27. In peas the trait for tall plants is dominant (T) and the trait for short plants is recessive (t). The trait for yellow seed color is dominant (Y) and the trait for green seed color is recessive (y). A cross between two plants results in 296 tall yellow plants and 104 tall green plants. Which of the following are most likely to be the genotypes of the parents? (A) TTYY x TTYY (B) TTyy x TTYY (C) TIYy x TrYy (D) TIYY x TTYy (E) TIYY x Ttyy
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- Still referring to Problem 1, what will be the possible genotypes of offspring from the following matings? With what frequency will each genotype show up? a. AABB aaBB b. AaBB AABb c. AaBb aabb d. AaBb AaBb(D) 1/6 (E) 1/64 26. In sheep, eye color is controlled by a single gene with two alleles. When a homozygous brown-eyed sheep is crossed with a homozygous green-eyed sheep, blue-eyed offspring are produced. If the blue-cyed sheep are mated with each other, what percent of their offspring will most likely have brown eyes? (A) 0% (В) 25% (C) 50% (D) 75% (E) 100% 27. In peas the trait for tall plants is dominant (T) and the trait for short plants is recessive (t). The trait for yellow seed color is dominant (Y) and the trait for green seed color is recessive (y). A cross between two plants results in 296 tall yellow plants and 104 tall green plants. Which of the following are most likely to be the genotypes of the parents? (A) TTYY x TTYY (В) Туу х ТТYу (C) TIYY x TIYy (D) TIYy x TTYY (E) TtYY x Ttyy 28. In humans, red-green color blindness is a sexlinked recessive trait. If a man and a woman produce a color-blind son, which of the following must be true? (A) The father is color-blind. (B)…type P = n! (p)* (q)"* х! (n - х)! Practice Problem: You cross a true-breeding pea plant with red flowers to a true-breeding pea plant with white flowers. All of your offspring have red flowers. Which gene is dominant? Why? What is the genotype of your offspring? You then cross the offspring to each other. What ratio do you expect? Why? You count 1000 plants and look at their flowers. Your results are as follows: 740 red 260 white Does this follow a simple Mendelian inheritance pattern? Why or why not? DADT 2 MEA SUDI ND D LUT IONS
- (C) 1/2 1/2 (D) 1/2 (E) 1/2 x 1/2 25. A couple has 5 children, all sons. If the woman gives birth to a sixth child, what is the probability that the sixth child will be a son? (A) 5/6 (В) 1/2 (C) 1/5 (D) 1/6 (E) 1/64 26. In sheep, eye color is controlled by a single gene with two alleles. When a homozygous brown-eyed sheep is crossed with a homozygous green-eyed sheep, blue-eyed offspring are produced. If the blue-cyed sheep are mated with each other, what percent of their offspring will most likely have brown eyes? (А) 0% (B) 25% (C) 50% (D) 75% (E) 100% 27. In peas the trait for tall plants is dominant (T) and the trait for short plants is recessive (t). The trait for yellow seed color is dominant (Y) and the trait for green seed color is recessive (y). A cross between two plants results in 296 tall yellow plants and 104 tall green plants. Which of the following are most likely to be the genotypes of the parents? (А) TTYҮ х TTҮҮ (В) Ттуу х ТТYу (C) TIYY x TIYy (D) TIYy x TTYY (E)…Three coat-color patterns that occur in some breeds of horses aretermed cremello (beige), chestnut (brown), and palomino (goldenwith light mane and tail). If two palomino horses are mated, theyproduce about 1/4 cremello, 1/4 chestnut, and 1/2 palomino offspring. In contrast, cremello horses and chestnut horses breed true.(In other words, two cremello horses will produce only cremellooffspring, and two chestnut horses will produce only chestnut offspring.) Explain this pattern of inheritance.A wild-type fruit fly (heterozygous for gray body color andred eyes) is mated with a black fruit fly with purple eyes. Theoffspring are wild-type, 721; black purple, 751; gray purple, 49;black red, 45. What is the recombination frequency betweenthese genes for body color and eye color? Using informationfrom problem 3, what fruit flies (genotypes and phenotypes)would you mate to determine the order of the body color, wingsize, and eye color genes on the chromosome?
