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Chapter 4: Problem 37

For a certain chromosomal region, the mean number of crossovers at meiosis is calculated to be two per meiosis. In that region, what proportion of meioses are predicted to have (a) no crossovers? (b) one crossover? (c) two crossovers?

Short Answer

Expert verified
The proportions are approximately: (a) 13.53% for no crossovers, (b) 27.07% for one crossover, (c) 27.07% for two crossovers.

Step by step solution

01

Understanding the Context

The problem involves finding the probabilities of certain numbers of crossovers during meiosis, where the average (mean) number of crossovers is given as 2 per meiosis.
02

Poisson Distribution Model

Since the problem deals with counting the number of crossovers per meiosis, we can model the distribution of these events using the Poisson distribution. The Poisson distribution is applicable because crossovers are rare, random, and independent events. The probability mass function for the Poisson distribution is given by \( P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \) where \( \lambda \) is the mean number of occurrences (2 in this case), \( k \) is the number of occurrences we are interested in, \( e \) is the base of the natural logarithm (approximately equal to 2.71828), and \( k! \) is the factorial of \( k \).
03

Calculating Probability for Zero Crossovers

To find the proportion of meioses with no crossovers (\( k = 0 \)), we use the formula: \( P(X = 0) = \frac{2^0 e^{-2}}{0!} = e^{-2} \). Calculating, we find that \( e^{-2} \approx 0.1353 \).
04

Calculating Probability for One Crossover

To find the proportion of meioses with one crossover (\( k = 1 \)), we use the formula: \( P(X = 1) = \frac{2^1 e^{-2}}{1!} = 2e^{-2} \). Calculating, we find that \( 2e^{-2} \approx 0.2707 \).
05

Calculating Probability for Two Crossovers

To find the proportion of meioses with two crossovers (\( k = 2 \)), we use the formula: \( P(X = 2) = \frac{2^2 e^{-2}}{2!} = \frac{4e^{-2}}{2} = 2e^{-2} \). Calculating, we find that \( 2e^{-2} \approx 0.2707 \).

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Key Concepts

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

Crossover Frequency
In genetics, crossovers during meiosis are crucial for genetic variation. Crossover refers to the exchange of genetic material between homologous chromosomes during meiosis. This process is essential as it increases genetic diversity among offspring. The frequency of these crossovers can vary significantly across different regions of the genome. Certain areas might experience frequent crossovers, while others remain unchanged. Crossover frequency is the average number of crossover events that occur per meiosis within a specific chromosome region. In our given example, this frequency is defined to be two per meiosis in the specified chromosomal region. This frequency is essential for understanding the likelihood of various genetic combinations appearing in offspring. Many factors can affect crossover frequency, including the specific genes involved, the organism's environment, and its genetic makeup.
Meiosis
Meiosis is a specialized type of cell division that reduces the chromosome number by half, resulting in the production of four genetically diverse gametes鈥攕perm in males and eggs in females. This process involves two consecutive divisions: meiosis I and meiosis II, each playing a different role in ensuring genetic variation. During meiosis I, homologous chromosomes pair up and exchange segments through crossover events. This ensures that the resulting gametes have different combinations of genes than the parent cell. Meiosis II then separates the sister chromatids, similar to the process seen in regular cell division, mitosis. The exchange and random assortment of chromosomes during meiosis are why siblings from the same parents are genetically unique. This biological process, by introducing genetic diversity, is essential for adaptation and evolution in populations.
Probability Calculation
In genetics, probability calculations are a fundamental tool used to predict the outcome of genetic crosses and events like crossover during meiosis. These calculations help scientists and researchers estimate the likelihood of specific genetic combinations or events occurring.For crossover events, the Poisson distribution is frequently used. This statistical distribution is particularly useful when dealing with rare and independent events, such as crossovers in a specific chromosomal region. The Poisson formula is expressed as:\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]where:
  • \( \lambda \) is the mean number of occurrences (2 in our problem).
  • \( k \) is the number of occurrences we are interested in (e.g., 0, 1, or 2 crossovers).
  • \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
  • \( k! \) is the factorial of \( k \).
Using the Poisson distribution, one can calculate the probability of having 0, 1, or 2 crossovers in our example. The previously calculated values for zero, one, and two crossovers further illustrate the practical application of this statistical method in real-world genetics problems.

