Concept explainers
a.
To determine: The genotypes that are probable for the mother of the color-blind man.
Introduction: The color blindness is carried on by a faulty color vision gene on the X chromosome as a recessive disorder. Also, this marks about one in ten men and also has two forms.
To determine: The probabilities that the first child will be a color-blind boy.
Introduction: Red or green color blindness is carried from mother to son at the 23rd chromosome, which is identified as the sex chromosome as it further illustrates as sex chromosome.
c.
To determine: The proportion that can be estimated to be color-blind of any girl.
Introduction: A color-blind boy cannot inherit a color-blind ‘gene’ from the father, yet his father is color blind as his father can pass an X chromosome to the daughters.
d.
To determine: The proportion that can be expected to have a normal color vision of all the children.
Introduction: The color blind 'gene' is conducted on one of the X chromosomes. As men possess only one X chromosome, and if his X chromosome transfers the color blind 'gene' that is 'X,' he will be color blind XY.
Want to see the full answer?
Check out a sample textbook solutionChapter 2 Solutions
Introduction to Genetic Analysis
- Questions a to e are answerable by yes or no. Indicate the possible parental genotypes if your answer is yes.a. Can a man with hairy ears have a hairy-eared daughter?b. Can two normal parents produce a colorblind son?c. Can two normal parents produce a colorblind daughter?d. Can a colorblind woman have a normal son?e. Can a bald man have a nonbald daughter?arrow_forwardBlack eye (B) is dominant over brown eye (b). A man that is a homozygous dominant black-eyed marries a woman that is a heterozygous black-eyed. The offspring resulted to 50% homozygous black and 50% heterozygous black. If one of their heterozygous children marries another heterozygous partner. Regardless of its gender, what are the genotypes of their children? a. 25% BB, 50% Bb, 25% bb b. 50% BB, 25% Bb, 25% bb c. 25% BB, 25% Bb, 50% bb d. 0% BB, 75% Bb, 25% bb In Law of Codominance, a pure line dominant trait crossed with a recessive trait will result to the appearance of both dominant and recessive trait in the offspring. What is the percentage of having a red-haired offspring if the parental genotype is both a heterozygous roan haired? Red hair is a dominant trait and white hair is a recessive trait. a. 100% b. 75% c. 50% d. 25% e. 0%arrow_forwardSickle cell anemia is an inherited red blood cell disorder in which there are not enough healthy red blood cells to carry oxygen throughout the body. The allele that causes sickle-cell anemia is autosomal recessive (s), and the dominant allele can be represented by S. How many offspring will be affected by the disorder if the mother is a carrier, and the father appears to be normal? (Include the gender) a. b. How many will become carriers? (include the gender) A- 三三三 四 四 II !!arrow_forward
- Duchenne muscular dystrophy is sex linked and usually affects only males. Victims of the disease become progressively weaker, starting early in life.a. What is the probability that a woman whose brother has Duchenne’s disease will have an affected child?b. If your mother’s brother (your uncle) had Duchenne’s disease, what is the probability that you have received the allele?c. If your father’s brother had the disease, what is the probability that you have received the allele?arrow_forwardIn man, the allele for normal color (A) is dominant to the allele for albinism (a). A normal man whose father was albino married a normal woman whose mother was albino. a. What are the chances that their first child will be albino? b. What are the chances their second child will be albino?arrow_forwardColor blindness in humans is controlled by an X-linked completely recessive allele (Xc), while breast cancer is controlled by an autosomal completely dominant allele, B. A color blind male, who is a heterozygote carrier for breast cancer has three children/n with a normal eyed female (whose mother was color blind), who is homozygote recessive for the breast cancer allele. What is the probability that out of three children, 2 will be color blind males, and not show breast cancer, and one will be a color blind female, who shows breast cancer?arrow_forward
- In humans, the genes for coloblindedness and hemophilia re both located on the X chromosome with no corresponding gene in the Y. These are both recessive alleles. a. If a man and a woman, both with normal vision, marry and have a colorblind son, draw the Punnet square that illustrates this. b. If the man dies and the woman remarries to a colorblind man, draw a Punnet Square showing the type of children could be expected from hre second marriage. How many/what percentages of each could ne expectedarrow_forwardTay–Sachs disease is caused by recessive alleles on anautosome. In which case(s) could two parents with anormal phenotype have a child with Tay–Sachs?a. Both parents are homozygous for a Tay–Sachs allele.b. Both parents are heterozygous for a Tay–Sachsallele.c. One parent is homozygous for a Tay–Sachs allele,and the other is heterozygous.arrow_forwardA couple who are about to get married learn from studying their family histories that, in both their families, theirunaffected grandparents had siblings with cystic fibrosis(a rare autosomal recessive disease).a. If the couple marries and has a child, what is theprobability that the child will have cystic fibrosis?b. If they have four children, what is the chance that thechildren will have the precise Mendelian ratio of 3:1 fornormal:cystic fibrosis?c. If their first child has cystic fibrosis, what is theprobability that their next three children will be normal?arrow_forward
- The condition phenylketonuria is caused by a recessive allele. There are two carriers who have progeny.a. Give the gene notation. b. Give the expected genotypic and phenotypic ratios. c. What is the probability that their child will be heterozygous if they have a normal child?d. What is the probability of having two affected children and one normal child if they have three children?arrow_forwardA woman with a rare autosomal recessive disorder was told that it was unlikely that her children would have the disorderas her husband did not have it. However, her first child has the disorder. a. What is the most likely explanation? b. Diagram the cross between the woman and her husband using a Punnett square, give the genotypic ratio (GR) and phenotypic ratio (PR) from the Punnett square. c. Based on the Punnett square results, what is the chance that her next child will have the disorder?arrow_forwardColorblindness is inherited as an X-linked recessive trait while pattern baldness is controlled by an autosomal gene that is dominant in males but recessive in females. A colorblind man who is also homozygous for baldness has children with a woman who carries normal genes for both traits. What is the probability that any of their child will be: a. Colorblind, bald male b. Colorblind, normal-haired male c. Female with normal sight and baldarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning