BIOLOGY
12th Edition
ISBN: 9781260169614
Author: Raven
Publisher: RENT MCG
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Chapter 13, Problem 1A
Summary Introduction
Introduction:
The genes, which are carried on the both of the sex chromosomes, are called sex-linked genes. The sex-linked genes not only determine the sex of an individual, they also affect some of the sex-linked traits.
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Hemophilia is a disease caused by a gene found on the X chromosome. Therefore, it is a sex-linked disease which is caused by the recessive allele. Suppose, a man with hemophilia marries a woman who is homozygous dominant for the trait. Predict the genotype of the man. (XH = normal; Xh= hemophiliac) *
a. XH Y
b. XHXh
c. XhXh
d. Xh Y
In order for a male to inherit a sex-linked (X-linked) disorder, he must receive the allele from: a. his mother only. b. either his mother or father. c. his father only. d. both his mother and father.
Skin color is a trait that is determined by a. strict dominant-recessive inheritance. b. incomplete dominant inheritance. c. codominant inheritance. d. polygenic inheritance.
Chapter 13 Solutions
BIOLOGY
Ch. 13.1 - Describe sex-linked inheritance in fruit flies.Ch. 13.1 - Prob. 2LOCh. 13.2 - Prob. 1LOCh. 13.2 - Explain the genetic consequences of dosage...Ch. 13.3 - Prob. 1LOCh. 13.4 - Prob. 1LOCh. 13.4 - Explain the relationship between frequency of...Ch. 13.4 - Prob. 3LOCh. 13.5 - Prob. 1LOCh. 13.5 - Prob. 2LO
Ch. 13.5 - Prob. 3LOCh. 13 - Inquiry question Mendel did not examine plant...Ch. 13 - What would Mendel have observed in a dihybrid...Ch. 13 - Prob. 2DACh. 13 - Why is the white-eye phenotype always observed in...Ch. 13 - In an organisms genome, autosomes are a. the...Ch. 13 - What cellular process is responsible for genetic...Ch. 13 - The map distance between two genes is determined...Ch. 13 - How many map units separate two alleles if the...Ch. 13 - How does maternal inheritance of mitochondrial...Ch. 13 - Which of the following genotypes due to...Ch. 13 - Prob. 1ACh. 13 - As real genetic distance increases, the distance...Ch. 13 - Down syndrome is the result of trisomy for...Ch. 13 - Genes that are on the same chromosome can show...Ch. 13 - The A and B genes are 10 cM apart on a chromosome....Ch. 13 - Prob. 6ACh. 13 - Color blindness is caused by a sex-linked,...Ch. 13 - Assume that the genes for seed color and seed...Ch. 13 - A low frequency of calico cats are male (about...
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- Muscular dystrophy A.Shows simultaneous presence of hypertrophic and atrophic muscle fibers in the same muscle. B.Shows degenerated muscles with intact nerves. C.Is an X-linked recessive trait. D.Causes failed nerve connection to muscles.arrow_forwardHemophilia is a disease caused by a gene found on the X chromosome. Therefore, it is a sex-linked disease which is caused by the recessive allele. Suppose, a man with hemophilia marries a woman who is homozygous dominant for the trait, what is the chance of having children who are hemophiliacs? (XH = normal; Xh= hemophiliac) * a. 0% b. 75% c. 100% d. 25%arrow_forwardWhich of the following is true of recessive sex-linked genes? A. A female showing the trait will pass it on to all of her sons who will all show the trait. B. A female showing the trait will pass it on to all of her daughters who will all show the trait. C. A male showing the trait will pass it on to all of his sons who will all show the trait. D. A male showing the trait will pass it on to all of his daughters who will all show the trait.arrow_forward
- In humans, normal skin color (A) is dominant over albino (a). An albino man marries a normal woman whose mother was an albino and whose father has normal skin color. What is the genotype of the man and the woman? What proportion of their children would be expected to be both normal skin pigmentation and have the trait for albino?arrow_forwardIn humans, hemophilia is an X-linked recessive condition characterized by the inability of blood to clot normally. A woman who is a carrier for hemophilia marries a man who has hemophilia. What are the genotypes of the woman and man, respectively? a. xH+xH+, xH+y b. xH+xH xH+y c. xH+xH xHy Od. XHXH, XH+y O e. xHxH xHyarrow_forwardHemophilia is a disease caused by a gene found on the X chromosome. Therefore, it is referred to as a sex-linked disease. The recessive allele causes the disease. A man with hemophilia (xhy) marries a woman who is homozygous dominant (XHXH. A. Illustrate using a Punnett square the probability that their children will have the disease. B. Will any of their children have the disease? C. Predict the probabilities of their children having the disease.arrow_forward
- Which of the following is NOT true of Y-linked traits? Select one: a. They are always expressed. b. They are extremely uncommon in females. c. They are passed from father to son. d. They carry both dominant and recessive alleles of a trait.arrow_forwardMany genetic disorders are inherited in a recessive manner Select one: a. False b. Truearrow_forwardWhen the phenotype of the offspring is determined by endoparasites in the father it is… a. epistasis b. due to genes located in the mitochondria. c. a maternal effect d. non-nuclear inheritance e. a non-additive genetic effectarrow_forward
- In humans, failure to synthesize melanin leads to a condition called albinism. This is a recessive condition. If parents heterozygous (Nn) for normal skin pigmentation give rise to 4 offspring, what is the probability that: a. they will have an albino child? b. the second child is albino? c. they will have all boys with normal pigmentation? d. they will have ONE albino child, no matter the order.arrow_forwardBaldness is a Sex Influenced Trait. A heterozygous bald man and a heterozygous nonbald woman plan to have two kids. What are the phenotypes for their kids? a.If both sons; they will all be bald. b.If both daughters; both will be bald. c.If both daughters; they will all be normal- nonbald. d.If male or female ; Either a bald or normal daughter and either a bald or normal sonarrow_forwardA boy is color-blind (X-linked recessive) and has a straight hairline (autosomal recessive). Which could be the genotype of his mother? Select one: А. Вbxwxw B. Bbww C. Xbxbww D. Xbywwarrow_forward
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