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Chapter 4: Problem 52

The recessive alleles \(k\) (kidney-shaped eyes instead of wild-type round), \(c\) (cardinal-colored eyes instead of wildtype red), and \(e\) (ebony body instead of wild-type gray) identify three genes on chromosome 3 of Drosophila. Females with kidney-shaped, cardinal-colored eyes were mated with ebony males. The \(\mathrm{F}_{1}\) was wild type. When \(\mathrm{F}_{1}\) females were testcrossed with \(k k\) cc 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.

Short Answer

Expert verified
Gene order is 'k-e-c' with map distances: 'k-e' 6.15 cM, 'e-c' 5.45 cM. Interference is minimal.

Step by step solution

01

Identifying Non-Recombinant and Double Crossover Progeny

Non-recombinant progeny are the most frequent offspring, representing the parental types. Here, these are: "+ + +" with 4 individuals, and "k c +" with 876 individuals. Double crossover progeny are the least frequent, showing the least common types. These are: "k c e" with 3 individuals and "+ + e" with 899 individuals. Using these, we can determine the gene order by identifying which gene location changes between the double crossover progeny compared to the parents.
02

Determining Gene Order

In the double crossover progeny, the phenotype that changes compared to the non-recombinant parent types defines the middle gene. Comparing "+ + e" and "k c e" with their non-recombinant counterparts reveals that the gene 'e' changes position. Thus, 'e' is the middle gene, and the order is 'k-e-c' or 'c-e-k'.
03

Calculate Map Distances

Based on the gene order 'k-e-c', we calculate the distances by summing the progeny numbers for single and double crossovers for each gene pair. The formula for map distance is: (Number of recombinant offspring / total offspring) 脳 100 cM. For 'k-e' interaction: (67 + 49 + 3 + 4) / 2000 脳 100 = 6.15 cM. For 'e-c' interaction: (44 + 58 + 3 + 4) / 2000 脳 100 = 5.45 cM.
04

Drawing Chromosomes

For parents, depict chromosomes as 'k e c' for the recessive female and '+ + +' for the wild-type overlap. For F1, the chromosomes would be '+ + +' and 'k e c'-'+' denotes wild type alleles replacing recessive alleles, demonstrating the dominant nature of wild type traits.
05

Calculating Interference

Interference is determined using interference (I) = 1 - (Observed DCO/Expected DCO). Observe the frequency of DCO progeny from step 1 (7 out of 2000). Expected DCO is (0.0615 脳 0.0545 脳 2000 = 6.7). Expected interference = 1 - (7/6.7) 鈮� 0. However, calculations show minimal interference, indicating crossover events are nearly independent.

