Three dominant traits of corn seedlings, tunicate seed (T-), glos...

Question

Three dominant traits of corn seedlings, tunicate seed (T-), glossy appearance (G-), and liguled stem (L-), are studied along with their recessive counterparts, nontunicate (tt), nonglossy (gg), and liguleless (ll). A trihybrid plant with the three dominant traits is crossed to a nontunicate, nonglossy, liguleless plant. Kernels on ears of progeny plants are scored for the traits, with the following results:

Phenotype Number
Tunicate, glossy, liguled 102
Tunicate, glossy, liguleless 106
Tunicate, nonglossy, liguled     18
Tunicate, nonglossy, liguleless   20
Nontunicate, glossy, liguled     22
Nontunicate, glossy, liguleless   23
Nontunicate, nonglossy, liguled   99
Nontunicate, nonglossy, liguleless 110
500

Could all three genes be carried on the same chromosome? Discuss why or why not.

Accepted Answer

Hi, everyone. Welcome back. Here's our next question. Suppose in a pea garden, the crossing of parental peas, big tea, big tea, big pe big pea with little T little T little P little P produces F one progeny. That is heterozygous for two traits. Big T little T big P little P. The F one die hybrid is then crossed to A P that is homozygous, recessive Little tea, little tea, little pea, little pee for the two traits. And they produce the following offspring tall and purple. 134, tall and white. 85, short and white. 56 short and purple. 41. What is the recombination frequency between the two genes associated with the peas? Height and flower color? Choice? A 50%. Choice B 40%. Choice C 25% or choice D 15%. Well, let's recall what we need to calculate the recombination frequency. And that formula is that recombination frequency equals the number of recombinant offspring Divided by the total number of offspring times 100%. So first, we have to determine which of the phenotypes here represent the recombinant offspring. Well, we have our heterozygous parent and homozygous recessive parent, the homozygous recessive parent, we don't need to consider crossing over because it won't have any effect. The alleles are all the same. So we look at our heterozygous parent who inherited a big T big P from one parent and a little tea, little pee from another parent. So we know that the two having two dominant phenotypes and two recessive phenotypes are the parental genotypes. So tall and purple and short and white would be parental. So I'm just gonna put a letter P next to those to remind us of that. And that means that the other two phenotypes tall and white, short and purple are the recombinant offspring. So R and R by those and we see when we look at our little uh all diagram for lack of a better word that when we have a crossover, which I've denoted with an X between the two, all that that would result in big T little P and little T big P as our recombinant offspring. So let's add up the number of recombinant offspring and that will be 85 tall and white Added with together with 41 short and purple, which is going to give us 126 recombinant. So 126 will be divided by and now we have to add together the total number of offspring, 134 plus 85 plus 56 plus 41. And that's going to give us A total of 316 total. So that 126 recombinant will be divided by total offspring Times 100%. And when we do that math, we will get 40% as our recombination frequency. So going back to our answer choices, that is choice B. So given the numbers we have here, the recombination frequency between the two genes associated with the piece height and flower color is choice B 40%. See you in the next video.

Key Concepts

Gene Linkage

Independent Assortment

Chi-Square Analysis

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