Concept explainers
a.
To determine: The genes that are linked.
Introduction: A gene is a sequence of
b.
To determine: The genotypes if two pure-breeding lines is crossed to produce the heterozygous individual.
Introduction: Pure-breeding creates a limited gene pool, and purebred animal breeds are also susceptible to a wide range of congenital health problems.
c.
To determine: The sketch of a linkage map of the linked genes, and the distances in map units.
Introduction: A linkage map is a table for a species or experimental population that shows the position of its known genes or genetic markers relative to each other in terms of recombination frequency, rather than a specific physical distance along each chromosome.
d.
To determine: The calculation of an interference value.
Introduction: The crossovers in adjacent chromosome regions independent or does a crossover in one area affect the likelihood of there being a crossover in a neighboring province. It turns out that often they are not separate, and the interaction is called interference.
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Introduction to Genetic Analysis
- Two pure-breeding lines of petunia plants are crossed. Line 1 plants grow to a height of 54 cm, and Line 2 plants grow to a height of 18 cm. Petunia plant height is controlled by three genes, A, B and C. Line 1 has the genotype A₁A₁B₁B₁C₁C₁, and line 2 has the genotype A2A2B₂B₂C₂C₂. Assume that genotype alone determines plant height under ideal growth conditions and that the alleles of the three genes are additive. If the F1 plants are self crossed, what is the expected proportion of F2 plants with the genotype A₁A₁B₁B₁C₁C₁ 1/8 1/32 1/16 1/4 1/64arrow_forwardFrom parents of genotypes A/A ⋅ B/B and a/a ⋅ b/b, adihybrid was produced. In a testcross of the dihybrid,the following seven progeny were obtained:A/a ⋅ B/b, a/a ⋅ b/b, A/a ⋅ B/b, A/a ⋅ b/b,a/a ⋅ b/b, A/a ⋅ B/b, and a/a ⋅ B/bDo these results provide convincing evidence of linkage?arrow_forwardAn individual heterozygous for four genes, A/a • B/b •C/c • D/d, is testcrossed with a/a • b/b • c/c • d/d, and 1000progeny are classified by the gametic contribution ofthe heterozygous parent as follows:a • B • C • D 42A • b • c • d 43A • B • C • d 140a • b • c • D 145a • B • c • D 6A • b • C • d 9A • B • c • d 305a • b • C • D 310a. Which genes are linked?b. If two pure-breeding lines had been crossed toproduce the heterozygous individual, what would theirgenotypes have been?c. Draw a linkage map of the linked genes, showing theorder and the distances in map units.d. Calculate an interference value, if appropriatearrow_forward
- In 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.arrow_forwardAssume that the trihybrid cross AABBrr x aabbRR is made in a plant species. Assume that A and B are dominant alleles, but there is no dominance effect of alleles at the R locus. a) How many different gametes are possible in the F1generation? What are the genotypes of these gametes? b) What is the probability of the parental aabbRR genotype in the F2 progeny? c) What proportion of the F2 progeny would be expected to be homozygous for all three genes?arrow_forwardIn the pearl-millet plant, color is determined by three alleles at a single locus: Rp1 (red), Rp2 (purple), and rp (green). Red is dominant over purple and green, and purple is dominant over green (Rp1 > Rp2 > rp). Give the expected phenotypes and ratios of offspring produced by the following crosses. a. Rp1/ Rp2 × Rp1/ rp b. Rp1/ rp × Rp2/ rp c. Rp1/ Rp2 × Rp1/ Rp2 d. Rp2/ rp × rp/ rp e. rp/ rp × Rp1/ Rp2arrow_forward
- Yellow guinea pigs crossed with white ones always produce cream-colored offspring. Two cream guinea pigs, when crossed, produce yellow, cream, and white offspring in the ratio of 1 yellow : 2 cream : 1 white. What principle of genetics is involved in this cross? (1 point) 2. The shape of radishes may be long, round, or oval. The following results were obtained in the different possible crosses: a. long x oval gave ½ long and ½ oval b. oval x round gave ½ oval and ½ round c. long x round gave all oval d. oval x oval gave ¼ long, ½ oval, and ¼ round Explain these results. Hint: Show genotypes of each cross) (2 points). a. b. c. d. 3. In human blood types, what are the genotypes of the following parents? (2 points). Phenotypes of ParentsPhenotypes of OffspringGenotypes of parents ABABO A x AB½ 0½ 0_______ x _______ A x AB½ ¼¼0_______ x _______ A x A¾00¼_______ x _______ A x O½00½_______…arrow_forwardAn individual with the genotype F/f • G/g • H/h is testcrossed. Among the progeny, the proportions of genotypes from this cross are shown in the following table: a) What gene is in the middle? Type you answer in this box: b) Calculate the genetic distances (in centimorgans, cM) between: genes g and f: genes f and h: genes g and h: c) What are the genotypes of the two parents? d) Calculate the interference and type your answer here:arrow_forwardIn corn, the cross WW ee FF × ww EE ff is made. The three loci are linked as follows:Assume no interference. a. If the F1 is testcrossed, what proportion of progeny will be ww ee ff? b. If the F1 is selfed, what proportion of progeny will be ww ee ff?arrow_forward
- In dogs, dark coat color phenotype (D) is dominant over albino (a) and short hair (S) is dominant over long hair (1). Assume that these phenotypes are caused by two independently assorting genes and write as much as you can, the genotypes of the parents in the cross below. D, a, S and I represent phenotypes Y indicates the presence of progeny with this phenotype , whereas N indicates the absence of progeny with this phenotype PARENTS PHENOTYPES TYPES of PROGENY PHENOTYPE D,S D,I a,S a,l D,S X a,l Y Yarrow_forwardPure-breeding parental strains of corn with the purple, smooth and yellow, wrinkled phenotypes were crossed to produce an F1 generation, which were then crossed to produce the F2 generation. Complete the diagram showing the results of the cross below. (I included an enlarged picture of the corn as well)arrow_forwardIn dogs, dark coat color is dominant over albino, andshort hair is dominant over long hair. Assume that theseeffects are caused by two independently assorting genes.Seven crosses were done as shown below, in which D andA stand for the dark and albino phenotypes, respectively,and S and L stand for the short-hair and long-hairphenotypes.Number of progenyParental phenotypes D, S D, L A, S A, La. D, S × D, S 88 31 29 12b. D, S × D, L 19 18 0 0c. D, S × A, S 21 0 20 0d. A, S × A, S 0 0 29 9e. D, L × D, L 0 31 0 11f. D, S × D, S 45 16 0 0g. D, S × D, L 31 30 10 10Write the genotypes of the parents in each cross. Use thesymbols C and c for the dark and albino coat-color allelesand the symbols H and h for the short-hair and long-hairalleles, respectively. Assume parents are homozygousunless there is evidence otherwise.arrow_forward
- Biology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning