c) heterozygous long x homozygous long Genotypes Phenotypes d) heterozygous long x heterozygous long Genotypes Phenotypes In man, normal pigmentation is due to a dominant allele "A" and albinism to its recessive allele "a". A normal man marries an albino woman and their first child is an albino. What are the genotypes of these three people?
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- This pedigree consist of cystic fibrosis that is an inherited disease caused by f a recessive allele. Find the genotypes of X and M? Given: O normal female □ normal male ◍ cystic fibrosis female ▨ cystic fibrosis male A) M= Ff X=Ff B) M=Ff X=ff C) M=ff X=ff D) M=ff X=FF6) For the pedigree shown below, answer the following questions. I II III a) What type of inheritance is shown? Explain how you know. b) The genotype of individual I -2 is The genotype of individual II -1 is1. Given the following pedigree below, use Punnett squares for each of the following possibilities: a) X- linked recessive and b) X-linked dominant in order to determine what is the mode of transmission of this trait. Disease allele = XA or Xa, depending on mode of transmission of the disease respectively. *Unaffected/No carrier-Normal Unaffected X chromosome = X I 11 * 1 2 1 2 3 a) X-linked recessive 11x12 III 2 genotype (circle one): XY 4 XAXA 2 3 5 Xaya *4 6 7 8 b) X-linked dominant 11x12 XAY Xay
- 6. a) Which individual (what number) in the pedigree below rules out X-linked dominant? b) Which individual in the pedigree below rules out autosomal recessive? Briefly explain why. c) Which individual in the pedigree rules out X-linked recessive? Briefly explain why. 1 2 II 1 2 II 1 2 3 4 5 6 7 8 IV 12) The length of your eyelashes follows an autosomal inheritance pattern. Long eyelashes (E) are dominant over short eyelashes (e). A woman with type AB blood who is heterozygous for eyelashes has children with a man who is heterozygous type-A blood and who has short eyelashes. a) Complete a Punnett square showing the possible genotypes of their offspring. b) What percent of their offspring would have short eyelashes and Type-A blood?Albinism in humans is inherited as a simple recessive trait.Determine the genotypes of the parents and offspring for the following families. When two alternative genotypes are possible,list both.(a) Two parents without albinism have five children, four withoutalbinism and one with albinism.(b) A male without albinism and a female with albinism havesix children, all without albinism.
- For the following problems, please choose from the following modes of inheritance: autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive 8080 50 OTO ㅁㅇㅇㅁ I. What is the most likely mode of inheritance portrayed in the pedigree above?In this case a family history revealed a genetic basis for the disorder. The pedigree is shown in Fig. 1 Below. Key Ø Female: affected Female: unaffected || IV V 5600 orize 077808 15 10 9 10 CHO વ Male: affected Male: unaffected Deceased Disease status not given Dizygotic twins Monozygotic twins Fig. 1 Disease pedigree. Five generations I, II, III, IV, V are shown. Females are represented by circles, males by squares, dizygotic (non-identical) twins by diagonal lines originating from the same point, Monozygotic (identical) twins by diagonal lines originating from the same point and joined symbols and deceased by a diagonal line through the symbol. Filled symbols indicate that the individual displays the disease phenotype. Unfilled symbols indicate that the individual does not display the disease phenotype. Carriers of the disease are not indicated. Information on disease status is not known for generation I and is omitted for the individuals represented by a symbol with an asterisk.…Consider the following pedigree. 하 3 10 (5 3 2 (a) What pattern of transmission is most consistent with this pedigree? (1) autosomal recessive, (2) autosomal dominant, (3) X-linked recessive, (4) X-linked dominant. (b) If individual V-2 marries a normal individual, and if the condition has a pene-trance of 85 percent, what is the probability that their second child will express the trait? (c) On the third line, what does the diamond with a 10 in the middle mean?
- Color blindness in humans is controlled by an X-linked completely recessive allele (Xc), while breast cancer is controlled by an autosomal completely dominant allele, B. A color blind male, who is a heterozygote carrier for breast cancer has three children/n with a normal eyed female (whose mother was color blind), who is homozygote recessive for the breast cancer allele. What is the probability that out of three children, 2 will be color blind males, and not show breast cancer, and one will be a color blind female, who shows breast cancer?Two autosomal mutations include albinism and dwarfism. Albinism (a) is recessive, and dwarfism (D) is dominant. Complete a dihybrid cross of two people that are heterozygous for normal pigmented skin. One person is normal height and has no genetic trace of dwarfism in the family. The other person has dwarfism but has a mother who is normal height. 21. Complete a full dihybrid Punnett square (6pts).For the following cross, show the P generation Genotypes and the Phenotypic ratio that would be seen in the F1 and F2. Remember, to produce the F2 generation you want to cross Heterozygotes from the F1. d) Genes 1 and 2 exhibit Epistasis (9:6:1) and Gene 3 is an Autosomal Dominant. In the P generation, the Male is Homozygous Recessive for the Genes showing Epistasis. Use E1, E2 and E3 to represent the Phenotypes shown by Epistasis. Report your results in the following format: P = aabb x AABB, F1 = 100%AaBb (Phenotype), and %3! F2 = 9/16 A_B_ (Phenotype), 3/16 aaB (Phenotype), 3/16 A_bb (Phenotype), 1/16 aabb (Phenotype)