For numbers 34-50. Given the following pedigree, answer the following questions.1st generation would be the grandparents, 2" generation would be the children of the grandparentsand the 3" generation would be the children of the 2"generation.Trait that would be track is autosomal dominant and recessive.Cleft chin (CC or Ce) = DominantNo cleft chin (ce) = RecessiveII1034-37. What is the expected Genotype of the grandfather?A. CCВ. СеС. сеD. Either A or B38-41. What is the expected genotype of number 4 given the following genes in the 1st generationA. CCВ. СеC. Either A or BD. Neither A nor 842-45. Who are the persons failed to inherit cleft chin from the grand mother?A. Number 5, 1, 8,9 and 7B. Number 2, 9, 7, 10 and 3C. Number 2, 5, 7, and 9D. Number 1, 3 and 10E. Nurmber 2 and 547-50. If number 10 married a person without a cleft chin and all the off-springs have cleft chin, what isthe genotype of number 10?A CВ. СсС. ссD. Either A nor B

Name: For numbers 34-50. Given the following pedigree, answer the following
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Description: For numbers 34-50. Given the following pedigree, answer the following questions.1st generation would be the grandparents, 2" generation would be the children of the grandparentsand the 3" generation would be the children of the 2"generation.Trait th

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For numbers 34-50. Given the following pedigree, answer the following questions. 1st generation would be the grandparents, 2" generation would be the children of the grandparents and the 3" generation would be the children of the 2"generation. Trait that would be track is autosomal dominant and recessive. Cleft chin (CC or Ce) = Dominant No cleft chin (ce) = Recessive II 10 34-37. What is the expected Genotype of the grandfather? А. СС В. Се C. cc D. Either A or B 38-41. What is the expected genotype of number 4 given the following genes in the 1st generation A. CC В. Се C Either A or 8 D. Neither A nor 8 42-45. Who are the persons failed to inherit cleft chin from the grand mother? A. Number 5, 1, 8,9 and 7 B. Number 2, 9, 7, 10 and 3 C. Number 2, 5, 7, and 9 D. Number 1, 3 and 10 E. Number 2 and 5 47-50. If number 10 married a person without a cleft chin and all the off-springs have cleft chin, what is the genotype of number 10? A C В. Сс С. сс D. Either A nor B

For numbers 34-50. Given the following pedigree, answer the following questions. 1st generation would be the grandparents, 2" generation would be the children of the grandparents and the 3" generation would be the children of the 2"generation. Trait that would be track is autosomal dominant and recessive. Cleft chin (CC or Ce) = Dominant No cleft chin (ce) = Recessive II 10 34-37. What is the expected Genotype of the grandfather? А. СС В. Се C. cc D. Either A or B 38-41. What is the expected genotype of number 4 given the following genes in the 1st generation A. CC В. Се C Either A or 8 D. Neither A nor 8 42-45. Who are the persons failed to inherit cleft chin from the grand mother? A. Number 5, 1, 8,9 and 7 B. Number 2, 9, 7, 10 and 3 C. Number 2, 5, 7, and 9 D. Number 1, 3 and 10 E. Number 2 and 5 47-50. If number 10 married a person without a cleft chin and all the off-springs have cleft chin, what is the genotype of number 10? A C В. Сс С. сс D. Either A nor B

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

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1. Construct a pedigree chart of the trait traced in the situation below.2. The pedigree chart must be constructed with the names and genotypes at the bottom of every individual in the chart.If it is autosomal: Always use the first letter of the dominant trait to represent the dominant allele. For example, if the dominant trait is red, so the allele R-red, while r-yellow. For example, if the person is homozygous dominant for curly hair, we write CC as his genotype. If sex-linked, do not forget to write the parents’ chromosomes (XX-female, XY-male), with their corresponding superscripts (dominant or recessive). The superscript must be based on the first letter of the disorder. For example, the mother is a carrier of hemophilia (which is x-linked recessive), so we write XHXh. If the father has hemophilia, we write XhY. Put the pedigree chart using code properly. The situation is given below: Trait: NeurofibromatosisThe dominant form is neurofibromatosis, caused by the production of…
1. Construct a pedigree chart of the trait traced in the situation below.2. The pedigree chart must be constructed with the names and genotypes at the bottom of every individual in the chart.If it is autosomal: Always use the first letter of the dominant trait to represent the dominant allele. For example, if the dominant trait is red, so the allele R-red, while r-yellow. For example, if the person is homozygous dominant for curly hair, we write CC as his genotype. If sex-linked, do not forget to write the parents’ chromosomes (XX-female, XY-male), with their corresponding superscripts (dominant or recessive). The superscript must be based on the first letter of the disorder. For example, the mother is a carrier of hemophilia (which is x-linked recessive), so we write XHXh. If the father has hemophilia, we write XhY. Put the pedigree chart using code properly. The situation is given below: Trait: NeurofibromatosisThe dominant form is neurofibromatosis, caused by the production of…
11. Below is a pedigree chart of an autosomal recessive disorder. Answer the following questions using the correct genetic terminology (do not just write letters like “Ee”). A. What is the genotype of individual 1 in generation II? B: What is the genotype of individual 2 in generation I?C: Is it true that individuals 6, 7, 8, 9, and 10 in generation III all have the same genotype? Why or why not?
TOPIC : Genetics Example's for Question GENE = Hair Color ALLELE = Straight, Curly DOMINANCE = BB RECESSIVE = bb GENOTYPE = BB, Bb, bb PHENOTYPE = Blue, Black HOMOZYGOTE = BB (Blue eye), bb (Black eye) HETEROZYGOTE = Bb (Blue eye) MONOHYBRID = Aa DIHYBRID = AaBb TRIHYBRID = AaBbCc POLIHYBRID = AaBbCcDd MENDELLIAN GENETICS = Hair Color NON MENDELLIA = Blood Type QUESTIONS: Give Example of the following fo differentiate, DO NOT GIVE THE MEANING! (Follow the instructions above!) 1. MONOHYBRID CROSS = ? DIHYBRID CROSS = ? 2. BACK CROSS = ? TEST CROSS = ?
5) Humans who have an abnormally high level of cholesterol are said to suffer from familialhypercholesterolemia.  The gene for this disorder is dominant (C). A man who isheterozygous for familial hypercholesterolemia marries a woman who is homozygousfor the recessive allele. What is the probability that they will have children that sufferfrom this disorder? please draw the puutnet square and also write the parents and gemtaics top the puutnet square thanks
PEDIGREE ANALYSIS ANSWER THE FOLLOWING QUESTIONS 1. The pedigree below shows a family's pedigree for colorblindness. Which sex can be carriers of colorblindness and not have it? 2. With this in mind, what kind of trait is colorblindness? 3. Why dies individual IV-7 have colorblindness? 4. Why do all the daughters in generation II carry the colorblind gene? 5. Name two (2) IV generation colorblind males.
1. Huntington’s disease is a dominant trait inheritance in humans. What is/are the possible genotype/s of the mother if her husband does not exhibit the disease, but their children unfortunately got the disease? A. hh only B. HH only C. HH or Hh D. Hh only 2. Huntington’s disease is a dominant trait inheritance in humans. What is/are the possible genotype/s of the children if both parents do not exhibit Huntington’s disease? A. Hh only B. hh only C. HH or Hh D. HH only 3. Huntington’s disease is a dominant trait inheritance in humans. What is the probability of having a normal child if the mother does not have the disease while her husband has the disease? A. 0% or 50% B. 100% C. 25% D. 0% or 25% E. 50% 4. Huntington’s disease is a dominant trait inheritance in humans. What is the probability of having a girl child with Huntington’s disease if both mother and father has the heterozygous genotype for the disease? A. 0% B. 50% C. 75%…
MODIFIED TRUE OR FALSE. In the following items, read each statement carefully. I. The law of segregation states that the two alleles for a heritable character separate during gamete formation and end up in a same gamete.II. The law of dominance states that one of the pairs of inherited traits will be dominant and the others recessive unless both the factors are recessive a. The first statement (I) is correct and the second statement (II) is wrong b. The first statement (I) is wrong and the second statement (II) is correct c. Both statements are correct d. Both statements are wrong MODIFIED TRUE OR FALSE. In the following items, read each statement carefully. I. Ecosystem is defined as the geographic area where plants, animals, and other organisms, as well as weather and landscape, work together to form a bubble of life.II. This system consists of a regularly interacting and interdependent components forming a unified whole. a. The first…
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Finger mid-digital hair (M) is dominant over the recessive trait of no mid-digital hair (m). Freckles (F) is dominant over the recessive trait of no freckles (f). (It is important to note that this example of mid-digital hair and freckles cannot be explained with simple genetics, but for the purpose of this question, assume that it can.)A woman that is heterozygous for freckles and has no mid-digital hair is married to a man with no freckles that is heterozygous for mid-digital hair. The couple is expecting a child. Draw a Punnett square of this cross. What is the probability that their offspring will have mid-digital hair? What is the probability that two offspring in a row will have no freckles? What is the probability of two offspring in a row having mid-digital hair and freckles?