regating loci are (P#1) AABbCcddEdFf and (P#2) AabbCcDDEdFf. F per case letters indicate dominant traits; lower case letters indicate ts. (That is, genotypes AA and Aa give phenotype "A"; genotype aa enotype "a".) P#1 means parent number 1, and P#2 means parent n (a) How many different types of gametes can each parent produ
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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=FFii) State the genotypes of individuals # 1-5 in the following table using the letter "A". Use the uppercase letter to represent the dominant allele and lowercase letter to represent the recessive allele. Genotype Individual #1 # 3 #4 #5 If individuals # 2 and 3 have another son what are the chances that this son will be affected? b) i) What is the most likely mode of inheritance for this pedigree? ii) State the genotypes of individuals # 6-8 in the following table using the letter "B". Use the uppercase letter to represent the dominant allele and lowercase letter to represent the recessive allele. Individual # 6 7 # 7 # 8 GenotypeFor 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)
- 1) A cross is made between two plants differing in four independently-assorting gene loci, AABBCCDD x aabbccdd, to produce an F, which is then self-fertilized. If the capital letters represent alleles with a completely dominant phenotypic effect, (a) how many different genotypes are possible in the F;? (b) what proportion of the F; will be homozygous dominant for all genes? (c) what proportion of the F; would have an ABCD phenotype? 2) Would your answers to (a), (b), and (c) be different if the initial cross were AAbbCCdd x aaBBccDD?A standard three-point mapping is conducted for recessive mutations in autosomal genes purple eye (pr), curved wing ( c) and black body (b). Their wild type alleles are also used for genetic mapping. An F1 Drosophila female heterozygous for purple eye (pr), curved wing (c) and black (b) is crossed to a triply homozygous mutant male. The observed numbers and phenotypes of the offspring are as follows: 360 pr c b 380 pr+ c+ b+ 104 pr c+ b 96 pr+ c b+ 30 pr c b+ 20 pr+ c+ b 6 pr c+ b+ 4 pr+ c b PROVIDE THE FOLLOWING: A) State the order of genes on this chromosome. B) Calculate map distances between the gene pairs: pr-c, pr-b, c-b. Show calculations, state the number of map units and which gene pairs they refer to.10 cM separates two hypothetical autosomal human genes. The dominant alleles are have complete penetrance and will result to Crossed eyes (e+) and short thumbs (th+). Four children are born to a normal guy and cross-eyed, small-thumbed woman. Two of the children have short thumbs and the other two have crossed eyes. She is carrying her fifth child. What is the probability that this fifth child will be cross-eyed and have short thumbs?
- 6) 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 isAlbinism 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.Consider the following pedigree. Solid symbols represent individuals affected by the trait. Assume complete penetrance and non-variable expressivity. II 3 4 III 1 2 3 5 6 a) what is the mode of inheritance of this trait? b) Does the ratio of affected to unaffected offspring in generation III-1 to 1II-4 match the expected ratio for this mode of inheritance? Explain your answer in terms of the expected ratio versus the ratio observed. Give a reason for your answer. No mark is assigned for yes or no)
- || AB ||| 1 AA 0₂ AA 0₁ In this pedigree A and B represent alleles at a marker locus very closely linked to the disease locus. Affected individuals are shown as shaded. The disease status in III 1 (a female of 10 years) is unknown as yet as the disease does not onset until teenage years or soon after. Which of the following is correct? 2 AB BB OA. The probable pattern of inheritance shown by the disease in this family is autosomal recessive. OB. If recombination does not occur the probability that III 1 will be affected if she has an AB marker genotype is 1. OC. If recombination does not occur, the probability that III 1 will be affected if she has a BB marker genotype is 1. OD. If the recombination fraction between the disease and marker loci equals 0.04, the probability that III 1 will be affected if she inherits an AB marker genotype equals 0.96. OE. The probable pattern of inheritance shown by the disease in this family is X-linked recessive.1-1 I-1I II II-1 II-2 П-3 П-4 II III-1 III-2 III-3 a) In humans, brown eyes are dominant and blue eyes are recessive. Determine which individuals in the pedigree above would have blue eyes. Justify your decision. b) Predict the phenotype of the possible offspring between III-1 and someone who has blue eyes.In mice, the wild-type coat color, agouti (AA) [left mouse], is dominant to solid-colored, black fur (aa) [middle mouse]. However, a separate gene (C) is necessary for pigment production. A mouse that is homozygous for a recessive c allele at this locus is unable to produce pigment and is albino [right mouse] regardless of the allele present at locus A. Thus, the following genotypes are listed with their associated phenotypes: ~ A/A; C/C or A/A; C/c or A/a; C/C or A/a; C/c ==> "agouti" ~ a/a; C/C or a/a; C/c ==> "black" ~ A/A; c/c or A/a; c/c or a/a; c/c or a/a; c/c ==> "albino" This is an example of recessive epistasis, in which the recessive c allele "stands upon" the possible genotypes for the A gene (A/A, A/a, or a/a). If two agouti mice with the A/a; C/c genotype are mated, what is the expected phenotypic ratio in their offspring? A.9 albino, 4 agouti, 3 black B. 9 agouti, 4 albino, 3 black C.9 black, 4 albino, 3 agouti D.9 agouti, 4 black, 3 albino