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Mendelian Problems Just a quick reminder - when solving genetic problems, always choose the first letter of the allele that is dominant. For instance, if the dominant trait is curly hair over straight hair, you would use CC to represent a homozygous curly haired person and cc to represent someone with straight hair. 1. (a) A homozygous violet flowered pea plant is crossed with a homozygous white flowered pea plant. If the violet colour is dominant (V), what is the genotype and phenotype of the F1 generation. P1 = VV P2 = vv P1/P2 V V v Vv Vv v Vv Vv Genotype of F1 = 100% Vv Phenotype for F1 = 100% violet (b) If 2 members of the F1 are crossed, what ratio of the F2 generation will have the same phenotype as the parents of the F1 and how many will have the same phenotype? F1 = Vv F1 = Vv P1/P2 V V VV v Vv v Vv vv Ratio of F2 generation with same phenotyoe: 3:1 or 75% Ratio of F2 generation with same genotype: 2:2 or 50% 2. Blue eyes are recessive to brown eyes. Show the expected children in a marriage between a blue eyed woman and a brown eyed man who had a blue eyed mother. P1 = bb P2 = BB or Bb – Parents of P2: Woman: bb Man: Bb or BB P1/P2 B B b Bb Bb b Bb Bb Therefore all children Bb, phenotype 100% brown eyes Or P1/P2 B b b Bb bb b Bb bb Therefore children either Bb or bb – only Bb will produce brown eyes Therefore P2 = Bb F1 generation: P1/P2 B b b Bb bb b Bb bb F1 generation: 50% brown eyes Bb 50% blue eyes bb 3. Two white sheep produce a black offspring. What must the parents’ genotypes for colour be? What is the probability that the next offspring will be black? (assume white is dominant - W) Assuming that white is dominant to black, possibly genotypes for parents are WW or Ww Cross #1: Parents WW and WW P1/P2 W W W WW WW W WW WW 100% WW, phenotype 100% white, therefore not black. Cross #2: Parents Ww and WW P1/P2 W w W WW Ww W WW Ww 50% WW = white 50% Ww = white, therefore not black Cross #3: Parents Ww and Ww P1/P2 W w W WW Ww w Ww ww 25% WW = white 50% Ww = white 25% ww = black Therefore the genotypes of the parents must be Ww and Ww. The ratio of white to black is 3:1, and the possibility that the next offspring will be black is 25%. 4. In guineas pigs, black coat colour is dominant to white. Suppose you had a black guinea pig whose parents were unknown to you, how could you find out whether the guinea pig was homozygous or heterozygous for black? Possible genotypes for my guinea pig: Bb or BB If I cross my guinea pig with a known heterozygous guinea pig, Bb then I will get the following crosses: Cross #1: Bb known, and Bb unknown P1/P2 B b B BB Bb b Bb bb 75% black, 25% white Cross #2: Bb known, and BB unknown. P1/P2 B b B BB B BB 100% black Bb Bb If I cross my guinea pig with a known heterozygote, then if I get 100% black offspring, I know it was homozygous, and if I get 75% black and 25% white, I know it’s heterozygous. 5. Suppose you were raising what you thought was a purebred line of cats, but found that some of the cats had spots of white fur at the tips of their tails, while others did not. How could you determine whether this trait is recessive or dominant? Cross a white tailed cat with a non-white tailed cat Possible crosses: Cross #1 WW x ww P1/P2 W W w Ww Ww w Ww Ww 100% Ww – which ever is dominant, 100% that colour Cross #2: Ww x ww P1/P2 W w w Ww ww w Ww ww 50% Ww – dominant colour 50% ww – non dominant colour Therefore if you cross a white tailed cat with a non-white tailed, depending on the ratio of kittens, you can determine whether or not it’s dominant 6. An extra finger in man is rare but is due to a dominant gene. When one parent is normal and the other parent has an extra finger, but is heterozygous for the trait, what is the probability that their first child will be normal? Their second child? P1 = FF P2 = Ff P1/P2 F F F FF FF f Ff Ff 50% FF – dominant 50% Ff – extra finger 50% chance their child will be normal, 50% chance their second child will be normal 7. Determine the probability that a plant of genotype CcWw will be produced from parental plants of the genotypes CcWw and Ccww. P1/P2 Cw Cw cw cw CW CCWw CCWw CcWw CcWw cW CcWw CcWw ccWw ccWw Cw CCww CCww Ccww Ccww cw Ccww Ccww Ccww ccww Probability of CcWw = 4/16 or 25% 8. If you thought we would stop at dihybrid crosses, you were wrong. Trihybrid crosses are those that involve 3 inherited characteristics where tri in Latin means “three”. What are the possible gametes for the following cross between AABBCC and aabbcc. What are the possible gametes of the F1 generation? Trihydrid cross: AaBbCc x AaBbCc P1/P2 ABC ABc AbC ABC AABBCC AABBCc AABbCC ABc AABBCc AABBcc AABbCc AbC AABbCC AABbCc AAbbCC Abc AABbCc AABbcc AAbbCc aBC AaBBCC AaBBCc AaBbCC aBc AaBBCc AaBBcc AaBbCc abC AaBbCC AaBbCc AabbCC abc AaBbCc AaBbcc AabbCc Abc AABbCc AABbcc AAbbCc AAbbcc AaBbCc AaBbcc AabbCc Aabbcc aBC AaBBCC AaBBCc AaBbCC AaBbCc aaBBCC aaBBCc aaBbCC aaBbCc aBc AaBBCc AaBBcc AaBbCc AaBbcc aaBBCc aaBBcc aaBbCC aaBbcc abC AaBbCC AaBbCc AabbCc AabbCc aaBbCc aaBbCc aabbCC aabbCc abc AaBbCc AaBbcc Aabbcc Aabbcc aaBbCc aaBbcc aabbCc aabbcc 9. Albinism (nn) in humans is inherited as a simple recessive trait where N represents normal. For the following families, determine the genotypes of the parents and offspring. When two genotypes are possible, list both. (a) Two normal parents have five children, four normal and one albino. (b) A normal male and an albino female have six children, all normal. (c )A normal male and an albino female have six children, three normal and three albino. a) 4:1 sounds like 3:1, which is the ratio for two heterozygous parents P1/P2 N n N NN Nn n Nn nn This cross shows a 3:1 ratio for normal to albino. b) Possible genotypes for normal parent: Nn or NN P1/P2 N n n Nn nn n Nn nn 50% normal, 50% albino, therefore not correct. Parents must be NN and nn. c) Possible genotypes for normal parent: Nn or NN P1/P2 N n n Nn nn n Nn nn 50% normal 50% albino Therefore the genotype of the parents are Nn and nn. 10. In human families it is often observed that certain characteristics may “skip” a generation or two and then reappear. How would you explain this in light of the facts learned in this unit? Phenotypes can “skip” a generation when a trait is dominant, and the middle generation is heterozygous. Then the dominant trait “covers” the recessive part of the gene, and then when the middle generation mates with someone else that is also heterozygous, the recessive trait can possibly show up when an offspring is made homozygous recessive. Rating scale: Assessment FOR learning 1 2 Ability to write the correct geno and phenotype (Knowledge) Ability to write complete conclusion sentence stating their answer in words (Communication) Ability to generate the correct crosses (Application) Ability to formulate the correct genotypes without being given them (Thinking/Inquiy) 3 4