The petals of the plant Collinsia parviflora are normally blue, giving the species its common na

The petals of the plant Collinsia parviflora are normally...

Key Concepts

Mendelian Genetics

Mendelian genetics provides the foundational principles of inheritance such as dominance, segregation, and independent assortment. These principles allow us to predict the distribution of alleles and resulting phenotypes in offspring when pure-breeding parents (homozygous for different traits) are crossed, as well as identify cases where non-Mendelian ratios arise due to additional factors.

Two-Locus Interaction Models

In two-locus interaction models, two separate genes together determine a single trait. One gene may be necessary for the production of a pigment while the other determines which pigment is produced, creating multiple phenotypic outcomes that do not follow simple single-gene inheritance. These interactions can generate multiple classes of offspring in the F2 generation and may involve modifier roles.

Epistasis

Epistasis occurs when the effect of one gene is masked or modified by one or more other genes. In cases where one locus controls pigment production and a second locus controls pigment type, a recessive condition at the pigment production locus can override the effect of the pigment type gene, resulting in a phenotype (such as white) regardless of the genotype at the other locus.

Genotype-Phenotype Relationship

The genotype-phenotype relationship describes how the combination of alleles at one or more genetic loci leads to the observed traits in an organism. Understanding this relationship is crucial in genetics because it involves deciphering how gene interactions, dominance, and epistasis combine to produce the range of observed phenotypes.

Test Cross Analysis

Test cross analysis is a technique used to determine an unknown genotype by crossing the individual with a homozygous recessive partner. Analyzing the phenotypic ratios of the offspring can reveal the presence of dominant and recessive alleles, especially in situations where interactions between multiple genes affect the trait under study.

Transcript

00:01 This question says, the petals of the plant called colincia parviflora are normally blue, given the species its common name, blue -eyed mary.

00:09 Two pure breeding lines were obtained from color variants found in nature.

00:13 The first line had pink petals and the second line had white petals.

00:18 The following crosses were made between pure lines with the following results shown.

00:23 The following results are shown in the following way.

00:42 In this case, let's try to find the phenotype and the weight.

00:47 For example, in this case you have 101 plus 33 and this is going to be equal to 134.

00:58 Now, what is the weight for blue in this case? you have to divide 101 by 134 and this is going to be equal to approximately 0 .75.

01:11 Okay, and in this case for white, you have to divide 33 by 134 and this is going to be equal to 0 .25.

01:21 So, this is like, or these weighters are like for 3 to 1 weighter when you cross a heterozygous with a heterozygous.

01:28 But this would give you, well, maybe three phenotypes in case there is incomplete dominance, but let's check it out anyway here.

01:36 For example, you have 192 plus 63 and this is going to be equal to 255.

01:44 So, 192 divided by 255 is equal to 0 .75 also.

01:54 And if you divide 63 by 255, this is going to be equal to 0 .25 also.

02:04 Okay, now, finally you have this one here where the three phenotypes are present.

02:11 So, in this case you have 272 plus 121 plus 89, this is going to be equal to 482.

02:22 So, if you divide 272 by 482, this is going to be equal to 0 .56.

02:34 Then 121 divided by 482, this is going to be equal to 0 .25.

02:43 And 89 divided by 482, this is going to be equal to 0 .19 approximately.

02:56 So, this ratio is known, okay? because if you do only this, you may think about incomplete dominance and only one gene like here.

03:05 But when you see this ratio, it means that there are actually two genes here.

03:09 And this is the phenotypic ratio for recessive epistasis, 9, 3, 4, okay? because you have here a total of 16.

03:18 And if you divide 9 by 16, you're going to get approximately 0 .56.

03:26 If you divide 3 by 16, you're going to get approximately 0 .18.

03:31 And if you divide 4 by 16, you're going to get 0 .25.

03:35 As you can see here, these ratios are the same as this one here.

03:38 So, practically, this is what is happening.

03:40 And in this case, this 0 .56 matches with this one here.

03:43 So, practically, 9, 16.

03:45 Remember that in this case, the 9 stands for dominant and dominant.

03:50 3, 16 for dominant a and recessive b.

03:55 And this one here is for recessive a and dominant b.

04:00 Okay? and also, well, recessive a, recessive b.

04:04 Because recessive a in this case, we're going to make it epistatic, okay? so, in this case, what is happening here is that when you have the homozygous dominant for a or the heterozygous for a, you're going to express the b gene.

04:19 And when you have the homozygous for b or the heterozygous for b, you're going to have blue.

04:24 When you have the homozygous recessive for b, you're going to get pink.

04:31 And when you have the homozygous recessive for a, it doesn't matter what you have here in because you're going to have directly white.

04:41 This is recessive epistasis.

04:43 So, in this case, the question number one says, explain this resource genetically and also show the genetic constitution of the parents, the f1 and the f2 in each cross.

04:55 So, in this case, this is what is happening here.

04:57 And about the genotypes, in this case, blue, in this case, well, these are going to be homozygous because you want to produce only one phenotype in the f1.

05:06 So, in this case, you have blue and white.

05:10 And you can also make blue in many forms.

05:15 So, first, in this case, this blue is going to be heterozygous for both genes.

05:21 It means like this.

05:26 This is the genotype for this blue here.

05:28 And in this case, this blue is going to be when you cross two with this genotype, you're going to have this panel square.

05:37 And from this panel square, you're going to have these ratios here.

05:39 For example, for the 9 dominant a and dominant b, you have 1, 2, 3, 4, 5, 6, 7, 8, and 9.

05:48 Okay.

05:48 So, these are the genotypes that are going to make up for these 9.

05:53 And the same for these 316s and for these 416s.

05:55 In the case of 416, you have 1, 2, 3, and 4.

05:59 All those that have the homozygous recessive for a.

06:02 Now, for this blue here, oh, and what about these two here for pink? in order to make pink, you need the homozygous recessive b.

06:12 And well, you're going to have the following.

06:14 Homozygous dominant b, a, and homozygous recessive for b.

06:17 This is pink.

06:18 And if you want white, you need the homozygous recessive for a.

06:24 And in this case, the homozygous dominant for b.

06:28 Okay.

06:28 Because dominant b with recessive b, you get heterozygous for b.

06:32 And the same is going to happen here.

06:34 Now, for the second cross, this one here, you have blue and if you have pink, it means that you have the homozygous recessive for b.

06:47 If you have the homozygous recessive for b, it means that one recessive b came from this parent and one recessive b came from this parent.

06:56 Okay.

06:57 These are the parents.

06:58 And now, you also have, well, you have pink and you have blue and you have zero white.

07:07 If you have zero white, it means that this is not happening.

07:11 Okay.

07:12 So, you are not going to have recessive a and recessive a here.

07:15 You're going to have at least only one recessive a but not two recessive a's in both parents.

07:23 Now, this phenotypic ratio is three to one.

07:27 And you have that the three is for blue and the one is for pink.

07:31 It means it is happening that for the b gene...

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