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- For each phenotype, list the genotypes (remember to use the letter of the dominant trait). Straight hair is dominant to curly straight curly Pointed heads are dominant to round heads pointed round Long tails are dominant over short tails Long hair is dominant over short hair long tail short tail long hair short hairParent one's genotype for a trait is Bb Parent two's genotype for a trait is Bb. What are the possible gametes for both parents. Example: Parent one's gametes can be: Parent two's gametes can be:People with recessive disorders are usually born to normal parents who are both heterozygotes which means that both parents are carriers of the dominant allele for the disorder but appear normal themselves. Select one: True False
- The allele for albinism is recessive to the allele for normal skin pigmentation. All individuals who are homozygous for this recessive allele (m) are unable to produce the enzyme needed for melanin production and are referred to as albinos. In the following statements, determine the correct genotypes: An albino male (genotype A) whose parents both have normal skin pigmentation (genotype B) marries a woman who does not have albinism (genotype C). This woman’s father is an albino (genotype D). The married couple has three children, two with normal skin pigmentation (genotype E) and one exhibiting albinism (genotype F)Hemophilia is a sex-linked trait. A person with hemophilia is lacking certain proteins that are necessary for normal blood clotting. Hemophilia is caused by a recessive allele so use "N" for normal and "n" for hemophilia. Since hemophilia is sex-linked, remember a woman will have two alleles (NN or Nn or nn) but a man will have only one allele (N or n). A woman who is heterozygous (a carrier) for hemophilia marries a normal man: a.What are the genotypes of the parents? b.What is the probability that a male offspring will have hemophilia? c. What is the probability of having a hemophiliac female offspring?The allele for albinism ( c ) is recessive to the allele for normal pigmentation ( C ). A normally pigmented woman whose father is an albino marries an albino man whose parents are normal. They have three children, two normal and one albino. Give the genotypes for EACH person listed.
- Dimples (D) are a dominant trait genetic trait. Complete a Punnett Square to determine the possible phenotypes and genotypes of a cross between a man who is homozygous dominant and a heterozygous woman. Choose the percentages correctly below. Dimples: 0% 25% 50% 75% 100% No dimples: 0% 25% 50% 75% 100% DD: 0% 25% 50% 75% 100% Dd: 0% 25% 50% 75% 100% dd: 0% 25% 50% 75% 100%In man, there is gradation in eye color and resulting to 9 phenotypes. These are (in order of the number of additive alleles): light blue, medium blue, dark blue, gray, green, hazel, light brown, medium brown, dark brown. Mr. A (dark brown eyes) and Mrs. A’s (light blue eyes) daughter, marries a man whose genotype is the same as herself. What is the probability that they could have: Dark brown eyed child? Hazel-eyed? What eye color is most likely to occur and what is its probability?The following pedigree shows the inheritance of Huntington’s disease, a fatal genetic disorder that causes neurodegeneration. Since signs and symptoms usually do not appear until adulthood, many who are carriers may not realize their risk of passing on the disease-causing allele. The following pedigree represents a family in which some people are affected by Huntington’s disease. Using just the information on this pedigree, is Huntington’s disease caused by a dominant allele or recessive alleles? What are the genotypes of the grandparents (I-1 and I-2)? What are the genotypes of the parents (II-6 and II-7)? If the parents above have another child, what is the chance that they will be affected by the Hungtington’s disease allele? What are the genotypes of the unaffected children (III-8, 9, 10)? What is the chance that the unaffected children above will pass on a Huntington’s disease allele to their children? What is the genotype of the affected child (III-11)? What is the chance…
- Tucker (Joe and Sue's child) married Bonnie, who also has Marfan Syndrome. Together they have three children and two of them do not have Marfan Syndrome. Based on the information above, fill in the blanks for the genotype of each person(s) listed below. HINT: type in "AA", "Aa", or "aa". If more than one genotype is possible for any person(s), then write only one of the possible genotypes. Bonnie's genotype is: Tucker's genotype is: Affected child's genotype is: Nonaffected children's genotype is:In a true breeding parental cross for an Autosomal recessive disease trait, the expected proportion of males in the F2 generation that are unaffected is:Hemophilia is an X-linked disorder that affects the body’s ability to create blood clots. The allele for normal blood clotting, XH, is dominant over the allele for hemophilia, Xh. An unaffected female that is not a carrier mated with an affected male. Which of the following rows identifies the possible genotypes of the offspring? Select one: a. Female Male XHXH and XHXh XHY and XhY b. Female Male XHXh XHY c. Female Male XHXh XHY and XhY d. Female Male XHXH and XHXh XHY