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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 \\[++;++?\\]

Short Answer

Expert verified
The proportion of prototrophic progeny is 56%.

Step by step solution

01

Understanding the Problem

In this exercise, we have two sets of genes located on different chromosomes. The goal is to determine the proportion of progeny that are prototrophic. The genes on chromosome 1, al-2 and arg-6, are 30 map units apart, indicating a recombination frequency of 0.30. The genes on chromosome 6, lys-5 and met-1, are 20 map units apart, indicating a recombination frequency of 0.20.
02

Finding Recombination Frequency

Convert map units into recombination frequencies. For genes on chromosome 1:\[\text{Recombination frequency} = 0.30\]For genes on chromosome 6:\[\text{Recombination frequency} = 0.20\]
03

Finding Proportion of Recombinant Progeny

Calculate the proportion of recombinant progeny. Since recombination frequency on chromosome 1 is 0.30, the non-recombining (or parental) progeny proportion is 0.70. Similarly, for chromosome 6, 0.80 of progeny are parental.
04

Calculating Prototrophic Progeny Proportion

To be prototrophic (\(++;++\)), the progeny must inherit both wild-type alleles from each chromosome without recombination. Thus, multiply the parental proportions from both chromosomes:\[0.70 \times 0.80 = 0.56\]
05

Final Answer

Thus, the proportion of progeny that are prototrophic is 0.56, meaning 56% of the progeny will be prototrophic.

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Key Concepts

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

Recombination Frequency
Recombination frequency is a measure of how often crossing over occurs between two genes during meiosis. It's a valuable tool in genetics as it helps to estimate the physical distance between genes on a chromosome. This frequency is usually expressed in map units or percentages. For example, if two genes are 30 map units apart, like in the case of al-2 and arg-6 on chromosome 1 from the exercise, the recombination frequency is 0.30 or 30%. This means that in 30% of the cases, crossing over will occur between these two loci during meiosis, mixing the parental alleles.

If the recombination frequency between two genes is low, like 20 map units for lys-5 and met-1 on chromosome 6, it indicates that the genes are relatively close on the chromosome, leading to a lower chance of separation during crossing over. This concept is fundamental in constructing genetic maps of chromosomes as it provides insight into the relative distances between genes.
Haploid Organisms
Haploid organisms have just one set of chromosomes in their cells. This is different from diploid organisms, which have two sets of chromosomes (one from each parent). Haploid organisms are commonly used in genetic studies because their single set of chromosomes simplifies the analysis of gene interactions and genetic mapping.

In haploid organisms, like the fungus in the given exercise, any mutation in a gene will directly manifest in the phenotype because there is no second allele to mask its expression. This characteristic makes haploid organisms excellent models for genetic research. By studying haplotypes, scientists can easily observe genetic recombination events, enabling them to map genes and determine recombination frequencies.
Chromosome Mapping
Chromosome mapping is the process of determining the relative positions of genes on a chromosome. This is achieved through the analysis of recombination frequencies between various genes. The closer two genes are, the less likely they are to be separated by crossing over, resulting in lower recombination frequencies. The further apart they are, the more likely they are to be separated, reflecting higher recombination frequencies.

Geneticists utilize these frequencies to create a map of the chromosome, where one map unit is equivalent to a 1% recombination frequency. In the exercise, for instance, the genes al-2 and arg-6 are 30 map units apart, providing information on their relative positions on chromosome 1. Similarly, lys-5 and met-1 are 20 map units apart on chromosome 6. Chromosome maps are crucial for researchers to understand genetic landscapes, analyze genetic linkage, and predict traits.
Prototrophic Progeny
Prototrophic progeny are offspring that can grow on minimal medium because they have all necessary enzymes to synthesize essential compounds from simple substances. They inherit all wild-type alleles necessary for survival without supplemental nutrients. In the exercise, we seek to determine the proportion of progeny that maintain this ability after crossing.

To be prototrophic, progeny must inherit the wild-type, non-recombinant alleles from both chromosomes. This exercise determines that 56% of the progeny exhibit prototrophy. This results from multiplying the proportion of non-recombinant progeny from chromosome 1 (0.70) by that from chromosome 6 (0.80). Understanding the genetics behind prototrophic progeny helps illustrate the effects of recombination and genetic linkage on phenotype expression.

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Most popular questions from this chapter

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 \(B r / b r ?\)

The Neurospora cross al- \(2^{+} \times a l-2\) is made. A linear tetrad analysis reveals that the second-division segregation frequency is 8 percent a. Draw two examples of second-division segregation patterns in this cross. b. What can be calculated by using the 8 percent value?

In the plant Arabidopsis, the loci for pod length \((L,\) long; \(l,\) short) and fruit hairs (H, hairy; \(h,\) smooth) are linked 16 map units apart on the same chromosome. The following crosses were made: (i) \(L H / L H \times l h / l h \rightarrow \mathrm{F}_{1}\) (ii) \(L h / L h \times l H / l H \rightarrow \mathrm{F}_{1}\) If the \(F_{1}\) 's from cross i and cross ii are crossed, a. what proportion of the progeny are expected to be \(l h / l h ?\) b. what proportion of the progeny are expected to be \(L h / l h ?\)

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

A female animal with genotype \(A / a \cdot B / b\) is crossed with a double- recessive male \((a / a \cdot b / b) .\) Their progeny include 442 A/a \(\cdot B / b, 458 a / a \cdot b / b, 46 A / a \cdot b / b,\) and 54 \(a / a \cdot B / b .\) Explain these results.

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