Fine spines (s), smooth fruit (t u), and uniform fruit color...

Fine spines $(s),$ smooth fruit $(t u),$ and uniform fruit color $(u)$ are three recessive traits in cucumbers, the genes of which are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross, and the following progeny are produced from this testcross: (TABLE CANNOT COPY) a. Determine the onder of these genes on the chromosome. b. Calculate the map distances between the genes. c. Determine the coefficient of coincidence and the interference among these genes. d. List the genes found on each chromosome in the parents used in the testcross.

Fine spines $(s),$ smooth fruit $(t u),$ and uniform fruit color $(u)$ are three recessive traits in cucumbers, the genes of which are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross, and the following progeny are produced from this testcross: (TABLE CANNOT COPY)
a. Determine the onder of these genes on the chromosome.
b. Calculate the map distances between the genes.
c. Determine the coefficient of coincidence and the interference among these genes.
d. List the genes found on each chromosome in the parents used in the testcross.

Key Concepts

Interference and Coefficient of Coincidence

Interference is a phenomenon where one crossover event reduces the probability of another nearby crossover occurring during meiosis. The coefficient of coincidence quantifies this by comparing the observed frequency of double crossover events to the expected frequency if crossovers were independent. A value less than one indicates positive interference, where fewer double crossovers occur than expected. These concepts help refine genetic mapping by accounting for non-independence in crossover events.

Recombination Frequency and Map Distance

Recombination frequency is the proportion of recombinant offspring produced during a cross, and it is used as an estimate for the physical distance between two linked genes on a chromosome. Map distance is measured in centimorgans (cM), where one centimorgan corresponds to a 1% chance of recombination occurring between two genes. These concepts are essential for constructing genetic maps that outline the order and spacing of genes.

Gene Order Determination

Gene order determination involves analyzing the pattern of recombinant and parental phenotypes produced in a cross. By comparing the frequency of various crossover events, especially the appearance of double recombinants, researchers can infer the sequential arrangement of genes on a chromosome. The least frequent classes often represent double crossovers, which are critical for positioning intermediate genes accurately.

Testcross

A testcross is a genetic cross between an individual showing the dominant phenotype but with an unknown genotype and an individual that is homozygous recessive for the traits in question. This mating is used to reveal the heterozygosity of the individual with the dominant phenotype by analyzing the phenotypic ratios of the progeny, which helps in detecting recombination events between linked genes.

Genetic Linkage

Genetic linkage refers to the phenomenon where genes that are located close together on the same chromosome tend to be inherited together. This occurs because the closer two genes are, the less likely a crossover event will occur between them during meiosis. The concept is fundamental when analyzing inheritance patterns that deviate from independent assortment, indicating physical proximity on the chromosome.

Transcript

00:01 All right, so the broad topic here is chromosome variation, and we're using cucumber, three different traits, and some math to take a look at that.

00:14 So we first want to determine the order of the gs on the chromosome from our chart.

00:22 And so what we can presume, this is part a here, is we can presume that the non -recombinence, will occur most frequently, right? because the recombination is a less often occurrence than not.

00:50 So what we have for the most occurring gene combinations is we have this little s, little u and then the dominant t .u at 70 occurrences.

01:08 And then the next one is the opposite.

01:10 It.

01:11 So big s, big u, and a little tlau.

01:16 So that one is an 82.

01:18 So these are going to be the non -uncombinant trios, i guess.

01:28 And so the next thing is that double crossovers should be least common.

01:43 Okay, so the things that show up the least is big s, big u, big tu at two, and then recessive everything, is the second least common at four.

01:58 Okay, so the thing that differs between these two is this last gene here, right? if we just switch those, we end up with the most and least common.

02:14 So that means that the gene for fruit, this t -u gene, must be in the middle.

02:30 Okay, so that means the order then is going to be of the s -u - or s -t -e.

02:39 Tu and then u.

02:46 So that's the order, and that is the solution to part a.

02:59 All right, so now i'm moving on to part b, and we want to calculate the map distances.

03:07 So to get the s, the spines, and the fruit distance, we want to take 2 plus 4.

03:23 So this is how many times we had the double crossover.

03:28 Show up plus 21 plus 21 okay and so this is the number of times that the the 21s come from from the chart and this is when we get a crossover of s and t u or the spines and fruit being different okay so we take all those and we're going to divide it by the total which we're given is 230 and this gives us 48 out of 230 which is equal to 20 .9 % or also to 20 .9 map units.

04:34 So then we're going to do the same thing of the u -tu, the smoothness of the fruit and the color distance.

04:45 So we're going to do basically a similar thing.

04:47 We want the double crossovers added to the single crossovers, divide it by a 2 .30.

04:57 This time we are getting 36 out of 230, which is 15 .6 % or 15 .6 map units.

05:14 Okay, so these two elements are part b.

05:22 If you want the remaining distance from stu, you just add those together, right? because by the order of going from s to u to u, and this first one is 20 .9, and the second one is 15 .6, and we add those together to get the total length, which looks like it'll be 36 .5...

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