WHAT IS LIFE? GDE.TO BIOLOGY W/PHYSIO.
5th Edition
ISBN: 9781319272531
Author: PHELAN
Publisher: MAC HIGHER
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Chapter 9, Problem 5MC
Summary Introduction
Introduction:
A test cross is useful in establishing the genotype of an organism exhibiting the dominant
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A. Consider this cross before answering the following questions:A pea plant heterozygous for the first trait and recessive for the second trait is crossed with a pea plant heterozygous for the first trait and homozygous-dominant for the second trait.
1. If the first gene is demonstrating dominant epistasis to the second gene, what is the probability that the F1 progeny will expressa. the dominant feature for the second trait?b. the recessive feature for the second trait?c. Neither dominant nor recessive features for the second trait?
B. Labrador retrievers have 3 varieties of fur colour: yellow, chocolate or black. Two genes
are involved, each with a dominant (Y and B) and a recessive allele (y and b). The allele Y
codes for the ability to produce hair pigment: Yy and YY dogs have dark pigmented fur but
all yx, individuals are lighter coloured and yellow. The B allele codes for the ability to make
hair with the darkest colouration: BB and Bb individuals have black fur, bb dogs have a
lighter shade which is chocolate. In the blank Punnet Square below, calculate and state
the ratio of phenotypes that occur with a cross between two individuals with the genotype
XyBb
Demonstrate how males are at an increased risk of sex-linked recessive traits by crossing a female who is a carrier for Hunter syndrome with a normal male. The genotypes involved are XH, Xh, and Y. State the genotypes and phenotypes of the offspring.
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WHAT IS LIFE? GDE.TO BIOLOGY W/PHYSIO.
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- . Assuming no involvement of the Bombay phenotype(in case you’ve already read ahead to Section 3.2):a. If a girl has blood type O, what could be the genotypes and corresponding phenotypes of her parents?b. If a girl has blood type B and her mother has bloodtype A, what genotype(s) and correspondingphenotype(s) could the other parent have?c. If a girl has blood type AB and her mother is alsoAB, what are the genotype(s) and correspondingphenotype(s) of any male who could not be thegirl’s father?arrow_forwarda. State a hypothesis explaining the inheritance of flower color in painted tongues. b. Assign genotypes to the parents, F₁ progeny, and F2 progeny for all five crosses. c. In a cross between true-breeding yellow and true-breeding lavender plants, all of the F1 progeny are bronze. If you used F₁ plants to produce and F2 generation, what phenotypes in what ratios would you expect? Are there any genotypes that might produce a phenotype that you cannot predict from earlier experiments, and if so, how might this alter the phenotypic ratios among the F2 progeny?arrow_forwardThis is all about sex linked inheritance in biology.arrow_forward
- B. In Minx cats, the alleles TT yield a cat with a normal tail, the heterozygous alleles Tt yield a cat with a short tail or no tail at all, and the recessive alleles tt are lethal to the embryo. Using Punnett squares, determine the genotypes and phenotypes of all offspring from parent Minx cats hat both have short-tails.arrow_forwardA. Color-blindness is a recessive, sex-linked disorder in humans. A color- blind man has a child with a woman who is a carrier of the disorder. KEY: X= normal vision XC = color-blindness 2. Illustrate using a Punnett square the probability of having children who will have normal vision and children who will be color-blind. Guide Questions: a. What is the genotype of the male?. b. What is the genotype of the female? c. What is the chance that the child will be color-blind? d. What is the chance that a daughter will be color-blind? e. What is the chance that a son will be color-blind?arrow_forwardB. In poinsettias, the gene for red bract color (R) is incompletely dominant over its allele for white bract color (r). Heterozygous individuals have pink bracts. What ratios of bract color would you expect among the following crosses? 1. red x red 2. red x pink 3. white x pink 4. pink x pinkarrow_forward
- Homozygous dominant parent (PP), and a Homozygous recessive or just simply say recessive parent (pp): a. Fill in the Punnett square. Each box represents a genotype possibility for an offspring. b. Place the allele donated by each parent in the corresponding box. Now list the possible genotypes and their corresponding phenotype. c. If an individual's genotype is heterozygous, the dominant trait will be expressed in the phenotype. Give the percent possible for the phenotypes. P P p p Genotype: ______________ Phenotype: ______________ Phenotype % probable: __________ 2. Cross between a Homozygous dominant parent (PP) and a Heterozygous parent (Pp). Fill in as in step one. 3. Cross between a Heterozygous parent (Pp), and another Heterozygous parent (Pp). Fill-in as before. 4. Test cross: A test cross is between a recessive parent (pp), and a Heterozygous parent…arrow_forwardDistinguish the following terms . A complete dominance and co - dominance B. genotype and phenotypearrow_forwarda. In a type of plant, pink flowers (P) is dominant over white flowers (p), and tall (T) is dominant over short (t). Determine the genotypes of the parents in the following cross: Pink, tall x pink, short O / pink, tall: /s pink, short: /s white, tall: / white, short Genotypes: () (ii) Phenotypes: Pink, tall Pink, short b. A woman has blood type A MM. She has a child with blood type O MN. Which of the following men could not be the child's father? Explain your reasoning. Man Blood type Peter O NN Rob AB MN Doug B MN Simon A NN Mike АВ Мarrow_forward
- b. What is the probability that the mating of Fifi with the described male will produce a kitten affected with the disease? Please helparrow_forwardch of the following best describes why males cannot be carriers of sex-linked traits? A. Males cannot be carriers because their Y chromosome makes them immune to sex-linked traits B. This is incorrect, males can be carriers of sex-linked traits O C. Males cannot be carriers because they only inherit one X chromosome, either having the trait or not O D. Males cannot be carriers because their mothers pass on an X chromosome without the sex-linked traitarrow_forwardA. Theoretical: The Punnett square will be used to examine the theoretical outcome of possible monohybrid crosses. 1. The first cross is between a Homozygous dominant parent (PP), and a Homozygous recessive or just simply say recessive parent (pp): a. Fill in the Punnett square. Each box represents a genotype possibility for an offspring. b. Place the allele donated by each parent in the corresponding box. Now list the possible genotypes and their corresponding phenotype. c. If an individual's genotype is heterozygous, the dominant trait will be expressed in the phenotype. Give the percent possible for the phenotypes. P Genotype: p Phenotype: Phenotype % probable: 1| Page 2. Cross between a Homozygous dominant parent (PP) and a Heterozygous parent (Pp). Fill in as in step one. 3. Cross between a Heterozygous parent (Pp), and another Heterozygous parent (Pp). Fill-in as before. 4. Test cross: A test cross is between a recessive parent (pp), and a Heterozygous parent (Pp). Again, fill-in…arrow_forward
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