The H B (normal) allele is dominant in females but recessive in males. A single H allele seems to cause pattern baldness only in the presence of the level of testosterone normally found in B N B B adult males. Males will develop pattern baldness with the genotypes H H or H H ; B whereas, for females to develop pattern baldness, they must inherit two H alleles. It appears that testosterone destroys or inhibits the production of an enzyme necessary for hair growth in the hair follicle. A. What is influencing the difference in expression of the pattern baldness gene in women and men? B. Parents who do not display pattern baldness have a son who exhibited pattern baldness by the age of thirty. What are his possible genotypes? C. If they also have a daughter, could she also display baldness. Explain. 7. In humans, the conditions for blood clotting dominates the condition for non-clotting (haemophilia). Both genes are linked on the X chromosome. A male haemophilic marries a woman who is a carrier for this condition. What are the chances that if they have a male child he will be normal for blood clotting? HUMAN PEDIGREES 1. Tay-Sachs disease is caused by a lethal autosomal recessive gene (t). Individuals with TaySachs disease first shown the condition at the age of 6 months and die at about 4 years old. a. Write the genotype for each individual by the circle or square. b. What is the probability that individuals II-1and II-2 could produce another Tay-Sachs child? 2. Sickle-cell anaemia is a disease that occurs in about 1 in 500 black children born in North America. Individuals with this disease can suffer from "sickle cell crises" when they are deprived of oxygen by exertion or respiratory ailment. When this happens, red blood cells collapse into a "sickle" shape that can block capillaries. This causes severe pain, and the lack of blood flow can result in oxygen deprivation that can make the condition even worse. a. How is this trait inherited? b. Write the genotype for each individual below the circle or square. c. About 1 in 10 North American blacks are heterozygous for sickle-cell anaemia. This number is very high for a trait that can be so detrimental for homozygous individuals. However, research has shown that heterozygous individuals have enhanced resistance to malaria. What effect might this factor have had on the frequency of sickle-cell gene in populations of North American blacks? d. Under what condition is the genotype of heterozygous individuals known? 3. Haemophilia is a recessive X-linked condition that used to be called the "bleeder's disease" because the blood of affected individuals takes a long time to clot. Normal blood usually clots within five minutes after being placed in a test tube, whereas haemophiliac blood may require several hours to clot. Without treatment, even common bruises often lead to serious internal bleeding in the haemophiliac individual. Pedigree Three: Haemophilia a. Why do more males have haemophilia than females? b. Explain the conditions required for female (III-3) to have haemophilia. 4. Below is a diagram showing the inheritance of blue-green colour blindness. A black symbol indicates colour blindness and a white symbol indicates normal colour vision. a) How is this trait inherited? b) Write in the genotype for each individual. c) What evidence is there that colour blindness is a sex-linked trait? d) Give the possible genotypes of all the individuals. 5. Marfan syndrome is an inherited condition that affects the connective tissue, resulting in unusually long bones and spinal curvature, as well as vision, cardiac, and respiratory problems. The syndrome tends to become increasingly severe over time. The following pedigree shows inheritance of Marfan syndrome in a multigenerational family. a) How is this syndrome inherited? b) Can you determine individual II4’s genotype? Explain. c) Individual II1 and II2 are considering having another child. What is the probability that this child will have Marfan syndrome? Explain using a Punnett square. 6. Examine the following pedigree showing the inheritance of straight hair in a four-generation family. a) Is straight hair a dominant or recessive trait? Explain. b) Identify the genotype of each individual in the pedigree. Whose genotype can you not be certain of? c) If individual V3 marries a man who is heterozygous, what is the probability that they will have a girl with straight hair? Explain using a Punnett square. 7. Humans may have a peaked or smooth hairline. If a man and a woman both have a smooth hairline, none of their children will have peaked hairlines. How is a peaked hairline most likely inherited? Draw a pedigree for a family where one parent and two of three children have a peaked hairline. One of the children with a peaked hairline marries an individual with a smooth hairline. Their children both have a peaked hairline. Identify the genotypes and phenotypes of each individual. 