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Most popular questions from this chapter

An individual heterozygous for four genes, \(A / a \cdot B / b \cdot C / c \cdot D / d\) is testcrossed with \(a / a \cdot b / b \cdot c / c \cdot d / d,\) and 1000 progeny are classified by the gametic contribution of the heterozygous parent as follows: $$\begin{array}{cc} a \cdot B \cdot C \cdot D & 42 \\ A \cdot b \cdot c \cdot d & 43 \\ A \cdot B \cdot C \cdot d & 140 \\ a \cdot b \cdot c \cdot D & 145 \\ a \cdot B \cdot c \cdot D & 6 \\ A \cdot b \cdot C \cdot d & 9 \\ A \cdot B \cdot c \cdot d & 305 \\ a \cdot b \cdot C \cdot D & 310 \end{array}$$ a. Which genes are linked? b. If two pure-breeding lines had been crossed to produce the heterozygous individual, what would their genotypes have been? c. Draw a linkage map of the linked genes, showing the order and the distances in map units. d. Calculate an interference value, if appropriate.

The recessive alleles \(k\) (kidney-shaped eyes instead of wild-type round), \(c\) (cardinal-colored eyes instead of wild-type red), and \(e\) (ebony body instead of wildtype gray) identify three genes on chromosome 3 of Drosophila. Females with kidney-shaped, cardinalcolored eyes were mated with ebony males. The \(\mathrm{F}_{1}\) was wild type. When \(\mathrm{F}_{1}\) females were testcrossed with \(k k\) \(c c\) ee males, the following progeny phenotypes were obtained: $$\begin{array}{cccr}k & c & e & 3 \\\k & c & \+ & 876 \\\k & \+ & e & 67 \\\k & \+ & \+ & 49 \\ \+ & c & e & 44 \\\\+ & c & \+ & 58 \\\\+ & \+ & e & 899 \\\\+ & \+ & \+ & 4 \\\\\text { Total } & & & 2000\end{array}$$ a. Determine the order of the genes and the map distances between them. b. Draw the chromosomes of the parents and the \(\mathrm{F}_{1}\) c. Calculate interference and say what you think of its significance

In a haploid fungus, the genes al-2 and arg-6 are 30 map units apart on chromosome \(1,\) and the genes lys- 5 and met-1 are 20 map units apart on chromosome 6. In a cross \\[a l-2+;+m e t-1 \times+\arg -6 ; 1 y s-5+\\] what proportion of progeny would be prototrophic \\[++;++?\\]

In the fungus Neurospora, a strain that is auxotrophic for thiamine (mutant allele \(t\) ) was crossed with a strain that is auxotrophic for methionine (mutant allele \(m\) ). Linear asci were isolated and classified into the following groups. $$\begin{array}{lcccccc}\text { Spore pair } & {}{} {\text { Ascus types }} \\\ \hline 1 \text { and } 2 & t+ & t+ & t+ & t+ & t m & t m \\ 3 \text { and } 4 & t+ & t m & +m & ++ & t m & ++ \\ 5 \text { and } 6 & +m & ++ & t+ & t m & ++ & t+ \\ 7 \text { and } 8 & +m & +m & +m & +m & ++ & +m \\ \hline \text { Number } & 260 & 76 & 4 & 54 & 1 & 5 \\ \hline\end{array}$$ a. Determine the linkage relations of these two genes to their centromere(s) and to each other. Specify distances in map units. b. Draw a diagram to show the origin of the ascus type with only one single representative (second from right

Consider the Neurospora cross \(+;+\times f ; p\) It is known that the \(+/ f\) locus is very close to the centromere on chromosome \(7-\) in fact, so close that there are never any second-division segregations. It is also known that the \(+/ p\) locus is on chromosome \(5,\) at such a distance that there is usually an average of 12 percent second- division segregations. With this information, what will be the proportion of octads that are a. parental ditypes showing \(M_{1}\) patterns for both loci? b. nonparental ditypes showing \(\mathrm{M}_{1}\) patterns for both loci? c. tetratypes showing an \(\mathrm{M}_{1}\) pattern for \(+/ f\) and an \(\mathrm{M}_{\mathrm{ll}}\) pattern for \(+/ p ?\) d. tetratypes showing an \(\mathrm{M}_{\mathrm{II}}\) pattern for \(+/ f\) and an \(\mathrm{M}_{\mathrm{I}}\) pattern for \(+/ p ?\)
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