1. A homozygous yellow pea plant is crossed with a homozygous green pea plant, Knowing that yellow is the dominant trait for pea plants:

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1 Genetics Homework Bio 120 Spring A homozygous yellow pea plant is crossed with a homozygous green pea plant, Knowing that yellow is the dominant trait for pea plants: Y = yellow y = green B) Genotype of yellow pea plant: YY C) Genotype of green pea plant: yy D) Draw the Punnett square of the cross Y Y y Yy Yy y Yy Yy E) what percentage of the offspring will be yellow. 100% 2. In lizards, stripes on the tail are dominant over not having stripes on the tail. Answer the following questions regarding the cross between a lizard that is heterozygous for stripes with a lizard with no stripes. S = stripes s= no stripes B) Genotype of lizard with strips: Ss C) Genotype of lizard with no strips: ss D) Draw a Punnett square of the cross S S s Ss Ss s Ss Ss E) What is the probability that they will produce an offspring that does not have stripes? 100%

2 3. In humans brown eyes are dominant and blue eyes are recessive. A brown-eyed couple already has a child with blue eyes. B = brown b = blue B) Genotype of parents: both are Bb C) Draw a Punnett square of the cross B b B BB Bb b Bb bb D) What is the probability that their next child will have blue eyes? 25% 4. Type I diabetes has been found to be inherited (in most cases) through a recessive allele. A non-diabetic woman, whose father was a diabetic, marries a diabetic man. D= normal d = diabetic B) Genotype of the woman: Dd C) Genotype of her husband: dd D) Draw a Punnetts of the cross D d d Dd dd d Dd dd E) What is the probability that their next child would be expected to have diabetes? 50%

3 5. What are the possible phenotypes of the children if the mother has Type A blood and the father has type AB blood? (Draw a Punnett square to verify your answer, you may need more than one Punnett square.) I A = A I B = B i = o B) Genotype(s) of the mother: I A I A or I A i C) Genotype of the father: I A I B D) Draw Punnetts squares: I A I A I A I A I A I A I A I B I A I B I A I B I A I A I A I A I A i I B I A I B I A i i E) Possible blood types (phenotypes): Type A, Type AB 6. Albinism is an autosomal recessive disease. A couple comes to you for advice. Both of their fathers had albinism. (the couple has normal pigmentation) A = normal a = albinism B) Genotype of the woman: Aa C) Genotype of the man: Aa D) Draw a Punnetts Square of the cross: A A AA Aa a Aa aa a E) What is the probability that their child will have albinism 25% F) What is the probability that their child will be carriers for the disease. 50%

4 7. Red-green color blindness is due to a sex-linked recessive allele on the X chromosome. Two normal visioned parents produce a color-blind son. X B = normal X b = color blind B) Genotype of the woman: X B X b C) Genotype of the man: X B Y D) Draw the Punnetts square of the cross X B Y X B X B X B X B Y X b X B X b X b Y E) What is the probability that their next child will be color blind? 25% 8. Huntingtons disease is a dominant autosomal disease. What is the probability that a couple will have a child that will develop huntingtons if one person is heterozygous for Huntingtons and the other person is homozygous recessive. 50% H = Huntingtons h = normal B) Genotype of the person with Huntingtons: Hh C) Genotype of the person without Huntingtons: hh D) Draw the Punnets square H h Hh hh h Hh hh h 9. An organism with the genotype of AaXx can produce gametes containing what alleles if the two genes are unlinked. A. either Aa or Xx B. either AX, Ax, ax, or ax C. AaXx D. AX or ax E. None of the above.

5 10. In jimson weed, the allele for violet flowers (V) is dominant to the allele for white (v) flowers. At another locus, the allele that makes prickly seed capsules is dominant (P) over the allele smooth ones (p). Two plants that are both heterozygous for both traits are crossed. Assume the two genes are unlinked. Using the probabilities from each trait, determine what ratios of the phenotypes are expected to be seen in the next generation: V = violet v = white P = prickly p = smooth VvPp x VvPp ¼ white ¾ violet ¼ smooth ¾ prickly V v P p V VV Vv P PP Pp v Vv vv p Pp pp white and smooth: ¼ * ¼ = 1/16 white and prickly: ¼ * ¾ = 3/16 violet and smooth: ¾ * ¼ = 3/16 violet and prickly: ¾ * ¾ = 9/16

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