8. As a top research scientist, you and your colleagues have discovered a new condition. In the course of your research, you have come across a pedigree (see below) for a family in which the condition occurs. a) On which chromosome is the allele for the condition found? b) Which individuals in the pedigree can you be certain are carriers of the allele? Explain your reasoning. c) If individual III3 has a son with a woman who is not a carrier of the allele, what is the probability that the son will have the condition? Explain. 9. In the pedigree below, different blood types are identified by the letters A, B, AB, and O. a) Individuals II4 and II5 have just had identical twin girls. List the possible blood types these infants may have based on the information provided in the pedigree. b) Individuals II6 and II7 have a second child with blood type O. What does this tell you about II6’s genotype? c) Could I1 and I2 have a child with the AB blood type? Explain why or why not. 10. Suppose you discovered that both you and your spouse are heterozygous for a lethal recessive gene. Would you have children? Explain. Would your decision change if the trait involved were debilitating but not lethal? Explain. 11. Many genetic diseases extract a high cost in terms of health care dollars. Since Medicare is funded by government funds, should the government be allowed to legislate who may and who may not have children? Defend your position. Genetics Exercise #7: Mixed Problems 1. How many different types of gametes could be produced by the individuals having the following genotypes, and what are the gametes? R a. X Y f. WwDd r b. LlGG g. X YLL c. ssBB h. X X Bb H d. X X N h n i. rwTT e. AATtRr j. RrYyAa *challenge 2. In pea plants, round seeds and tall stems are dominant to wrinkled seeds and short stems. Determine the phenotypes and their proportions for the following crosses. a. RRtt x rrTT b. RrTt x RrTt c. Rrtt x rrTt 3. In peas, yellow seed colour is dominant to green. What will be the colours and their proportions in the offspring of the following crosses? a. b. c. d. homozygous yellow x green heterozygous yellow x green heterozygous yellow x homozygous yellow heterozygous yellow x heterozygous yellow 4. In Drosophila wing length is controlled by autosomal genes in which the normal long-wing allele is dominant over the vestigial-wing allele. Vestigial wings are very small and non-functional. a. If a female fruit fly having white eyes and homozygous long wings is crossed with a male having red-eyes and vestigial wings, what will be the appearance of the F1? b. What will be the appearance of the F2? c. If an F1 female is mated with a male, which has the same genotype as her father, what will be the appearance of the offspring? d. If an F1 male is mated with a female, which has the same genotype as his mother, what will be the appearance of the offspring? 5. In humans, the condition for normal blood clotting (H) dominates the condition "bleeder's disease". The gene controlling blood clotting is carried on the X chromosome. a. A male haemophiliac marries a woman who is a carrier for this condition. What are the chances that if they have a male child he will be normal for blood clotting? b. A male who has normal blood clotting marries a woman who is a carrier for haemophilia. What are the chances that if they have a son he will be normal for blood clotting? c. Rarely a female haemophiliac is born. How could this happen? 6. In cats, short hair is dominant over long and solid black coat is dominant over he Siamese coat pattern of black and tan. A cat breeder wants to know if a certain black, shorthaired cat is homozygous or heterozygous. How can this be determined? 7. A cross between two sweet pea plants produced 41 plants with pink flowers, 18 with white flowers, and 19 with red flowers. a. How is flower colour inherited in sweet peas? b. What is the phenotype of the parents? Explain how you determined this. 8. Using the information from the previous question, determine the expected phenotypes and their proportions among the offspring of the following crosses. a. white x pink b. red x red c. pink x pink r 9. Plumage colour in ducks is dependent on a set of three alleles: M produces restricted mallard r plumage, M produces mallard, and m produces dusky mallard. The dominance hierarchy is M > M > m. Determine the genotypic and phenotypic ratios expected in the F1 from the following crosses. r r r a. M M x M M r r r b. M M x M m r r c. M M x M m d. r Mmx 10. A young woman with type O blood gave birth to a baby with type O blood. In a court case she claims that a certain young man is the father of her child. The man has type A blood. Could he be the father of her child? Can it be proven on this evidence alone that he is the father? 11. Normal pigmentation (A) dominates no pigmentation (albino = aa). Brown-eyed colouring (B) dominated blue-eyed colouring (bb). Two people with normal skin pigmentation produce one brown-eyed, three blue-eyed, and one albino child. What are the parents' possible genotypes? 