Cengage Advantage Books: Biology: The Dynamic Science, Loose-leaf Version
4th Edition
ISBN: 9781305655911
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Brooks Cole
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Chapter 12, Problem 16TYK
Having malformed hands with shortened fingers is a dominant trait controlled by a single gene; people who are homozygous for the recessive allele have normal hands and fingers. Having woolly hair is a dominant trait controlled by a different gene; homozygous recessive individuals have normal, nonwoolly hair. Suppose a woman with normal hands and nonwoolly hair marries a man who has malformed hands and woolly hair. Their first child has normal hands and nonwoolly hair. What are the genotypes of the mother, the father, and the child? If this couple has a second child, what is the probability that it will have normal hands and woolly hair?
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Hair texture is determined by gene c in an incomplete dominance inheritance pattern, where the heterozygous condition (c1c2) results in wavy hair. Hairline is controlled by gene H, where a widow's peak is dominant to a straight hairline.
If a woman with curly hair and a straight hairline has a child with a man with a widow's peak and straight hair (his mom had a straight hairline), what is the chance they would have a kid that has exactly the same hair as mom?
Draw the Punnett square and demonstrate the possible genotypes and phenotypes that could result from these two parents' mating.
Eye Color is a trait that involves a number of genes. To simplify our example, we can break down the traitinto two phenotypes: dark color and light color eyes. Dark coloration is considered dominant, or isexpressed in a heterozygous condition (when the alleles are different).Q: If you have one light eyed parent and one parent that is homozygous for dark eye color, what are thepotential eye colors of their offspring (young)? To start, what do the Parents’ alleles look like? What istheir Genotype? (Remember, you pick which letter you want to use)Light eye colored parent:Dark eye colored parent:One of the things Mendel designed was a way to provide information about how the genes are passed onfrom parent to offspring. He designed something called a Punnett Square. This square places the parents’alleles outside the box, then uses those alleles to fill in the squares.Let’s start with a reminder of what we know about the parents:
Parent
Phenotype(appearance)
Genotype(alleles)
1
2
Since the Titanic sank Jack and Rose never got the chance to have children together. Rose spent the rest of her life wondering what their children would have looked like. Jack had golden-brown hair and Rose has red hair. Would their children have hair like the mother or father? Jack is heterozygous dominant for brown hair and Rose is homozygous recessive for red hair. Use your own letters for the alleles.
Chapter 12 Solutions
Cengage Advantage Books: Biology: The Dynamic Science, Loose-leaf Version
Ch. 12.1 - Two pairs of traits are segregating in a cross....Ch. 12.1 - If, instead, the four phenotypes in question 1...Ch. 12.2 - Palomino horses have a golden coat color, with a...Ch. 12.2 - A true-breeding rabbit with agouti (mottled,...Ch. 12 - The dominant C allele of a gene that controls...Ch. 12 - In peas, the allele Le produces tall plants and...Ch. 12 - Prob. 3TYKCh. 12 - What genotypes, and in what frequencies, will be...Ch. 12 - In addition to the two genes in problem 4, assume...Ch. 12 - Prob. 6TYK
Ch. 12 - In cats, the genotype AA produces tabby fur color;...Ch. 12 - Prob. 8TYKCh. 12 - The ability of humans to taste the bitter chemical...Ch. 12 - A man is homozygous dominant for alleles at 10...Ch. 12 - In guinea pigs, an allele for rough fur (R) is...Ch. 12 - Prob. 12TYKCh. 12 - Some recessive alleles have such a detrimental...Ch. 12 - In chickens, feathered legs are produced by a...Ch. 12 - A mix-up in a hospital ward caused a mother with O...Ch. 12 - Having malformed hands with shortened fingers is a...Ch. 12 - Prob. 17TYKCh. 12 - Design an Experiment Imagine that you are a...Ch. 12 - Prob. 19TYKCh. 12 - Prob. 1ITDCh. 12 - Prob. 2ITDCh. 12 - Half of the worlds population eats rice at least...
