Campbell Biology: Concepts & Connections, Books a la Carte Plus Mastering Biology with eText -- Access Card Package (8th Edition)
8th Edition
ISBN: 9780133909029
Author: Jane B. Reece, Martha R. Taylor, Eric J. Simon, Jean L. Dickey, Kelly A. Hogan
Publisher: PEARSON
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Textbook Question
Chapter 9, Problem 6TYK
Tim and Jan both have freckles (see Module 9.9), but their son Mike does not. Show with a Punnett square how this is possible. If Tim and Jan have two more children, what is the probability that both will have freckles?
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Mary and Dai both have freckles (a dominant trait). Their first child, Mariko, does not have freckles. Show using a Punnett square show how this is possible. If they have two more children who DO freckles (making for one without and two with) what is the probability that the NEXT child will NOT have freckles?
Frank and Susan both have freckles but they're firstborn son Lawrence does not demonstrate the characteristics. What are the most likely genotype of Frank Susan and Lawrence?
Unattached earlobes (E) are dominant over attached earlobes (e) [Figure (a) and (b)]. If a woman with unattached earlobes and a man with attached earlobes have children, what percentage of their children has the possibility of having unattached earlobes if the mother is homozygous for the trait?
Chapter 9 Solutions
Campbell Biology: Concepts & Connections, Books a la Carte Plus Mastering Biology with eText -- Access Card Package (8th Edition)
Ch. 9 - Connecting the Concepts 1. Complete this concept...Ch. 9 - Whether an allele is dominant or recessive depends...Ch. 9 - Edward was found to be heterozygous (Ss) for...Ch. 9 - Two fruit flies with eyes of the usual red color...Ch. 9 - A man with type B blood and a woman who has type A...Ch. 9 - Tim and Jan both have freckles (see Module 9.9),...Ch. 9 - Both Tim and Jan (problem 6) have a widows peak...Ch. 9 - In rabbits, black hair depends on a dominant...Ch. 9 - A fruit fly with a gray body and red eyes...Ch. 9 - A series of matings shows that the recombination...
Ch. 9 - A couple are both phenotypically normal, but their...Ch. 9 - Why do more men than women have colorblindness?Ch. 9 - In fruit flies, the genes for wing shape and body...Ch. 9 - Adult height in humans is at least partially...Ch. 9 - Heather was surprised to discover she suffered...Ch. 9 - In 1981, a stray black cat with unusual rounded,...Ch. 9 - SCIENTIFIC THINKING The breakthrough that led...
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- Both parents love strawberry blonde hair but hate the fact they can touch their noses with their tongues. What are the odds their child will be a strawberry blonde non-nose touching individual?arrow_forwardIf the woman (phenotypically normal) in question 5(a) married to a color-blind man and already had a color-blind son, what is the probability that they will have a color-blind daughter? Short Answer. Please provide your final answer as a fractionarrow_forwardGilbert's genetic makeup is homozygous for dark hair. Jan's is homozygous for blond hair. What proportion of their children are likely to be dark-haired? Explain.arrow_forward
- A couple has three children, all of whom have brown eyes and blond hair. Both parents are heterozygous for brown eyes (Bb), but one is blond (rr) and the other is a redhead (Rr). What is the probability that the next child will be a brown-eyed redhead? For full credit, you must show all workarrow_forwardJan and Michael are both carriers for the cystic fibrosis allele. If Jan and Michael have three children with the disease and then Jan becomes pregant with fratnernal twins, what is the probability that both twins will also have the disease?arrow_forwardThe Eastwoods are going to have a child. She has chestnut hair (MBdMBk), and he has dark brown hair. Find their child's possible hair colors and the probabilities of each. (Enter exact numbers as integers, fractions, or decimals.) blond strawberry blond bright red light brown reddish brown dark red medium brown chestnut auburn dark brown shiny dark brown glossy dark brown black shiny black glossy blackarrow_forward
- Jan and Michael are both carriers for the cystic fibrosis allele. If Jan and Michael have five children, what the probability that three or more of the children will have the disease?arrow_forwardBob and Joan know from a blood test that they are each heterozygous (carriers) for the autosomal recessive gene that causes sickle cell disease. If their first three children are healthy, what is the probability that their fourth child will have the disease?arrow_forwardMark has an autosomal recessive condition called sickle cell anemia, a serious blood disorder that is associated with painful complications. His spouse has not experienced symptoms of sickle cell anemia but their first child developed the disease. What conclusions can be drawn about the mother's genotype and the probability of having a second child with the condition? Answer Choices: A: her genotype is heterozygous and there is a 25% percent chance of having another child with the disorder B: her genotype is heterozygous and there is a 50% percent chance of having another child with the disorder C: her genotype is heterozygous and there is a 75% percent chance of having another child with the disorder D: she lacks the allele for sickle cell anemia in that she lacks symptoms of the disease. There is a 0% percent chance of having another child with the disorderMark has an autosomal recessive condition called sickle cell anemia, a serious blood disorder that is associated with…arrow_forward
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