Question

You have created a mutant that is streptomycin resistant (s+), has the ability to photosynthesize (p+), and is resistant to a particular phage, P5 (r+). Your PI would like you to try and move these traits into E. coli by means of generalized transduction.The donor phage has the genotype s+ p+ r+, while the recipient bacterium has the genotype s- p- r-. You carry out selection for the phage resistance trait and then screen all of the r+ transductants for the other two traits. The table below shows the observations from your experiment.

Phenotype Offspring Genotype Phage resistance, photosynthesizing 244 Phage resistance, streptomycin resi stance 37 Phage resi

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Please look in the image for solution:

Phenotype Genotype Offspring p+r+s- Phage resistance, photosynthesizing 244 Phage resistance, streptomycin resi stance 37 p-r

Cotronscuctien freguency of 3 123 1o0- 402 r and s 540 AA C-tansduction tregueny of 244123 xioo6796 540 r and p AP co- transd

If you find my answer is useful, then please hit the like button ☺

Add a comment
Know the answer?
Add Answer to:
You have created a mutant that is streptomycin resistant (s+), has the ability to photosynthesize (p+), and is resistant...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • You have three genes on the same chromosome - A, B and C. Each gene has...

    You have three genes on the same chromosome - A, B and C. Each gene has two alleles in a dominant/recessive relationship. For these genes the homozygous recessive has the mutant phenotype for that trait, the dominant phenotype = wild type for that trait. allele A is dominant to a; phenotype a = mutant for trait a; phenotype A = wild type for trait A allele B is dominant to b; phenotype b = mutant for trait b; phenotype B...

  • You have three genes on the same chromosome - A, B and C. Each gene has...

    You have three genes on the same chromosome - A, B and C. Each gene has two alleles in a dominant/recessive relationship. For these genes the homozygous recessive has the mutant phenotype for that trait, the dominant phenotype = wild type for that trait. allele A is dominant to a; phenotype a = mutant for trait a; phenotype A = wild type for trait A allele B is dominant to b; phenotype b = mutant for trait b; phenotype B...

  • Which of these statements is incorrect? Syntenic genes are located on the same chromosome. Independent assortment...

    Which of these statements is incorrect? Syntenic genes are located on the same chromosome. Independent assortment results in recombinant chromosomes. You can reliably predict the relative genetic distance from genes’ physical distance on a chromosome. Linked genes are always syntenic. What is the relative genetic distance between two linked genes if the recombination frequency is 0.49? 0.49 cM 4.9 cM 49 cM 490 cM What statement best explains the distortion in Mendelian ratios observed by Bateson & Punnett in 1905?...

  • are to determine the probability of inheriting a widow's In this activity you will ea Penne...

    are to determine the probability of inheriting a widow's In this activity you will ea Penne peak or a straight hairline Procedure Read the instructions for using a Punnet square (Figure 11.1). 1. Discuss the instructions with your lab partners, 2. Fill in the blank spaces and solve problem #1. Using a Punnett Square Gametes • Place possible gametes along two opposite sides of the square Record genotypes of potential offspring in appropriate cells of the square Gametes P PP...

  • 3. (7) Suppose that you have selected E. coli that can grow only when the medium...

    3. (7) Suppose that you have selected E. coli that can grow only when the medium is supplemented with cysteine. You isolate one mutant that you believe is defective in a new gene require for cysteine biosynthesis - cysX . You have two strains of bacteria: Hfr cysX - gal+ his+ lac+ pur+ trp+ ton s F - cysX+ gal - his - lac - pur - trp - ton R A) What supplements would you need to add to...

  • can you please help me with number 4? An example of linked genes in Drosophila The...

    can you please help me with number 4? An example of linked genes in Drosophila The genes for wing shape and body color are linked (they are on the same chromosome) Drosophila and linked genes In the example shown left, wild type alleles are dominant and are given an upper case symbol of the mutant phenotype (Cu or Eb). This notation used for Drosophila departs from the convention of using the dominant gene to provide the symbol. This is necessary...

  • 5.) You have discovered mutants in two new gene mutants in Drosophila. One mutant, curl, has...

    5.) You have discovered mutants in two new gene mutants in Drosophila. One mutant, curl, has small, curly, nonfunctional wings. The other mutant, big, has enormous eyes. You decide to test if these two new genes are on chromosome #2, so you perform a three-gene linkage testcross using a gene that you know is on chromosome #2, trp, a mutant that cannot make its own tryptophan (and so must get it from its food). In all three cases, these mutant...

  • 3. (7) Suppose that you have selected E cofthat can grow only when the medium is...

    3. (7) Suppose that you have selected E cofthat can grow only when the medium is supplemented with cysteine. You isolate one mutant that you believe is defective in a new gene require for cysteine biosynthesis- sX You have two strains of bacteria: Hitr cysX- galehislac+ purtrpefort cysX+gal-his-tac-pur-trp-tort A) What supplements would you need to add to minimal media to recover all cysX lach, fon recombinants? Would you add glucose? B) What recombinant genotypes would you recover on plates containing...

  • i need some help with this lab ASAP please! HUMAN GENETICS It to study because of...

    i need some help with this lab ASAP please! HUMAN GENETICS It to study because of the relatively long life span and the limited number In addition, the number of chromosome pairs (23) increases the possible number of genetic combinations. It is possible, however, to take a sample from human frequency of a trait and the possible ways a given trait is inherited. populations to estimate the Objectives .Investigate the inheritance of some human traits. Estimate the frequency of selected...

  • Question 48 (1 point) A guinea pig with the genotype GgBBcc is heterozygous for two genes....

    Question 48 (1 point) A guinea pig with the genotype GgBBcc is heterozygous for two genes. O homozygous for one gene. heterozygous for three genes. O homozygous for two genes. Question 49 (1 point) Which cells in animals and plants are typically haploid? epithelial cells digestive cells ribosomes gametes Question 50 (1 point) Asexual reproduction and sexual reproduction are different in that: individuals reproducing asexually transmit 100% of their genes to their of MacBook Pro Q E Question 50 (1...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT