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In the plant Arabidopsis, the loci for pod length (L, long; 1, short) and fruit hairs ( \(H,\) hairy; \(h,\) smooth) are linked 16 m.u. apart on the same chromosome. The following crosses were made: (i) \(L H / L H \times l h / l h \rightarrow F_{1}\) (ii) \(L h / L h \times l H / l H \rightarrow F_{1}\) If the \(\mathrm{F}_{1}\) '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 \(L h / l h ?\)

Short Answer

Expert verified
a. 0.64% b. 6.72%

Step by step solution

01

Understanding Linkage and Map Units

Linkage refers to the tendency of genes close together on a chromosome to be inherited together. Map units (m.u.) quantify this closeness and in this problem, the genes are 16 m.u. apart. This implies a 16% recombination frequency.
02

Cross i Analysis: Purebreds

In cross (i), both homozygous parents are given: one is homozygous dominant (\(LH/LH\)) and the other is homozygous recessive (\(lh/lh\)). The \(F_1\) offspring from this cross will all have the genotype \(LH/lh\), showing the dominant phenotypes.
03

Cross ii Analysis: Single Pair

In cross (ii), both parents are heterozygous but in repulsion (i.e., \(Lh/Lh\) is crossed with \(lH/lH\)). The \(F_1\) offspring will each have the genotype \(Lh/lH\). Their phenotypes can vary due to recombination.
04

Set Up the Test Cross for F1

When \(F_1\)'s from cross i (\(LH/lh\)) and cross ii (\(Lh/lH\)) are crossed, the possible gametes from each are involved. Compute the possible gametes considering the recombination rate of 16%.
05

Calculate Gamete Frequencies

The non-recombinant gametes from \(LH/lh\) are \(LH\) and \(lh\) (42% each), and recombinant gametes are \(Lh\) and \(lH\) (8% each). From \(Lh/lH\), similarly, non-recombinant gametes \(Lh\) & \(lH\) are 42% each, and recombinant \(LH\) & \(lh\) are 8% each.
06

Determine Progeny Genotypes

Combine gametes from both \(F_1\):- \(lh/lh\): occurs with \(lh\) from both, each is 8%: total probability = 0.08 * 0.08 = 0.0064.- \(Lh/lh\): \(Lh\) from first (42%) and \(lh\) from second (8%): total = 0.42 * 0.08 = 0.0336 (and vice versa: 0.08 * 0.42 = 0.0336).

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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 concept used to describe how often crossover occurs between two linked genes on a chromosome. It is expressed as a percentage and provides an estimate of how frequently genetic recombination will happen during meiosis. In the context of genetics, recombination refers to the process by which offspring receive a mix of genes different from that found in either parent. For example, in our case, the genes for pod length and fruit hairs are located 16 map units apart on the same chromosome.
This 16 m.u. indicates a recombination frequency of 16%, meaning there is a 16% chance that recombination will occur between these genes in the offspring's gametes. The significance of recombination frequency extends beyond its numerical value; it also helps illustrate genetic diversity. Genes that are closer together tend to have lower recombination frequencies, which implies they are more likely to be inherited together.
However, those that are further apart have higher frequencies, indicating a higher chance of being separated during recombination events.
Map Units
Map units, abbreviated as m.u., are a unit of measure used in genetic linkage maps to illustrate the distance between genes on a chromosome. One map unit is equivalent to a 1% recombination frequency. This convenient way of measuring genetic distance helps researchers and students alike understand the relative positions of genes on a chromosome. In our exercise, the genes for pod length and fruit hairs are stated to be 16 m.u. apart.
This means they are expected to crossover or recombine in 16% of the meiosis cases. Because of map units, we can visualize how likely two genes are to be inherited together – the closer they are, the more likely they are to be linked.
Map units serve an educational purpose by simplifying complex genetic concepts into a digestible form. They convert percentages into a scale that can many times make abstract genetic ideas more tangible for learners trying to grasp genetic mapping and inheritance patterns.
Gamete Frequencies
Gamete frequencies give us a quantitative insight into how specific gametes are produced from given genotypes during sexual reproduction. Considering recombination rates, different gamete frequencies emerge, reflecting the proportions of recombinant and non-recombinant gametes. In our particular case, we look at the gamete frequencies produced by the F1 offspring from two different crosses. From cross (i), F1 gametes LH/lh show non-recombinant gametes LH and lh each forming 42% of the time, while recombinant gametes Lh and lH occur 8% of the time each. Meanwhile, from cross (ii), F1 gametes Lh/lH have non-recombinant gametes Lh and lH at 42% each, with recombinant gametes LH and lh also at 8% each.
Understanding these frequencies is essential in predicting the genetic composition of the offspring. It allows us to calculate probabilities for specific offspring genotypes. In our example:
  • The probability of progeny being lh/lh is the sum product of these frequencies (0.08 * 0.08 for each gamete type from the respective crossings).
  • Similarly, for obtaining Lh/lh, calculations based on these frequencies give its expected proportion in the offspring population.
In conclusion, gamete frequencies pave the way for comprehending the genetic variation and possible outcomes of genetic crosses.

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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 \(B r / b r ?\)

R. A. Emerson crossed two different pure-breeding lines of corn and obtained a phenotypically wild-type \(\mathrm{F}_{1}\) that was heterozygous for three alleles that determine recessive phenotypes: an determines anther; br, brachytic; and \(f,\) fine. He testcrossed the \(\mathrm{F}_{1}\) with a tester that was homozygous recessive for the three genes and obtained these progeny phenotypes: 355 anther; 339 brachytic, fine; 88 completely wild type; 55 anther, brachytic, fine; 21 fine; 17 anther, brachytic; 2 brachytic; 2 anther, fine. a. What were the genotypes of the parental lines? b. Draw a linkage map for the three genes (include map distances). c. Calculate the interference value.

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?

In a haploid fungus, the genes \(a l-2\) and \(a r g-6\) ar apart on chromosome \(1,\) and the genes \(l y s-5\) \(a l-2+;+m e t-1 \times+\arg -6 ; l y s-5+\) what proportion of progeny would be prototrophic \(++\) \(++?\)

Use the Haldane map function to calculate the corrected map distance in cases where the measured \(\mathrm{RF}=5 \%\) \(10 \%, 20 \%, 30 \%,\) and \(40 \% .\) Sketch a graph of RF against corrected map distance, and use it to answer the question, When should one use a map function?

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