QUESTION 1 Consider three linked genes: A, B, C. Percent recombination between gene B and C is 11% and no crossing over occurs between A and B. If a heterozygote in cis conformation is test crossed, what is the probability of an offspring with the genotype AaBbcc? A. 0.011 B. 0.055 C.0.022 D.0.089 E. 0.045

QUESTION 1 Consider three linked genes: A, B, C. Percent recombination between gene B and C is 11% and no crossing over occurs between A and B. If a heterozygote in cis conformation is test crossed, what is the probability of an offspring with the genotype AaBbcc? A. 0.011 B. 0.055 C.0.022 D.0.089 E. 0.045

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter12: Mendel's Experiments And Heredity

Section: Chapter Questions

Problem 18RQ: The forked line and probability methods make use of what probability rule? test cross product rule...

See similar textbooks

Similar questions

Extra Question Chapter 4 1. Consider the following cross concerning 4 different gene loci: AaBbCcDd (x) AabbCcdd From this cross, what is the probability of getting a progeny (offspring) with genotype AABbccdd? b. From this cross, what is the probability of getting a progeny (offspring) with genotype AabbCcDd? c. From this cross, what is the probability of getting a male progeny (offspring) with genotype aaBbCcdd? 2. Your neighbor has twelve children. One is blue eye color and short. Two are brown eye color and short. Two are blue eye color and tall. Seven look just like the parents; brown eye color with tall. What can you discover about the genetics of eye color and height of the children? a. How many traits are you dealing with? Each trait has phenotypes: Specify the phenotypes. b. What is the probability of the height of the children? What is the probability of the eye color of the children? (Refer to monohybrid punnett square slides 17-19) c. What are recessive traits based on the…
Assignment 3 Linkage and Recombination 1. In corn, the genes an (anther ear), br (brachytic), and f (fine stripe) are linked. Testeross data are as follows: Number Number 355 2 Progeny Progeny +++ 88 an ++ ++f + br + + br f 21 an +f an br + an br f 2 17 399 55 Determine the linkage map and the genotype of the homozygous parents used to obtain the heterozygote for testcross.
QUESTION 6 Genes a, b, and c are linked, and pairwise crosses gave the following distances: a-b 20 cM b-c 2 cM a-c 22 cM Which gene is in the middle? O a O It's not possible to tell without additional information. QUESTION 7 A man and woman both have tpe AB blood and are carriers for the Bombay mutation. What is the probability their first child will be phenotypically O (Bombay)? O 1/4 O 1/16 O 1/2 O 3/16 QUESTION 8 The Bombay phenotype occurs because: O Rh antibodies can travel across the placenta and cause an incompatibility reaction. O an individual is homozygous recessive at the Rh locus. O the H substance is not present and so no A or Bantigens can be produced. O an individual does not haveor alleles, and so cannot produce the corresponding antigens.
Question: Black hair of the mouse is produced by dominant gene B and white hair by recessive allele b.Female with black hair crossed with male black hair (both genotypes are heterozygous). Produced individuals 112 and 113, female with white hair and male with black hair, then mating occur between 114 and 113, and another mating occur between 112 and 111. assume that 111 and 114 individuals do not carry the recessive allele, represents the above in a pedigree chart and determine the genotype of each individual.
Question: This is a normal 3 point test cross, except that instead of regular phenotypes, you are looking at DNA markers on a gel. One parent, according to the gel, is heterozygous at each marker. The other parent is homozygous for each marker. (Again, this means it is a test cross: AaEeHh x AAEEHH --but don't be confused by that, because these are not "dominant" and "recessive" per se; the phenotype is just a band on a gel). For each offspring, figure out its genotype (homozygous or heterozygous for each gene. Then, figure that one parent made only AEH gametes, so you can cross that out if it helps.) Then treat it pretty much the same as a 3 point test cross.
Question- 1. In onions, male sterility is due to the interaction of a chromosomal allele pair hh and “sterile” (S) cytoplasm. All other combinations (i.e. HH/Hh and “sterile” cytoplasm, HH/Hh or hh “fertile” (F) cytoplasm) result in male-fertile plants. The male-sterile trait is incorporated into inbred lines to produce hybrid F1 seed on a commercial scale.a) How would you perpetuate the male-sterile line? Show the cross.b) Briefly outline the method of producing hybrid seed (heterozygote) for the commercial crop. Show the cross.c) Does it matter whether the cytoplasm is fertile or sterile in the male-fertile inbred? Explain.
Questions and Problems: Show your solutions. 3. If two gene pairs A and a and B and bare assorting independently with A dominant to a and B dominant to b, what is the probability of obtaining: а. an AB gamete from an AaBb individual? b. an AABB zygote from an AaBb x AaBb cross? С. an aaBb offspring from an AaBB x aabb cross? d. a phenotype aB from an Aabb x aaBb cross? е. an ab gamete from an aaBB individual?
GENETICS Question; 1. Give the genotypes and its ratio of the dihybrid cross of FIGURE 7. Regarding this problem, you may refer the genotype of the monohybrid cross of pea flower color in Figure 6.
3-4. Two strains of Neurospora crassa are crossed, one of genotype o'm g* and the other of genotype o m g". All three genes are on the same linkage group, in that particular order. 3. In which octad(s) did crossing over event(s) occur? a. o'm*g* o*m*g* o'm*g¯ o*m*g¯ omīg* om°g* o m¯g¯ o°m¯g¯ b. o'm*g* o`m*g* omīg* om¯g* o°m*g¯ o°m*g¯ o°m¯g¯ omīg c. o'm*g* o*m*g* o'mg¯ o*m¯g¯ om*g* o¯m*g* o m°g¯ o°m°g¯ d. all of the above e. none of the above 4. In which octad(s) did gene conversion event(s) occur? a. om*g* oʻm*g* oʻm*g o'm*g omg* om¯g* om¯g¯ o°m°g¯ b. o'm'g* o'm°g* o`m°g* om°g* o°m°g¯ o°m°g° o°m°g omīg c. O'm*g* o'm*g* o'mīg¯ o'm¯g¯ o'm*g* o°m*g* o m°g¯ o°m¯g¯ d. all of the above e. none of the above
The forked line and probability methods make use of what probability rule? test cross product rule monohybrid rule sum rule
Calibri 12 - BIU A Section 4: Monohybrid test cross A monohybrid cross is a mating where alleles of just one gene are tracked in the offspring. A test cross is a mating in which two parents (P generation) are crossed to give F1 (first filial generation) offspring, and then two F1 parents are crossed to give F2 (second filial generation) offspring. By tracking offspring through two generations, a test cross can reveal patterns of inheritance for particular genes. 9. In a cross between a black and a white guinea pig, all members of the F1 generation are black. The F2 generation is made of % black and % white guinea pigs. Diagram this cross, and list the genotypes and phenotypes for each generation.
QUESTION 6 Flower diameter in sunflowers is a quantitative trait. A plant with 6-cm flowers, from a highly inbred strain, is crossed to a plant with 30-cm flowers, also from a highly inbred strain. The F1 have 18-cm flowers. F₁ x F₁ crosses yield F2 plants with flowers ranging from 6 to 30 cm in diameter, in approximately 4-cm intervals (6, 10, 14, 18, 22, 26, 30). The number of different genes influencing flower diameter in this plant is O a. 5 O b. 6 O c. 4 O d. 3 O e. 7