Problem-Set-1_Agri13_Genetics.pdf

Agri 13 – Principles of Genetics Name: _______________________ Course and Section: ___________ Date: ________________________ Instruction. Please answer the following questions completely and neatly. Write your answers in one whole sheet of yellow paper. Use black or blue ink only. Do not use CORRE...

Agri 13 – Principles of Genetics Name: _______________________ Course and Section: ___________ Date: ________________________ Instruction. Please answer the following questions completely and neatly. Write your answers in one whole sheet of yellow paper. Use black or blue ink only. Do not use CORRECTION TAPE. It is an ALL OR NOTHING PROBLEM SET. Problem Set on Gene Interaction (Allelic and Non-allelic Interactions) 1. What types and frequencies of gametes are formed by the following genotypes; all gene pairs are segregating independently? (20 pts) a. AABbCC b. AABBCcDD c. AaBbCc d. MmNnOoPp 2. Consider blue eyes in man as recessive to brown eyes. Show the expected children of a marriage between a blue-eyed woman and brown-eyed man who had a blue-eyed mother. Please provide the genotypes of the parents (3pts), the completed cross (Punnett Square) (5 pts), and the phenotypic and genotypic ratios (2 pts). 3. The presence of a dimple on the cheek is governed by a dominant gene. A couple has already four children with dimpled cheeks. What would be the probability of having the next child as another dimpled if both parents are heterozygous for the trait? Please provide the genotypes of the parents (3pts), the completed cross (Punnett Square) (5 pts), and the phenotypic and genotypic ratios (2 pts). 4. A woman has an abnormality of the eyelids called ptosis, which makes it impossible for her to open her eyes completely. The condition has been found to depend on a single dominant gene (P). The woman’s father had ptosis, but her mother had normal eyelids. Her father’s mother had normal eyelids. a. What are the probable genotypes of the woman, her father, and her mother? (3 pts) b. What proportion of her children will be expected to have ptosis is she marries a man with normal eyelids? Please provide the genotypes of the parents (3pts), the completed cross (Punnett Square) (5 pts), and the phenotypic and genotypic ratios (2 pts). 5. In peas, gray seed color (G) is dominant to green (g). The following data were collected. Indicate the genotype of each set of parents (8 pts). Parents Offsprings Gray x Gray 59 20 Gray x Gray 100 0 Green x Green 0 105 Gray x Green 67 60 6. In tomatoes, two pairs of genes affect the color of the ripe fruit as follows: R red flesh Y yellow skin R yellow flesh y colorless skin Dominance is complete for red flesh and yellow skin. If the genes are independently segregating, calculate the expected phenotypic and genotypic ratios from the following crosses (40 pts). a. Rryy x rrYy b. RrYy x rrYy c. RrYY x Rryy d. RrYy x Rryy 7. What proportion of the offspring of the following crosses involving independent segregation will be homozygous for all gene pairs? Show the result of each cross (7 pts each). a. AaBb x AaBb b. AABBCC x AAbbcc c. AaBBCc x AABbcc 8. In man, two abnormal conditions, cataracts in the eyes and excessive fragility in the bones, seem to develop on separate dominant genes. A man with cataracts and normal bones, whose father had normal eyes, married a woman free from cataracts but with fragile bones. Her father had normal bones. What is the probability that their first child will: Please provide the genotypes of the parents (3pts), the completed cross (Punnett Square) (5 pts), and the probability of the following (2 pts each). a. free from both abnormalities b. have cataracts c. have fragile bones but not cataracts d. have both cataracts and fragile bones 9. In fowl, the genotype rrpp gives single comb, R_P_ walnut comb, rrP_ pea comb, and R_pp rose comb. a. What comb types will appear and in what proportions in the F1 and in the F2 if single-comb birds are crossed with birds of a true-breeding walnut-combed strain? (5 pts) b. What are the genotypes of the parents in a walnut x rose mating from which the progeny are 3 rose:3 walnut:1pea:1 single? (5 pts) c. What are the genotypes of the parents in a walnut x rose mating from which all the progeny are walnut? How many genotypes will produce a walnut phenotype? Please write them out. (5 pts) 10. A certain plant may have either single or double flowers; single flowers when at least one dominant gene of each of the two different pairs of genes DD and SS are present, double flowers in the presence of either or both pairs of the recessive genes. If plants with genotypes DDss and ddSS were crossed, what would be the chance in F2 to get double flowers? single flowers? (10 pts) 11. In mouse, pigment will not be produced in the hair unless the gene C is present. If the individual is CC or Cc, its coat colouration depends upon its genotype with respect to another gene A on a different chromosome. Aa or AA individuals are gray; aa individuals are black. Two gray mice produce litter with gray, albino, and black mice in a ratio of 9:4:3. What was the genotype of the parents? What type of gene action involved? (12 pts) 12. White chicken may be the result of two different genetic combinations: the homozygous state of the recessive gene c or the homozygous state of the recessive gene o. Color production requires the action of both gene C and O. What progeny might be expected in the F2 of the following cross and in what proportion: ccOO and CCoo? (10 pts) 13. Using the Chi-squared test, determine whether the following F2 results fit better a 3:1 or a 13:3 ratio (40 pts, 5 pts for each test). a) 126 red, 48 white b) 145 red, 32 white c) 128 red, 46 white d) 136 red, 100 white 14. In cattle, “short spine” is lethal shortly after birth; it is caused by the homozygous recessive genotype, ss. Heterozygotes are normal. A series of mating between roan animals heterozygous for the short spine gene will produce what phenotypic ratio (12 pts): a) At birth b) After several days 15. In fruit flies, gray body (G) is dominant over the ebony body (g), and red eyes (R) are dominant over brown eyes (r). A fly is heterozygous for both body colour and eye colour is mated with a fly heterozygous for the body colour and with brown eyes. The cross resulted in the following offspring: 15 flies with ebony bodies, brown eyes 31 flies with gray bodies, brown eyes 12 flies with ebony bodies, red eyes 38 flies with gray bodies, red eyes a. Formulate the null hypothesis (3 pts). b. Calculate the Chi-square value using the formula for Chi-square (and the table below) (10 pts). c. Do you reject or not reject your null hypothesis? Explain your answer. Include what the P value means in your answer (5 pts).