/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A cross is made between two \(E\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A cross is made between two \(E\) colistrains: Hfr arg \(+\) bio \(l e u^{+} \times \mathrm{F}^{-}\) arg \(^{-}\) bio \(^{-}\) leu'. Interrupted mating studies show that \(\arg ^{+}\) enters the recipient last, and so arg' recombinants are selected on a medium containing bio and leu only. These recombinants are tested for the presence of \(b i o^{+}\) and \(l e u^{+} .\) The following numbers of individuals are found for each genotype: $$\begin{array}{lrlrl} \arg ^{+} b i o^{+} \operatorname{leu}^{+} & 320 & \arg ^{+} b i o^{-} l e u^{+} & & 0 \\ \arg ^{+} b i o^{+} l e u^{-} & 8 & \arg ^{+} b i o^{-} l e u^{-} & 48 \end{array}$$ a. What is the gene order? b. What are the map distances in recombination percentages?

Short Answer

Expert verified
Gene order: arg - bio - leu. Map distances: 14.89% each.

Step by step solution

01

Analyze the genetic profile

Given two strains are crossed, with the recipient (\( \mathrm{F}^{-} \)) initially deficient in all three markers: arg, bio, leu. The donor (Hfr) is introduced with positive markers: arg\(^{+}\), bio\(+\) and leu\(+\). Since arg\(+\) is the last to enter, any arg\(+\) recombinants must have integrated donor DNA starting from arg.
02

Determine gene order from recombinants

The frequency of genotype appearances in the recombinants indicates the proximity of genes on the chromosome. \(\arg\), when \(b i o^{+}\) and \( l e u^{+}\) are also present, has 320 individuals, suggesting \(bio\) and \(leu\) are near \(arg\). The rare configuration \(\arg^{+} bio^{+} leu^{-}\) (8) suggests \(leu\) is closest to \(arg\), as seen by its reduced presence when disrupted.
03

Establish gene order

Based on findings, gene order is \(arg - bio - leu\). This is because \(bio^{+}\) appears less frequently removed than \(leu\), showing \(leu\) as the closest marker to \(arg\) on its integration track.
04

Calculate map distances by recombination frequency

Recombination percentages show map distances. First, calculate recombination for \(bio - leu\) using the \( \arg^{+} bio^{+} leu^{-} \) and \( \arg^{+} bio^{-} leu^{-} \) recombinants. Together (8 + 48 =) 56 out of (320 + 8 + 48)= 376 possible configurations.\[\text{Recombination Frequency (bio-leu)} = \frac{56}{376} \times 100 = 14.89\%%\]
05

Recombination for brakes

Assume \(arg - bio\) as the remaining possible recombination. Calculate:\[\arg^{+} bio^{-} = 48 \,and \, \arg^{+} bio^{+} leu^{-} = 8, together = 56 \text{ reconfigs out of } 376\]\[\text{Recombination Frequency (arg-bio) } = \frac{56}{376} \times 100 = 14.89\%\]
06

Final validation of distances

The established order is \( arg - bio - leu \) with measured recombination distances.Each calculated as map distance:\[\arg \to bio \approx 14.89\%\text{ and } \text{ bio } \to leu = 14.89\%\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interrupted Mating
Interrupted mating is a technique used in genetics to map gene order on bacterial chromosomes. By interrupting the mating process at different intervals, scientists can determine the sequence in which genes are transferred between bacteria. Here’s how it generally works:
  • An Hfr strain (a donor bacterial cell with high frequency of recombination) is mated with an F- strain (recipient).
  • The mating process begins and is interrupted at various time points to see which genes have been transferred.
  • The last gene detected in the transferred sequence indicates it was the furthest from the origin of transfer.
In this exercise, scientists found that the gene 'arg extsuperscript{+}' enters the recipient last. This information helps us deduce the gene order by thinking about proximity and sequence of entry.
Recombination Frequency
Recombination frequency explains how likely two genes are to be separated during recombination, providing insight into their physical distance on the chromosome. It's a staple in genetic mapping!
To calculate recombination frequency:
  • Count the number of offspring showing a new mix of parental traits (recombinants).
  • Divide this by the total number of offspring to get a fraction.
  • Multiply the fraction by 100 to get a percentage.
In this example, scientists calculated a recombination frequency between 'bio' and 'leu' genes as well as between 'arg' and 'bio':
  • For bio-leu: \(\frac{56}{376}\times 100 = 14.89\%\)
  • For arg-bio: also \(\frac{56}{376}\times 100 = 14.89\%\)
This percentage directly relates to map distances, essentially showing how far apart genes are on the chromosome.
Gene Order
Determining the correct order of genes is crucial for understanding how traits are inherited. In bacterial genetics, gene order on a chromosome can be figured out by analyzing the sequence in which genes enter a recipient cell during interrupted mating.
Here, the gene order is thought to be 'arg - bio - leu'. Here’s how scientists figured it out:
  • The most common recombinant was 'arg extsuperscript{+}bio extsuperscript{+}leu extsuperscript{+}', suggesting these genes travel closely together.
  • The rare recombinant 'arg extsuperscript{+}bio extsuperscript{+}leu extsuperscript{-}' showed that 'leu' is closer to 'arg' since disrupting its presence was less frequent.
  • Slight disruptions in combination patterns (frequencies) highlight gene proximity, with lesser frequencies indicating further distances away from the main entry point.
Understanding gene order like this helps to draw genetic maps and is invaluable in studies of inheritance and mutation.
Map Distances
Map distances give a numerical value to the space between genes on a chromosome. They're derived from recombination frequencies and provide a visual arrangement for gene placement. Here's their significance:
  • The closer together two genes are, the lower the recombination frequency, meaning they're less often separated during genetic recombination.
  • This is reflected in map distances – smaller distances indicate genes are closely situated.
  • In the given problem, the calculated map distances between 'arg' and 'bio' and between 'bio' and 'leu' are both 14.89%.
These distances express how frequently the genes recombine. Essentially, it's about translating recombination outcomes into metric distances to map them on the genetic landscape of a bacteria.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A microbial geneticist isolates a new mutation in \(E\) coli and wishes to map its chromosomal location. She uses interrupted-mating experiments with Hfr strains and generalized-transduction experiments with phage P1. Explain why each technique, by itself, is insufficient for accurate mapping.

A generalized transduction e xperiment uses a metE' \(p y r D^{+}\) strain as donor and \(m e t E^{-} p y r D^{-}\) as recipient. \(m e t E^{+}\) transductants are selected and then tested for the \(p y r D^{+}\) allele. The following numbers were obtained: $$\begin{array}{lr} m e t E^{+} p y r D^{-} & 857 \\ m e t E^{+} p y r D^{+} & 1 \end{array}$$ Do these results suggest that these loci are closely linked? What other explanations are there for the lone "double"?

\(\mathrm{F}^{\prime}\) strains in \(E\). coli are derived from Hfr strains. In some cases, these \(\mathrm{F}^{\prime}\) strains show a high rate of integration back into the bacterial chromosome of a second strain. Furthermore, the site of integration is often the site occupied by the sex factor in the original Hfr strain (before production of the \(\mathrm{F}^{\prime}\) strains). Explain these results.

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 \(\mathrm{M}=\) many recombinants, \(\mathrm{L}=\) low numbers of recombinants, and \(0=\) no recombinants: $$\begin{array}{cccc} & 1 & 2 & 3 & 4 \\ \hline 5 & 0 & \mathrm{M} & \mathrm{M} & 0 \\ 6 & 0 & \mathrm{M} & \mathrm{M} & 0 \\ 7 & \mathrm{L} & 0 & 0 & \mathrm{M} \\ 8 & 0 & \mathrm{L} & \mathrm{L} & 0 \end{array}$$ On the basis of these results, assign a sex type (either Hfr, \(F^{+},\) or \(F^{-}\) ) to each strain.

A bacterial transformation is performed with a donor strain that is resistant to four drugs, \(A, B, C,\) and \(D,\) and a recipient strain that is sensitive to all four drugs. The resulting recipient cell population is divided and plated on media containing various combinations of the drugs. The following table shows the results. $$\begin{array}{lccc} \begin{array}{c} \text { Drugs } \\ \text { added } \end{array} & \begin{array}{c} \text { Number } \\ \text { of colonies } \end{array} & \begin{array}{c} \text { Drugs } \\ \text { added } \end{array} & \begin{array}{c} \text { Number } \\ \text { of colonies } \end{array} \\ \hline \text { None } & 10,000 & \text { BC } & 50 \\ \text { A } & 1155 & \text { BD } & 48 \\ \text { B } & 1147 & \text { CD } & 785 \\ \text { C } & 1162 & \text { ABC } & 31 \\ \text { D } & 1140 & \text { ABD } & 43 \\ \text { AB } & 47 & \text { ACD } & 631 \\ \text { AC } & 641 & \text { BCD } & 35 \\ \text { AD } & 941 & \text { ABCD } & 29 \\ \hline \end{array}$$ a. One of the genes is distant from the other three, which appear to be closely linked. Which is the distant gene? b. What is the likely order of the three closely linked genes?

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.