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- Tanya has a family history of colorblindness but can see colors, and her partner Tim is affected by colorblindness. Colorblindness is an X-linked recessive condition with the alleles: XB (unaffected and dominant) and Xb (affected and recessive). Tanya’s mother was affected, but her father was not. Her partner Tim is also affected. What is the probability that if Tanya and Tim had a daughter that she would be affected? How about a Son?arrow_forwardArmin and Annie are going to have a baby. Annie has dimples in her cheeks (a dominant trait), while Armin does not. You know that Annie's father has dimples in both cheeks, while her mother does not. Her mother must have the recessive trait. Annie's father has the dominant trait, but you don’t know if he is a homozygote or heterozygote. But you still know what Annie’s genotype must be because Annie must have a recessive allele since that is all she could have inherited from her mother. Since Annie has a dimples you know she inherited a dominant allele from her father. What are the chances Armin and Annie’s baby will have dimples? Determine the Phenotype and Genotype and its probability. Show your solution in a clean sheet of paper.arrow_forwardIn humans, cystic fibrosis and albinism are both inherited as independently assorting autosomal recessive traits. Use f to represent the recessive allele for cystic fibrosis and F for the allele for the unaffected condition. Use A to represent the dominant allele for normal melanin and a for the recessive allele for albinism. A man is albino and heterozygous for cystic fibrosis. His wife has the unaffected phenotype for both traits, although her father had cystic fibrosis and her mother is albino. 4a. What is the genotype of the wife 4b.What is the probability that a child of this couple is albino and has cystic fibrosis? 4c.What is the probability that a child of this couple is a girl who is albino and has cystic fibrosis? 4d.What is the probability that a child of this couple is normal for both traits?arrow_forward
- Stefanie is a young farmer growing new and heritage pumpkins. In her rows of pumpkins she finds a dark orange pumpkin with warts and a green pumpkin with smooth skin. She knows that orange skin color is dominant in pumpkins, but she cannot find any information on the inheritance of warty skin in pumpkins. She decides to cross the pumpkins to determine if warty skin is a dominant or recessive trait. If you hypothesize that warty skin is the dominant trait... -Then, based on your hypothesis, predict the results of the F1 and F2 dihybrid crosses between the orange warty pumpkin snd the green smooth pumpkin. (Assume these pumpkins are homozygous for both traits.arrow_forwardAlbinism is the absence of pigment throughout the body. Animals with albinism have white fur and reddish eyes. Albinism is inherited as a recessive trait, so animals with two a alleles have albinism, whereas animals with at least one dominant A allele have normal coloring. Using the Punnett square below, determine the probability of two albino parents (with aa genotypes) producing a child with albinism.arrow_forwardThe shape of a heliodor is an X-linked trait. Male heliodors inherit two X chromosomes, while the females inherit an X and a Y chromosome. The recessive allele codes for the common bell-shape while the dominant phenotype is a round shape . Hudson is a round male and has offspring with Heidi, a bell-shaped female. They already have one daughter, Hannah, who is shaped like her mother. If their next child is a boy, there is a _____% chance it will be round.arrow_forward
- Sickle cell anemia is a blood disorder that is expressed with incomplete dominance. The homozygous recessive phenotype has sickle shaped red blood cells that cause anemia and often death if untreated. If an unaffected father and mother move to the United States from Sub-Saharan Africa and several of their offspring have sickle cell anemia, what do we know about both parents' genotypes? One is homozygous recessive, the other is heterozygous They are both heterozygous They are both homozygous recessive One is homozygous dominant, the other is heterozygousarrow_forwardGregor Mendel discovered the basis of heredity with his sweet pea plant experiments. In his studies, he determined that certain traits, such as pod color and pea shape, express complete dominance. Trait Dominant Recessive Pod color Yellow Green Pea shape Round Wrinkled A cross occurs between a plant with heterozygous yellow pods and wrinkled peas and a plant with green pods and heterozygous round peas. What is the probability that the offspring will exhibit recessive genes for both traits? a) 1/2 b) 1/4 c) 1/8 d) 1/16arrow_forwardPeople with polydactyly have more than five fingers or toes on their hands or feet. Polydactyly results from an autosomal dominant genetic disorder. Draw a pedigree for a family with polydactyly, using the following information: Zack has polydactyly but his wife, Zoe does not. Zack and Zoe’s daughter Yolanda has polydactyly but their son Yogi does not. Yolanda has two children with her husband, Xavier, who does not have polydactyly: Wilma, who is normal and Wade, who has polydactyly. Use your pedigree to determine what is Zoe's genotype. A) DD B)dd C)Ddarrow_forward
- The shape of a heliodor is an X-linked trait. Male heliodors inherit two X chromosomes, while the females inherit an X and a Y chromosome. The recessive allele codes for the common bell-shape while the dominant phenotype is a round shape . Hudson is a round male and has offspring with Heidi, a bell-shaped female. They already have one daughter, Hannah, who is shaped like her mother. What percentage of their offspring will be carriers for the bell-shape trait? a.0% b.25% c.50% d.75% e.100%arrow_forwardImagine that a couple is planning to have children. The male is heterozygous for Huntington’s disease and homozygous dominant for Tay-Sachs. The female is homozygous recessive for Huntington’s disease and heterozygous for Tay-Sachs. The couple is curious about the possibility and probability of their offspring inheriting Tay-Sachs and/or Huntington’s. For humans, Huntington’s disease is dominant (H) over the “normal” condition (h), and the “normal” condition is dominant (T) over Tay-Sachs (t). Complete a Punnett square for this cross and record the probabilities for genotypes and phenotypes of the offspring as ratios.arrow_forwardThe shape of a heliodor is an X-linked trait. Male heliodors inherit two X chromosomes, while the females inherit an X and a Y chromosome. The recessive allele codes for the common bell-shape while the dominant phenotype is a round shape . Hudson is a round male and has offspring with Heidi, a bell-shaped female. They already have one daughter, Hannah, who is shaped like her mother. There is [blank] % that their next child will also be bell-shaped female like Hannah.arrow_forward
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