The Hardy-Weinberg Equation 2. Sickle-cell disease (SCD) is a blood disorder. It causes an abnormality in the blood's cells shape-the cell assumes the shape of a crescent. This also inhibits the cell's ability to carry oxygen, leading to painful symptoms and even death. The person inherits the alleles for SCD from their parent. A person who is heterozygous for the sickle cell allele (Ss) often shows no symptoms of the disease and is considered a carrier. A person who is homozygous recessive for the sickle-cell allele (ss) will exhibit signs and symptoms of sickle-cell anemia. If 9% of the population exhibits signs and symptoms of the disease, what percent of the population is a carrier (heterozygous individuals)? Frequency of the dominant allele Frequency of the recessive allele % homozygous dominant % homozygous recessive % heterozygous 3. A large population of moths contains 35% white moths caused by the double recessive genotype, bb. What are all the allelic and genotypic frequencies for this population? Frequency of the dominant allele Frequency of the recessive allele % homozygous dominant % homozygous recessive % heterozygous

The Hardy-Weinberg Equation 2. Sickle-cell disease (SCD) is a blood disorder. It causes an abnormality in the blood's cells shape-the cell assumes the shape of a crescent. This also inhibits the cell's ability to carry oxygen, leading to painful symptoms and even death. The person inherits the alleles for SCD from their parent. A person who is heterozygous for the sickle cell allele (Ss) often shows no symptoms of the disease and is considered a carrier. A person who is homozygous recessive for the sickle-cell allele (ss) will exhibit signs and symptoms of sickle-cell anemia. If 9% of the population exhibits signs and symptoms of the disease, what percent of the population is a carrier (heterozygous individuals)? Frequency of the dominant allele Frequency of the recessive allele % homozygous dominant % homozygous recessive % heterozygous 3. A large population of moths contains 35% white moths caused by the double recessive genotype, bb. What are all the allelic and genotypic frequencies for this population? Frequency of the dominant allele Frequency of the recessive allele % homozygous dominant % homozygous recessive % heterozygous

Biology Today and Tomorrow without Physiology (MindTap Course List)

5th Edition

ISBN:9781305117396

Author:Cecie Starr, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Christine Evers, Lisa Starr

Chapter9: Patterns Of Inheritance

Section: Chapter Questions

Problem 3DID

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The Xb locus on Alouatta genus, the howler monkey X chromosome has 2 alleles, b+ and b. The b+ allele causes the presence of the Xb surface protein on their granulocyte cells, while the recessive allele b which does not allow the antigen to appear. The Xb locus is at 15 centimorgans, from the GRG locus. The GRG locus produces a normal enzyme activity. Individual howler monkeys with one form of an eye condition have the grg/grg genotype, with the males grg/Y presenting the condition. A male howler monkey with the eye condition caused by this enzyme mutation and no Xb antigen has a normal female howler monkey daughter with Xb antigen. This howler monkey female daughter is expecting a little howler monkey from the mating from a male howler monkey from a different tree. A) Determine the genotype of the howler monkey daughter, from the problem. B) Determine the types and percentages of each gametes in oogenesis produced of the howler monkey daughter female, based on the map distance given.…
The woolly hair case. An inherited trait among Norwegians causes affected individuals to have very wavy hair, not unlike that of a sheep. The trait, called woolly, is very evident when it occurs in families: no child possess woolly hair unless at least one parent does. Imagine you are a judge, and you have before you a woolly-haired man suing his normal-haired wife for divorce because their first child has woolly hair but their second child has normal hair. The husband claims this constitutes evidence of his wife’s infidelity. Do you accept his claim? How is the trait inherited in the family? Justify your decision with your explanation of the inheritance of the condition in this family.
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selves in the offspring. The crosses are lled ssed differently in homozy- heterozygous individuals. This produces a con- dition called incomplete dominance. In such cases, both worked out in the same manner as indicated previously, but heterozygous offspring exhibit a phenotype interme- diate between that of the homozygous individuals. Some examples follow. Activity 2 Working Out Crosses Involving Incomplete Dominance 1. The inheritance of flower color in snapdragons illustrates the principle of incomplete dominance. The geno-type RR is expressed as a red flower, Rr yields pink flowers, and rr produces white flowers. Work out the following crosses to determine the expected phenotypes and both genotypic and phenotypic percentages. % of each phenotype: 2. In humans, the inheritance of sickle cell anemia/trait is deter- mined by a single pair of alleles that exhibit incomplete domi- nance. Individuals homozygous for the sickling gene (s) have sickle cell anemia. In double dose (ss), the…
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i). pronde. clear- cut. explenations.in differences/Similarīties, if ony, between." Extensnons of mendelren. mhentonce Vs. Non.- mendellen. nhentnce.
% of each phê 2. In humans, the inheritance of sickle cell anemia/trait is deter- mined by a single pair of alleles that exhibit incomplete domi- nance. Individuals homozygous for the sickling gene (s) have sickle cell anemia. In double dose (ss), the sickling gene causes production of a very abnormal hemoglobin, which crystallizes and becomes sharp and spiky under conditions of oxygen deficit. Heterozygous individuals (Ss) have the sickle cell trait; they make both normal and sickling hemoglobin. Usually these individuals are healthy, but prolonged decreases in blood oxygen levels can lead to a sickle cell crisis. Individuals with the genotype SS form normal hemoglobin. Work out the following crosses: a. Parental genotypes: SS × SS % of each genotype: 41 % of each phenotype: b. Parental genotypes: ss X Ss % of each genotype: % of each phenotype:
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