In humans, the alleles for blood type are designated IA (A-type blood), IB (B-type blood) and i (O-type blood). What are expected frequencies of the phenotypes in the following matings? • Heterozygote A x Heterozygote B: •IA IB x IA I • IA IA. x IB IB • AB x O
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- Heterozygous Cc chickens express a condition called creeper, in which the leg and wing bones are shorter than normal (cc). The dominant C allele is lethal when homozygous. Skin color is determined by the alleles W and w. W_ chickens have white skin and ww chickens have yellow skin. In a mating between 2 chickens heterozygous for both of these genes: 1) What phenotypes will be observed in the progeny? 2) What fraction of the offspring will have each phenotype? Assign alleles. List possible phentoypes. Make gametes, draw Punnett square, and list fractions of genotypes and/or phenotypes of offspring,The ABO blood groups in humans are expressed as the I A , I B , and i alleles. The I A allele encodes the A blood group antigen, I B encodes B, and i encodes O. Both A and B are dominant to O. If a heterozygous blood type A parent (I A i) and a heterozygous blood type B parent (I B i) mate, one quarter of their offspring are expected to have the AB blood type (I A I B ) in which both antigens are expressed equally. Therefore, ABO blood groups are an example of: a. multiple alleles and incomplete dominance b. codominance and incomplete dominance c. incomplete dominance only d. multiple alleles and codominanceDraw a Punnet square indicating the F2 genotypes from the mating of an F1 male (QQBbHh) with an Fi female (QQBÜHH). Note: Make sure to write all the heterozygous genotypes as Gg, not gG.
- In cocker spaniels, the following genotypes and phenotypes are found: AABB = white A-bb = red aabb = lemon AaB- = black aaB- = liver %3D AABB = grey If two cocker spaniels of the genotypes AaBb x AABB are mated, and eight pups are born, what is the most likely distribution of coat colors? A black male is mated with a liver-colored female, and they produce the following pups: 3/8 black 1/8 red 3/8 liver-colored 1/8 lemon-colored. What are the genotypes of the two parents? A red female is mated with a liver-colored male, and one of the pups produced is lemon- colored. What are the genotypes of the parents? What proportion of these offspring would be expected to be black? A red female is mated with a liver-colored male, and one of the pups produced is lemon- colored. What proportion of these offspring would be expected to be black?Still referring to Problem 1, what will be the possible genotypes of offspring from the following matings? With what frequency will each genotype show up? a. AABB aaBB b. AaBB AABb c. AaBb aabb d. AaBb AaBbWhat is the probability that the following parental mating AABBCC x AaBbCc will produce an offspring of genotype AaBbcc (assume independent assortment of all gene pairs)? O 1/4 O 1/8 O 1/16 O 1/32 O 1/64
- Homozygous Dominant First Offspring First Two Offspring Use dihybrid cross to solve the following problems. Genotype НА На hA ha НА НН АА НН Аа Hh AA Hh Аа На НН Аa НН аа Hh Aa Hh aа hA Hh AA Hh Aa hh AA hh Aa ha Hh Aa Hh aa hh Aa hh aa Key: H = long hair h = short hair A = brown a = green What is the probability of having offspring that are homozygous dominant for both traits? O A. 1/16 ОВ. 1/8 о с. 1/4 O D. 1/2 DELL & #3 6. 7 9. * C0n the Asiatic Black Bear (a normal XX:XY mammal XX: species) there are two traits under investigation. One trait is a white chest spot which can be either circular (+) or oval (ov). The second trait relates to the fur color with two phenotypes: black (+) and a reddish brown (b). A reddish brown female with a circular chest spot was crossed to a black male also with a circular chest spot. Use this information and the data below on their offspring to answer the following questions: Males: All reddish brown colored with 75% circular spot and 25% oval spot. Females: All black colored with 75% circular spot and 25% oval spot a) Which general type of chromosome does each trait reside on? Fur color gene ________________ Chest spot gene _________________ b) What are the genotypes of the two individuals that were crossed? c) What would the phenotypes and phenotypic ratio be if the original female was crossed with a reddish brown male with an…Heterozygous Cp cp chickens express a condition called creeper, in which the leg and wing bones are shorter than normal (cp cp). The dominant Cp allele is lethal when homozygous. Two alleles of an independently segregating gene determine white (W-) versus yellow (ww) skin color. From matings between chickens heterozygous for both of these genes, what phenotypic classes will be represented among the viable progeny and what are their expected relative frequencies?
- A wild-type fruit fly (heterozygous for gray body color andred eyes) is mated with a black fruit fly with purple eyes. Theoffspring are wild-type, 721; black purple, 751; gray purple, 49;black red, 45. What is the recombination frequency betweenthese genes for body color and eye color? Using informationfrom problem 3, what fruit flies (genotypes and phenotypes)would you mate to determine the order of the body color, wingsize, and eye color genes on the chromosome?Suppose the mating had occurred between homozygous recessive male for a disease and a normal female known to be heterozygous. What would the probabilities be for the following outcomes? 3 offspring in the following order: normal, with disease, normal Use the factorial equation n!/x!(n-y)! * (p)x (q)n-x n = number of events (offspring) x = number of normal / dominant offspring n-y =total - number of abnormal / recessive offspring p = probability of producing a normal/dom q = probability of producing an abnormal/recIn onion, male sterility is produced when the nuclear genotype is aa and the mitochondrial gene S (sterile) are present. Any other combination of nuclear genotype and mitochondrial gene (including gene F for fertile) will result in a male fertile plant. Give the genotypic ratio and the phenotypic ratio or the percentage of male sterile and male fertile offspring that will be produced in the following crosses. 1. Aa + S male x aa + F female 2. Reciprocal cross of number 1. (Note that when we do reciprocal cross, we interchange/swap the genotypes of the parents (if there is a nuclear gene involved, you interchange the nuclear genotype as well). 3. Aa + S female x Aa + F male 4. Reciprocal cross of number 3.