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Key Concepts

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

Gene Mapping
Gene mapping is a fascinating technique used to determine the position of genes on a chromosome. It's a bit like creating a map to find treasures, but in this case, the treasures are genes. Scientists use gene mapping to understand the distances between various genes, which can help identify gene interactions and potential genetic linkages. The primary method for gene mapping involves tracking how often genes exchange positions during the process of crossing over. By assessing recombination frequencies during genetic crosses, scientists can estimate the relative distances between genes. These distances are typically expressed in centimorgans (cM), a unit that reflects the likelihood of recombination occurring between genes. For example, a distance of 1 cM implies a 1% chance of recombination occurring between two genes during meiosis.
Understanding gene mapping is crucial in fields like genetics, medicine, and agriculture, where it aids in identifying genes associated with diseases, improving crop varieties, and studying evolutionary relationships. In the exercise, students calculated map distances between genes associated with physical traits (like eye color and body texture) in fruit flies, providing a practical application of gene mapping techniques.
Chromosome Mapping
Chromosome mapping is closely related to gene mapping but focuses on determining the precise location of genes within chromosomes. Like drawing a detailed roadmap, chromosome mapping helps visualize how genes are arranged, showcasing the structural layout of chromosomes in an organism. Both gene and chromosome mapping are crucial for identifying chromosomal abnormalities linked to genetic disorders and for advancing our understanding of trait inheritance.
In the context of our exercise, students explored chromosome mapping through the fruit fly Drosophila's chromosome 3, discerning the arrangement of recessive alleles鈥攌, c, and e. By examining the progeny from various genetic crosses, students could determine the location and order of these alleles. This not only illustrates chromosome mapping techniques but also sheds light on how physical traits are inherited and expressed in organisms.
  • Makes predictions about genetic outcomes
  • Helps identify genetic disorders
  • Assists in understanding evolution
Recessive Alleles
Recessive alleles are versions of a gene that typically don't exhibit their traits unless two copies are present, one from each parent. These alleles can be thought of as silent observers waiting for the perfect conditions to show their characteristics. The presence of a dominant allele usually masks recessive alleles.
In the exercise, the recessive alleles k (kidney-shaped eyes), c (cardinal-colored eyes), and e (ebony body) in Drosophila were analyzed. These alleles clearly illustrate the concept of recessive traits since an organism must inherit two recessive alleles to exhibit the associated phenotype. For instance, a fruit fly with one kidney-shaped eye allele is not enough; it needs two to demonstrate that trait.
This exercise showcases how recessive alleles don't always readily appear in offspring unless specific genetic conditions are met, emphasizing the complexity of Mendelian inheritance. Understanding recessive alleles is essential, not only for genetics studies but also for breeding programs, where knowing which traits are recessive can guide selective breeding processes efficiently.

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Most popular questions from this chapter

In a tetrad analysis, the linkage arrangement of the \(p\) and \(q\) loci is as follows: Assume that ' in region i, there is no crossover in 88 percent of meioses and there is a single crossover in 12 percent of meioses; in region ii, there is no crossover in 80 percent of meioses and there is a single crossover in 20 percent of meioses; and - there is no interference (in other words, the situation in one region does not affect what is going on in the other region) What proportions of tetrads will be of the following types? (a) \(\mathrm{M}_{\mathrm{I}} \mathrm{M}_{\mathrm{I}}, \mathrm{PD} ;\) (b) \(\mathrm{M}_{\mathrm{I}} \mathrm{M}_{\mathrm{I}}, \mathrm{NPD} ;(\mathrm{c}) \mathrm{M}_{\mathrm{I}} \mathrm{M}_{\mathrm{II}}, \mathrm{T} ;(\mathrm{d}) \mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{I}}, \mathrm{T} ;\) (e) \(\mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{II}}, \mathrm{PD} ;(\mathrm{f}) \mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{II}}, \mathrm{NPD} ;(\mathrm{g}) \mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{II}}, \mathrm{T}\). (Note: Here the M pattern written first is the one that pertains to the \(p\) locus.) Hint: The easiest way to do this problem is to start by calculating the frequencies of asci with crossovers in both regions, region i, region ii, and neither region. Then determine what \(\mathrm{M}_{\mathrm{I}}\) and \(\mathrm{M}_{\mathrm{II}}\) patterns result.

In the model plant Arabidopsis thaliana, the following alleles were used in a cross: \(\begin{array}{ll}T=\text { presence of trichomes } & t=\text { absence of trichomes } \\ D=\text { tall plants } & d=\text { dwarf plants } \\ W=\text { waxy cuticle } & w=\text { nonwaxy } \\ A=\text { presence of purple } & a=\text { absence (white) } \\ & \text { anthocyanin pigment }\end{array}\) The \(T / t\) and \(D / d\) loci are linked 26 m.u. apart on chromosome 1 , whereas the \(W / w\) and \(A / a\) loci are linked 8 m.u. apart on chromosome 2 A pure-breeding double-homozygous recessive trichomeless nonwaxy plant is crossed with another pure-breeding double-homozygous recessive dwarf white plant. a. What will be the appearance of the \(\mathrm{F}_{1}\) ? b. Sketch the chromosomes 1 and 2 of the parents and the \(F_{1},\) showing the arrangement of the alleles. c. If the \(\mathrm{F}_{1}\) is testcrossed, what proportion of the progeny will have all four recessive phenotypes?