12. In rabbits, short hair is due to a dominant gene L, and long hair to its recessive allele l. Black fur is due to a dominant gene B and white hair to its recessive allele b. Two rabbits were crossed several times and they produced a total of 88 shorthaired black and 29 longhaired black offspring. What are the probable genotypes of the parents? 13. Below is a pedigree showing the inheritance of the common condition of myopia (shortsightedness). A square represents a male and a circle represents a female. A white symbol indicates normal vision and a black symbol represents myopia. a. Is the allele controlling myopia dominant or recessive? Sex-linked or autosomal? How do you know? b. Give the genotypes of all individuals. 14. You have seen a pedigree can be used to trace the inheritance pattern of a single-allele trait. A pedigree can also be used to study multiple-allele traits. In the pedigree below, different blood types are identified by the letters A, B, AB, and O. Examine the pedigree below, and then answer the questions that follow. a. Neither individual I 4 or I 6 has ever had their blood tested. What are their blood types? b. Write the genotypes of as many individuals as you can. Whose genotypes can you not be sure of? Explain. c. Individual III 3 marries a man with blood type AB, and they have four children. Will any of these children have blood type O? Explain. d. Individuals II 1 and II 2 have a second child. After you see the child’s blood-test results, you know that both parents have the genotype IAi. What blood type is the child? e. Using examples from the pedigree, describe some of the limitations that apply to human genetic analysis. In what ways could these limitations affect genetic research? f. Blood typing has often been used as an aid in paternity disputes. Can a blood test ever prove that a man is definitely the father of a particular child? Explain. Laboratory Investigation: The Genetics of Corn Purpose: to investigate the inheritance of traits using corn Background Information: The genetics of corn are well understood. The popular strains of corn are produced by carefully planned crosses between a number of strains, each bred for particular traits. Corn is a fairly easy crop to use in genetics programs. Because male and female flowers grow in large, separate clusters on a plant, it is easy to carry out desired pollination. Each corn "ear" contains hundreds of offspring - the individual kernels result from random pollination of single female flowers. Each piece of "silk" on an ear of corn is the remains of a female flower. And since corn is an annual plant, the results of the crosses can be observed within a few months. Seed colour in corn ears is controlled by a single pair of alleles. The amount of sugar or starch contained in corn is also controlled by a single pair of alleles on a different chromosome. If a kernel contains a high concentration of sugar, it wrinkles when it is dried. A kernel containing a high quantity of starch remains smooth when it is dried. Materials: 3 different samples of corn ears (monohybrid sample, dihybrid sample, unknown sample) Procedure: 1. Obtain a sample A or B corn ear. Identify the trait or traits that differ in the ear. Predict the dominant and recessive phenotypes for corn. 2. Code the alleles for the trait or traits observed in the sample of corn. Use a Punnett square to predict the expected phenotypic ratio for the sample. 3. Prepare a data table indicating the different phenotypes present in the corn sample. This table will be used to tally and total the observed phenotypes of 100 seeds. 4. Count 100 of the kernels in sequence and record the observed phenotypes in the data table. Start with any row of seeds. When you come to the end of a row, continue your count on the next row. Continue until you have counted the required number of seeds. 5. Repeat steps 1 to 4 with the sample (A or B) not yet observed. 6. Obtain a sample C corn ear. This ear was produced from a test cross. Identify the trait or traits that differ in the ear. Repeat steps 3 and 4. Analysis: 1. Calculate the observed phenotypic ratios in the three samples. Compare the actual ratios for samples A and B with the expected ratios predicted. 2. What effect might the use of class data have on the observed numbers and ratios? 3. Prepare a table illustrating the phenotypes and the possible genotypes for samples A and B. 4. Identify the phenotypes and genotypes of the parental plants for samples A and B. 5. Calculate the observed phenotypic ratios in the sample C. Predict the phenotypes and genotypes of the parental plants for this sample. Application: 1. Why are test crosses important to plant breeders? 2. A dihybrid cross can produce 16 different combinations of alleles. Explain why 100 seeds were counted rather than 16. 3. A corn plant with starchy, yellow kernels is cross-pollinated with a corn plant with sweet, purple kernels. One hundred kernels from the hybrid are counted and the following results obtained: 52 starchy, yellow kernels and 48 starchy, purple kernels. What are the genotypes of the parents and F1 generation? 4. Under what genetic conditions can you immediately recognize whether an organism is heterozygous for a particular trait?