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Chapter 4: Problem 22

You have a Drosophila line that is homozygous for autosomal recessive alleles \(a, b,\) and \(c,\) linked in that order. You cross females of this line with males homozygous for the corresponding wild-type alleles. You then cross the \(\mathrm{F}_{1}\) heterozygous males with their heterozygous sisters. You obtain the following \(F_{2}\) phenotypes (where letters denote recessive phenotypes and pluses denote wild-type phenotypes): \(1364+++, 365 a b c, 87 a b+\) \(84++c, 47 a++, 44+b c, 5 a+c,\) and \(4+b+\) a. What is the recombinant frequency between \(a\) and \(b ?\) Between \(b\) and \(c ?\) (Remember, there is no crossing over in Drosophila males. b. What is the coefficient of coincidence?

Short Answer

Expert verified
Recombinant frequency: a & b is 6.7%, b & c is 4.4%. Coefficient of coincidence is 1.526.

Step by step solution

01

Organize the Data

We start by classifying the given F2 phenotypes into parental and recombinant categories. The parental types are the most frequent and resemble the original parental types: - Parental phenotypes: 1364 (+++) and 365 (abc). - Recombinant phenotypes: 87 (ab+), 84 (++c), 47 (a++), 44 (+bc), 5 (a+c), 4 (+b+).
02

Calculate the Recombinant Frequency between a and b

Recombinant offspring between a and b are those that have combinations different from parental types abc and +++ involving a and b genes. These are ab+ and a++. - Number of recombinant offspring: 87 (ab+) + 47 (a++) = 134.- Total progeny: 1364 + 365 + 87 + 84 + 47 + 44 + 5 + 4 = 2000.- Recombinant frequency = \( \frac{134}{2000} = 0.067 \) or 6.7%.
03

Calculate the Recombinant Frequency between b and c

Recombinant offspring between b and c are those with different combinations involving b and c genes from the original parental types. These are ++c and +b+.- Number of recombinant offspring: 84 (++c) + 4 (+b+) = 88.- Total progeny: 2000.- Recombinant frequency = \( \frac{88}{2000} = 0.044 \) or 4.4%.
04

Calculate the Coefficient of Coincidence

The coefficient of coincidence (COC) is the ratio of observed double crossovers to expected double crossovers:- Double recombinant phenotypes are a+c and +b+: 5 (a+c) + 4 (+b+) = 9.- Expected double recombinants = Total progeny \( \times \) (Rate between a and b \( \times \) Rate between b and c) \[ \text{Expected double recombinants} = 2000 \times 0.067 \times 0.044 = 5.896 \]- Coefficient of coincidence (COC) = \( \frac{9}{5.896} \approx 1.526 \).

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Key Concepts

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

Recombinant Frequency
Recombinant frequency is a measure used in genetics to determine how often two genetic loci are separated during meiosis due to crossing over. To calculate it, you consider the number of offspring that are recombinants and divide by the total number of offspring.

In the context of Drosophila genetics, recombinant frequency helps geneticists establish the distance between genes on a chromosome. For example, in the exercise provided, the recombinant frequency between genes \( a \) and \( b \) is calculated with 134 recombinant offspring out of 2000 total progeny. This gives a recombinant frequency of \( \frac{134}{2000} = 0.067 \) or 6.7%.

Similarly, for the genes \( b \) and \( c \), the calculation involves 88 recombinant offspring out of 2000, resulting in a recombinant frequency of \( \frac{88}{2000} = 0.044 \) or 4.4%.

Recombinant frequencies are used to construct genetic maps, as they relate directly to the distances between genes. A smaller frequency typically indicates closer proximity on the chromosome.
Coefficient of Coincidence
In genetics, the "coefficient of coincidence" is used to measure the actual occurrence of double crossover events relative to how often they are expected to occur. It's an essential concept in understanding genetic linkage and mapping.

The exercise highlights this concept by showing the calculation of the coefficient of coincidence. Double crossover events are rare and typically provide a more accurate estimate of genetic distances. Observing 9 double crossover recombinants (5 \( a+c \) and 4 \( +b+ \)) against an expectation of approximately 5.896 suggests a coefficient of coincidence around 1.526. This value is calculated by taking the ratio of observed double recombinants (9) to the expected double recombinants (around 5.896).

A coefficient of coincidence greater than 1, as seen here, indicates that more double crossovers occurred than expected, potentially suggesting interference has boosted crossover events in these intervals.
Drosophila Genetics
Drosophila, commonly known as fruit flies, have long been a staple of genetic research due to their simple genome, rapid lifecycle, and prolific breeding. Their genetics are fundamental for understanding basic genetic principles such as linkage and recombination, as demonstrated in the exercise.

One key factor in Drosophila genetics is the absence of crossing over during male meiosis, which simplifies genetic analysis. This characteristic means that any recombination observed must have occurred in the female's meiosis, making it easier to interpret mapping crosses.

In the outlined exercise, Drosophila is used to study linked genes, where researchers can trace which genes are inherited together more frequently than by chance alone. Understanding the principles from experiments with such organisms has enormously contributed to unraveling more complex genetic systems in higher organisms.

Drosophila serves as a model organism not only because of ease of maintenance and numerous offspring generation but also due to the broad applicability of its genetic insights to other species, including humans.

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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.

From parents of genotypes \(A / A \cdot B / B\) and \(a / a \cdot b / b, a\) dihybrid was produced. In a testcross of the dihybrid, the following seven progeny were obtained: \\[\begin{array}{c}A / a \cdot B / b, a / a \cdot b / b, A / a \cdot B / b, A / a \cdot b / b \\ a / a \cdot b / b, A / a \cdot B / b, \text { and } a / a \cdot B / b\end{array}\\] Do these results provide convincing evidence of linkage?

In corn, the cross \(W W\) ee \(F F \times w w E E f f\) is made. The three loci are linked as follows: Assume no interference. a. If the \(\mathrm{F}_{1}\) is testcrossed, what proportion of progeny will be ww ee ff? b. If the \(\mathrm{F}_{1}\) is selfed, what proportion of progeny will be ww ee ff?

The mother of a family with 10 children has blood type \(\mathrm{Rh}^{+}\). She also has a very rare condition (elliptocytosis, phenotype \(\mathrm{E}\) ) that causes red blood cells to be oval rather than round in shape but that produces no adverse clinical effects. The father is \(\mathrm{Rh}^{-}\) (lacks the \(\mathrm{Rh}^{+}\) antigen) and has normal red blood cells (phenotype e). The children are \(1 \mathrm{Rh}^{+} \mathrm{e}, 4 \mathrm{Rh}^{+} \mathrm{E},\) and \(5 \mathrm{Rh}^{-}\) e. Information is available on the mother's parents, who are \(\mathrm{Rh}^{+} \mathrm{E}\) and \(\mathrm{Rh}^{-}\) e. One of the 10 children (who is \(\mathrm{Rh}^{+} \mathrm{E}\) ) marries someone who is \(\mathrm{Rh}^{+} \mathrm{e}\), and they have an \(\mathrm{Rh}^{+} \mathrm{E}\) child. a. Draw the pedigree of this whole family. b. Is the pedigree in agreement with the hypothesis that the \(R h^{+}\) allele is dominant and \(R h^{-}\) is recessive? c. What is the mechanism of transmission of elliptocytosis? d. Could the genes governing the \(\mathrm{E}\) and \(\mathrm{Rh}\) phenotypes be on the same chromosome? If so, estimate the map distance between them, and comment on your result.

A Neurospora cross was made between a strain that carried the mating-type allele \(A\) and the mutant allele arg-1 and another strain that carried the matingtype allele \(a\) and the wild-type allele for \(\arg -1(+)\) Four hundred linear octads were isolated, and they fell into the seven classes given in the table below. (For simplicity, they are shown as tetrads.) a. Deduce the linkage arrangement of the mating-type locus and the arg-1 locus. Include the centromere or centromeres on any map that you draw. Label all intervals in map units. b. Diagram the meiotic divisions that led to class 6 Label clearly. $$\begin{array}{ccccccc} \mathbf{1} & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline A \cdot \text {arg} & A \cdot+ & A \cdot \arg & A \cdot \text {arg} & A \cdot \text {arg} & A \cdot+ & A \cdot+ \\ A \cdot \text {arg} & A \cdot+ & A \cdot+ & a \cdot \arg & a \cdot+ & a \cdot \text {arg} & a \cdot \arg \\ a \cdot+ & a \cdot \arg & a \cdot \arg & A \cdot+ & A \cdot \arg & A \cdot+ & A \cdot \arg \\ a \cdot+ & \frac{a \cdot \arg }{127} & \frac{a \cdot+}{100} & \frac{a \cdot+}{36} & \frac{a \cdot+}{2} & \frac{a \cdot \arg }{4} & \frac{a \cdot+}{6} \end{array}$$
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