TLS Online TPP Program

#Question id: 15152


In this problem we will explore some of the many ways that mutations in two different genes can interact to produce different Mendelian ratios. Consider a hypothetical insect species that has red eyes. Imagine mutations in two different unlinked genes that can, in certain combinations, block the formation of red eye pigment yielding mutants with white eyes. In principle, there are two different possible arrangements for two biochemical steps responsible for the formation of red eye pigment. The two genes might act in series such that a mutation in either gene would block the formation of red pigment. Alternatively, the two genes could act in parallel such that mutations in both genes would be required to block the formation of red pigment.
Further complexity arises from the possibility that mutations in either gene that lead to a block in enzymatic activity could be either dominant or recessive. If the crosses is made between a wild type insect with red eyes and a true breeding white eyed strain with mutations in both genes. Such considerations yield the Pathways in series with dominant mutations in both genes, determine the phenotype of the F1 progeny and the expected phenotypic ratio of red to white eyed insects in the F2.

#Unit 8. Inheritance Biology
  1. F1 will have red eyes

    Phenotypic ratio in F2 will be 3 white: 13 red

  2. F1 will have red eyes

    Phenotypic ratio in F2 will be 1 white:15 red

  3. F1 will have white eyes

    Phenotypic ratio in F2 will be 9 white: 7 red

  4. F1 will have white eyes

    Phenotypic ratio in F2 will be 15 white: 1 red

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TLS Online TPP Program

#Question id: 5298

#Unit 8. Inheritance Biology

Four E. coli strains of genotype a+ b− are labeled 1, 2, 3, and 4. Four strains of genotype a− b+ are labeled 5, 6, 7, and 8. The two genotypes are mixed in all possible combinations and (after incubation) are plated to determine the frequency of a+ b+ recombinants. The following results are obtained, where M = many recombinants, L = low numbers of recombinants, and 0 = no recombinants. On the basis of these results, assign a sex type (either Hfr, F+, or F−) to each strain

TLS Online TPP Program

#Question id: 5297

#Unit 8. Inheritance Biology

In a generalized transduction experiment, phages are collected from an E. coli donor strain of genotype cys+ leu+thr+ and used to transduce a recipient of genotype cys_ leu_ thr_. Initially, the treated recipient population is plated on a minimal medium supplemented with leucine and threonine. Many colonies are obtained. These colonies are then replica-plated onto three different media: What is order of gene?

TLS Online TPP Program

#Question id: 5296

#Unit 8. Inheritance Biology

A transformation experiment is performed with a donor strain that is resistant to four drugs: A, B, C, and D. The recipient is sensitive to all four drugs. The treated recipient cell population is divided up and plated on media containing various combinations of the drugs. The table below shows the results. What is the probable order of the three tightly linked genes?  

TLS Online TPP Program

#Question id: 5295

#Unit 8. Inheritance Biology

A cross is made between an Hfr that is met+ thi+ pur+ and an F-that is met- thi- pur-. Interrupted mating studies show that met+ enters the recipient last, so met+ recombinants are selected on a medium containing supplements that satisfy only the pur and thi requirements.  These recombinants are tested for the presence of the thi+ and pur+ alleles. The following numbers of individuals are found for each genotype:

A) Gene Order

I

Thi Pur  Met

II

Pur  Met  Thi

III

Pur   Thi  Met

B) Distance between Pur and Thi

IV

15.4 map unit

V

1.8 map unit

What is the following correct combination?

TLS Online TPP Program

#Question id: 5294

#Unit 8. Inheritance Biology

Interrupted mating experiments were performed using three different Hfr strains (1-3). The three strains have different combinations of selectable markers. The time of entry for markers for each strain is shown is shown in the table given.

Hfr 1

Phe

(5)

Trp

(35)

Met

(50)

Ala

(52)

Hfr 2

Trp

(2)

Val

(12)

His

(57)

Pro

(87)

Hfr 3

Ala

(10)

Met

(12)

Val

(37)

Phe

(57)

Using the above data, predict the correct sequence of markers on the E. coli chromosome

TLS Online TPP Program

#Question id: 5293

#Unit 8. Inheritance Biology

The two strains a and b are tested for being auxotrophic for the following nutritional mutations: pyridoxine (pdx), biotin (bio), methionine (met), adenine (ade), and para-aminobenzoic acid (pab). Growth is indicated by ( + ) and lack of growth by ( - ). From these tests, what is genotype of strain a and b respectively (+ indicate prototroph while other auxotroph)?