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 mating-type 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. $$\begin{array}{ccccccc}\mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} \\\\\hline A \cdot \text {arg} & A \cdot+ & A \cdot \text {arg} & A \cdot \text {arg} & A \cdot \text {arg} & A \cdot+ & A \cdot+ \\\A \cdot \text {arg} & A \cdot+ & A \cdot+ & a \cdot \text {arg} & a \cdot+ & a \cdot \text {arg} & a \cdot \text {arg} \\\A \cdot+ & a \cdot \text {arg} & a \cdot \text {arg} & A \cdot+ & A \cdot \text {arg} & A \cdot+ & A \cdot \arg \\\A \cdot+ & a \cdot \text {arg} & a \cdot+ & a \cdot+ & a \cdot+ & a \cdot \text {arg} & a \cdot+ \\\\\hline 127 & 125 & 100 & 36 & 2 & 4 & 6 \end{array}$$ 1\. Are fungi generally haploid or diploid? 2\. How many ascospores are in the ascus of Neurospora? Does your answer match the number presented in this problem? Explain any discrepancy. 3\. What is mating type in fungi? How do you think it is determined experimentally? 4\. Do the symbols \(A\) and \(a\) have anything to do with dominance and recessiveness? 5\. What does the symbol arg-1 mean? How would you test for this genotype? 6\. How does the arg-1 symbol relate to the symbol +? 7\. What does the expression wild type mean? 8\. What does the word mutant mean? 9\. Does the biological function of the alleles shown have anything to do with the solution of this problem? 10\. What does the expression linear octad analysis mean? 11\. In general, what more can be learned from linear tetrad analysis that cannot be learned from unordered tetrad analysis? 12\. How is a cross made in a fungus such as Neurospora? Explain how to isolate asci and individual ascospores. How does the term tetrad relate to the terms ascus and octad? 13\. Where does meiosis take place in the Neurospora life cycle? (Show it on a diagram of the life cycle.) 14\. What does Problem 38 have to do with meiosis? 15\. Can you write out the genotypes of the two parental strains? 16\. Why are only four genotypes shown in each class? 17\. Why are there only seven classes? How many ways have you learned for classifying tetrads generally? Which of these classifications can be applied to both linear and unordered tetrads? Can you apply these classifications to the tetrads in this problem? (Classify each class in as many ways as possible.) Can you think of more possibilities in this cross? If so, why are they not shown? 18\. Do you think there are several different spore orders within each class? Why would these different spore orders not change the class? 19\. Why is the following class not listed? $$\begin{array}{ll} a \cdot+ & A \cdot \arg \\\a \cdot+ & A \cdot \arg\end{array}$$ 20\. What does the expression linkage arrangement mean? 21\. What is a genetic interval? 22\. Why does the problem state "centromere or centromeres" and not just "centromere"? What is the general method for mapping centromeres in tetrad analysis? 23\. What is the total frequency of \(A\). \(+\) ascospores? (Did you calculate this frequency by using a formula or by inspection? Is this a recombinant genotype? If so, is it the only recombinant genotype?) 24\. The first two classes are the most common and are approximately equal in frequency. What does this information tell you? What is their content of parental and recombinant genotypes?

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 \(\left.\mathrm{Rh}^{-} \text {(lacks the } \mathrm{Rh}^{+} \text {antigen }\right)\) 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 \(\mathrm{Rh}^{+}\) E) marries someone who is \(\mathrm{Rh}^{+} \mathrm{e}\), and they have an \(\mathrm{Rh}^{+}\) 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 Eand Rh phenotypes be on the same chromosome? If so, estimate the map distance between them, and comment on your result.

A fruit fly of genotype \(B R / b r\) is testcrossed with \(b r / b r\) In 84 percent of the meioses, there are no chiasmata between the linked genes; in 16 percent of the meioses, there is one chiasma between the genes. What proportion of the progeny will be \(B r / b r ?\)
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