- 2 (a) The trait of hen- versus cock-feathering is a sex-limited trait controlled by a single gene. HH and Hh males always exhibit the hen-featuring trait. Only hh males show the cock-feathering trait. Starting with two heterozygous fowls that are hen- feathered, describe how you would obtain a true-breeding line that always produces cock-feathered males. [Ciri berhulu avam hatin3) A wild type fruit fly (dihybrid for gray body color and normal wings) is mated with a black fly with vestigial wings. Determine the expected phenotypic ratio resulting from this cross if we initially assume the alleles for these traits are unlinked and exhibit complete dominance. The offspring of the mating described above actually exhibited the following phenotypic ratio: Wild type: 778 Black vestigial: 785 Black normal: 158 Gray vestigial: 162 a) What does the observed phenotypic ratio suggest regarding the alleles for body color and wing shape? b) What could account for the existence of the observed recombinant phenotypes in the ratios shown above? c) Calculate the recombination frequency between the genes for body color and wing shape. What is the relative distance between these alleles on the same chromosome?With regard to scurs in cattle ( a sex influenced trait) a cow with no scurs whose mother had scurs had offspring with a bull with scurs whose father had no scurs. What are their probabilities of having the following combinations?a. their first offspring will not have scursb. their first offspring will be a male with no scursc. their first three offspring will be females with no scurs step by step explain in details please!!!!
- In cattle, the gene for hornless (H) is dominant to the gene for horned (h), the gene for black (B) is dominant to that of red (b), and the gene for white face, or Hereford spotting, (S) is dominant to that for solid color (s). A cow (female) that is heterozygous for all traits is inseminated by a bull (male) of the genotype bbhhSs. What is the probability of obtaining a calf that is a black, hornless bull with Hereford spotting?#2 a) If sex and eye colored were viewed as two different phenotypes with male being dominant over female (or vice versa), then Mendelian ratios of a dihybrid cross may explain the ratios that Morgan got. Here is some hypothetical data based on the ratios gotten by Morgan and the expected Mendelian ratios for a dihybrid F1 cross. Perform a Chi-square test (separate scrap paper) with the null hypothesis being that the "Morgan observed ratios" are within chance from the Mendelian ratios. State whether the Chi-square test supports or rejects the null hypothesis and explain. (Be sure to include the X2 value, degrees of freedom, critical value.) Hint: what did Morgan discover? Think about this for your degrees of freedom.* Expected Mendelian Ratios Morgan Observed Ratios 900 Red, male 800 Red, male 300 Red, female 400 Red, female 300 white, male 400 white, male O white, female 100 white, female Terms of Use Support | Schoology Blog | PRIVACY POLICY IN# 2 a) If sex and eye colored were viewed as two different phenotypes with male being dominant over female (or vice versa), then Mendelian ratios of a dihybrid cross may explain the ratios that Morgan got. Here is some hypothetical data based on the ratios gotten by Morgan and the expected Mendelian ratios for a dihybrid F1 cross. Perform a Chi-square test (separate scrap paper) with the null hypothesis being that the "Morgan observed ratios" are within chance from the Mendelian ratios. State whether the Chi-square test supports or rejects the null hypothesis and explain. (Be sure t include the X2 value, degrees of freedom, critical value.) Hint: what did Morgan discover? Think about this for your degrees of freedom. * Expected Mendelian Ratios Morgan Observed Ratios 900 Red, male 800 Red, male 300 Red, female 400 Red, female 400 white, male 300 white, maile O white, female 100 white, female Schoology Support | Schoology Blog | PRIVACY POLICY | Terms of Use INTL