One gene has alleles A and a: another gene has alleles B and b. For each of the following genotypes, what types of gametes will be produced? a. AABB b. АаBB c. Aabb d. AaBb с.
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For each of the following genotypes, what tyoes of gametes will be produced?
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- Earlobe attachment is controlled by a single gene. The free (F) condition is dominant over the attached (f) condition. A woman with free earlobes produced several children with a man having attached earlobes. The following results were obtained with the F1 offspring: 8 free, 7 attached. What are the parents’ genotypes? a. A heterozygous free-lobed man is mated to a heterozygous free-lobed woman. What are the genotypes and phenotypes of their possible offspring?A one-eyed flying people-eater usually has purple skin. One-eyed flying people-eaters who are recessive for skin color are albino. A heterozygous purple one-eyed people-eater has offspring with an albino female one-eyed people-eater. Value: 1 Which of the following are the genotypes of the parents? (select one for each parent.) [mark all correct answers] a. Father - PP b. Father - Pp c. Father - pp d. Mother - PP e. Mother - Pp f. Mother - pp g. Not enough information is given to determine the father's genotype. h. Not enough information is given to determine the mother's genotype.In one of his experiments, Mendel crossed homozygous yellow plants with homozygous green plants. The resulting F1 generation was allowed to self-fertilize. The F2 generation produced 930 yellow seeds and 305 green seeds. What are the genotypes of this F2 generation? O YY : yy O Yy : yy O Yy: yy O YY: Yy : yy
- #1 a) An allele for coat color in mice, Y, produces a yellow coat color when present. The phenotypes of the F1 generation are shown. A cross between a Non-yellow mouse and yellow mouse produces half yellow and half non-yellow progeny (offsprings). Non-yellow happens to be white in this example, but it does not need to be; it can be black, grey, or some other mouse color. Identify the genotype of the "Dead" phenotype. Describe the rationale behind your genotypic selection. * Pgeneration Yellow Non-yellow Yellow Yellow Gametes Y Meiosis Fertilization 1/2 1/2 Yellow Non-yellow Fi generation Yellow Yellow Non-yellow Dead Support | Schoology Blog | PRIVACY POLICY | Terms of Use INTLMendelian Genetics Consider blue eyes in a man as recessive to brown eyes. Show the expected children of a marriage between a blue-eyed woman and brown-eyed man who had a blue-eyed mother. Determine the genotypic ratio (GR) and phenotypic ratio (PR) of the F1 using Punnett Square Method. Hint: Determine the genotypes of each individual first.The following cross is set: DDEeFFggHh X DdEEFfGGhh a. What is the maximum number of additive alleles possible in their offspring? b. What is the maximum number of non-additive alleles possible in their offspring?
- AsapIn humans, there are three alleles for blood type: A, B, and O. A and B are codominant over O. A male with type AB blood and a female with type A blood have a child. The male’s parents both had type AB blood. The female’s mother had type A blood and her father had type B blood. What is the potential phenotype of the child? A. type A (25%), type B (25%), type AB (25%), type O (25%) B. type A (25%), type B (50%), type AB (25%) C. type A (25%), type B (25%), type AB (50%) D. type A (50%), type B (25%), type AB (25%)If body-color B is dominant to green body color b what would the genotype be for a heterozygous individual be? Using the blue B and a green b body color from above create a Punit square that represents the offspring of two heterozygous parents? what is the probability that they will have a child with a green body-color If the child is green what is its genotype? Cross a green parent with a homozygous blue parent. What color will their kids be?
- Consider the following cross examining four gene in two parental line: Parent 1: A/a; B/B; D/d; E/e Parent 2: A/a; B/b; d/d; e/e Assuming independent assortment for the four genes, what fraction of progeny will be phenotypically identical to either parent 1 or parent 2? (Hint: first figure out the fraction of progeny that resembles parent 1 and parent 2 separately, then get the overall fraction.) 9/16 1/16 3/4 3/8 3/16Consider the following genotype: (P1) Ala BC/be D/d E/e EgH/fGh X (P2) Ala BC/bɛ D/d E/e fgh/fGh There are 10 mu. between B and C genes There are 5 mu, between F and G genes There are 10 m.u. between G and H genes A) What is the probability that an offspring will be of F and H phenotype? B) What is the probability that P1 produces a gamete that is A B c ? C) What would be the phenotypic ratio associated with the B and C genes in this cross.In rabbits, black hair depends on a dominant allele, B, and brown hair on a recessive allele, b. short hair is due to a dominant allele, S, and long hair to a recessive allele, s. If a true-breeding black short-haired male is mated with a brown long-haired female, describe their offspring. What will be the genotypes of the offspring? If two of these f1 rabbits are mated, what phenotypes would you expect among their offspring? In what proportions?