In humans, the glucose-6-phosphate dehydrogenase (G6PD) gene involved in sugar metabolism is X-linked. The dominant allele (G) codes for fast enzymes for normal cell functioning while the recessive allele (g) codes for slow enzymes, which are responsible for the reduced enzymatic activity of cells. However, the presence of slow enzymes has minimal effect in the overall functioning of heterozygotes, thus normal. Recessives, however, are more prone to oxidative reactions that speed up the cell damaging process, thus weak (but not lethal). If random inactivation occurs during dosage compensation, what would be the genotypes and phenotypes of the children of a heterozygous mother and a recessive father?

In humans, the glucose-6-phosphate dehydrogenase (G6PD) gene involved in sugar metabolism is X-linked. The dominant allele (G) codes for fast enzymes for normal cell functioning while the recessive allele (g) codes for slow enzymes, which are responsible for the reduced enzymatic activity of cells. However, the presence of slow enzymes has minimal effect in the overall functioning of heterozygotes, thus normal. Recessives, however, are more prone to oxidative reactions that speed up the cell damaging process, thus weak (but not lethal). If random inactivation occurs during dosage compensation, what would be the genotypes and phenotypes of the children of a heterozygous mother and a recessive father?

Human Biology (MindTap Course List)

11th Edition

ISBN:9781305112100

Author:Cecie Starr, Beverly McMillan

Publisher:Cecie Starr, Beverly McMillan

Chapter19: Introduction To Genetics

Section: Chapter Questions

Problem 9CT

See similar textbooks

Similar questions

Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?
In humans, the genetic disease cystic fibrosis is caused by a recessive allele (a). The normal (healthy) allele is dominant (A). What is the genotype of someone who has cystic fibrosis? What are the two different genotypes that a healthy person could have? If two people were both heterozygous for the cystic fibrosis gene, what fraction of their children would be likely to have this disease? Hint: Draw a Punnett square to figure it out.
Tay Sachs is a rare autosomal recessive disorder that causes mental and physical disabilities leading to death in infants. Affected individuals are lacking the enzyme hexosaminidase, causing lipids to build up in the brain.The HEXA gene on chromosome 15 codes for hexosaminidase, and a four base pair insertion in the gene results in an altered reading frame and non-functional enzyme being produced. Individuals who are carriers (heterozygotes) of the Tay-Sachs allele are not affected by the disease but appear to have increased protection against tuberculosis.The incidence of Tay-Sachs disease is much higher among Ashkenazi Jews originating from Eastern Europe than the general population of the United States. About 1 in 3 500 babies of Ashkenazi Jewish heritage are born with Tay-Sachs disease and about 1 in 30 Ashkenazi Jews are carriers compared to about 1 in 320 000 babies born with the disease and about 1 in 300 carriers in the general United States population. Ashkenazi Jews living in…
In humans the allele coding for normal metabolism of the amino acid phenylalanine (P) is dominant over the recessive autosomal allele (p) that causes the disease phenylketonuria, or PKU. Without treatment, this disease is characterized by brain damage due to the failure of the gene to code correctly for the enzyme phenylalanine hydroxylase. If a man marries a woman and both are heterozygous for hemophilia and heterozygous for PKU, what is the chance that they will have a child afflicted by both diseases? Your response MUST include:(a): list the traits involved (example: Color: Blue- B (dominate) Green - b (recessive) -Do not use this example as this is not part of the problem(b): identify if its a monohybrid or dihybrid cross and why (use prompt above) (c): create a punnet square that represents the problem above
In mice, the trait for high cholesterol is specified by a dominant allele designatedHC, whereas the wild-type allele for normal cholesterol levels is designated hc.Black fur is specified by a recessive allele designated bl, whereas the wild-typeallele which gives brown fur is designated BL. The genes for both of these traitsare 30cM apart on the same autosome. A brown female (#1) with high cholesterolis mated to a black male (#2) with normal cholesterol. The progeny from this crossinclude a brown male (#3) with high cholesterol and a black female (#4) withnormal cholesterol. Mouse #3 is mated to a brown female (#5) with normalcholesterol. If 4 mice are produced from this cross, what is the probability that all 4will have brown fur?
In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: IA, IB, and i. The IA allele codes for the A sugar, The IB allele codes for the B sugar, and the i allele doesn't code for any sugar. IA and IB are codominant, and i is recessive to both IA and IB. If an individual with type AB blood has a child with an individual with type O blood, what blood types could their children possibly have?
In a particular country in sub-Saharan Africa, a medical study revealed that 0.075% of the country's population are suffering from sickle-cell anemia. Sickle cell disease is caused by a mutation in the haemoglobin-Beta gene found on chromosome 11. Sickle cell anemia is controlled by a pair of alleles, H and h, where the individuals with the illness have the homozygous recessive genotype; those with normal red blood cells have the alternative genotype. Using the Hardy-Weinberg's equation, calculate the percentage of the population of the two genotypes for normal blood cells. In a few remote areas in that country, the percentage of sickle cell anemia reaches as high as 5%. Calculate the percentage of homozygous and heterozygous individuals with the normal blood cells and find the ratio for both phenotypes.
In rabbits, the color of body fat is controlled by a single genewith two alleles, designated Y and y. The outcome of this traitis affected by the diet of the rabbit. When raised on a standardvegetarian diet, the dominant Y allele confers white body fat,and the y allele confers yellow body fat. However, when raisedon a xanthophyll-free diet, a homozygote yy rabbit has whitebody fat. If a heterozygous rabbit is crossed to a rabbit withyellow body fat, what are the proportions of offspring withwhite and yellow body fat when raised on a standard vegetariandiet? How do the proportions change if the offspring are raisedon a xanthophyll-free diet?
Take the example of B-thalassemia, an autosomal recessive genetic disease that particularly affects people from around the Mediterranean. This disease is associated with an anomaly of hemoglobin, a protein essential for the transport of oxygen, which is composed of four chains: two alpha (a) and two beta (B). In case of B-thalassemia, the ẞ chains are produced in insufficient or no quantity in an individual homozygous recessive resulting in insufficient production of overall hemoglobin leading to anemia and other physiological challenges. The gene that controls the synthesis of the ẞ chains is located on chromosome 11. Here is part of the coding portion of this gene (which controls a total of 146 amino acids and of which you only see the portion 36 to 41) and one of the targeted mutations: 1. Give the sequence of amino acids from the template and mutated strands. 2. What type of point mutation is it? 3. Using the principles of the theory of evolution, explain briefly and generally why…
The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyA. Calculate the map distances separating the three genes.B. Is…
The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyWhat information do you know based on the question and your…
The following X-linked recessive traits are found in fruit flies:vermilion eyes are recessive to red eyes, miniature wings are recessiveto long wings, and sable body is recessive to gray body. A cross wasmade between wild-type males with red eyes, long wings, and graybodies and females with vermilion eyes, miniature wings, and sablebodies. The heterozygous female offspring from this cross, whichhad red eyes, long wings, and gray bodies, were then crossed tomales with vermilion eyes, miniature wings, and sable bodies. Thefollowing data were obtained for the F2 generation (including bothmales and females):1320 vermilion eyes, miniature wings, sable body1346 red eyes, long wings, gray body102 vermilion eyes, miniature wings, gray body90 red eyes, long wings, sable body42 vermilion eyes, long wings, gray body48 red eyes, miniature wings, sable body2 vermilion eyes, long wings, sable body1 red eyes, miniature wings, gray bodyWhat topic in genetics does this question address?