Please help with this question! Based on the pedigrees of the Finnishfamilies, what type of inheritance doeslactose-intolerance have?Autosomal dominantO Autosomal recessiveSex limitedO Y-linkedO X-linked recessive In this activity, you will analyze some pedigrees and answer some questions to determine how the lactosetolerance/intolerance trait is inherited. You will also analyze DNA sequences. Assume the trait assorts independently and it isfully penetrant.Family BFamily AIIIIIIIV DIVVVFamily CFamily DIIIII610 11 12IVIVVVO Lactose tolerant femaleO Lactose tolerant male? Unknown phenotypeLactose intolerant femaleLactose intolerant maleDeceased

Answer: Introduction: Autosomal recessive inheritance is a type inheritance in which genetic…

Answered: Based on the pedigrees of the Finnish…

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Based on the pedigrees of the Finnish families, what type of inheritance does lactose-intolerance have?

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 5QP: Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of...

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A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
d/1n5NtidRwTwUzcDkDPi5Z9P SHPZ9IA-XH-pfftLbhNc/edit 1) @ Is Add-ons Help Last edit was 15 minutes ago | Calibri в I UA 12 + 3I | II 6 1 I 7. Construct a Punnett square for a cross between two heterozygous pea plants with violet flower color. a. What genotypes would you expect in the offspring? b. What percentage or ratio of each genotype would you expect in the offspring? !!!
There are six types of agglutinogen named C,D, E and c,d,e.the first three are dominant and last three are recessive.discuss
Genotype HA На hA ha НА НН АА НН Аа Hh AA Hh Aa На НН Аа НН аа Hh Аа Hh aa hA Hh AA Hh Aa hh AA hh Aa ha Hh Aa Hh aa hh Aa hh aa Key: H = long hair h = short hair A = brown a = green What is the probability that the fırst offspring will be either long-haired and brown-eyed or long- haired and green-eyed? 80% O5% 25% 50%
Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?
Use Chi-square to test which of 2 possible ratios is the most probable ratio of progeny kernels. Use this information to determine the cross type and parental genotype. Show your work here or on a separate sheet to receive full credit. Cross type does not refer to the type of epistasis!
For each gentype below indicate whether it is heterozygous (he)or homozygous (ho)
Niemann Pick Type C disease is a recessive disorder that causes the accumulation of cholesterol and other lipids in lysosomes, ultimately affecting both the liver and the nervous system. Below are the genotypes and phenotypes of offspring of a family with a history of Niemann Pick. 7 NN ( all normal phenotype) 3 Nn (all normal phenotype) 4 nn (1 early onset dementia, 1 mid-life onset dementia, 2 late-onset dementia). From this information, Niemann-Pick disease is an example of: A) variable expressivity B) incomplete dominance C) incomplete penetrance D) variable expressivity and incomplete penetrance E) multiple alleles
In man, muscular dystrophy is a condition in which the muscles waste away during early life and may result in a shorter life expectancy. It is due to a sex-linked, recessive gene. A certain couple has five children – three boys (ages 1yr, 3yrs, and 10yrs old) and two girls (ages 5yrs and 7yrs old). The oldest boy shows the symptoms of this disease. You are their family physician and they come to you for advice. What would you tell them about the chances of their other children developing the disease?
Which of these predictions would you make for a gene that is haplosufficient and essential for viability? A diploid heterozygous for a null allele and the wild-type allele will be viable. A diploid heterozygous for a null allele and a wild-type allele will be nonviable. A diploid homozygous for a null allele will be nonviable. A null allele would lead to nonviable haploid cells.