Summary and Interpretation of Data 1. Compile the data for...

Summary and Interpretation of Data 1. Compile the data for all the baby dragons produced by all the students in the following chart. Mother (WwHh) wh wH Wh WH Father (wwhh) wh Genotype of baby = ______ Number of babies with this genotype = ____ Phenotype: Wings ___ or no wings ___ Horns big ___ or small ___ Genotype of baby = ______ Number of babies with this genotype = ____ Phenotype: Wings ___ or no wings ___ Horns big ___ or small ___ Genotype of baby = ______ Number of babies with this genotype = ____ Phenotype: Wings ___ or no wings ___ Horns big ___ or small ___ Genotype of baby = ______ Number of babies with this genotype = ____ Phenotype: Wings ___ or no wings ___ Horns big ___ or small ___ 2. Do any of the baby dragons with wings have small horns? Does either parent have the combination of wings and small horns? (See page 1.) How did this new combination of characteristics (wings and small horns) arise in some of the baby dragons? (Your answer will include events during meiosis and fertilization, so you may find it helpful to review the diagram of meiosis on page 2 and the chart of fertilization on page 3.) 3. On page 3 you used your understanding of meiosis and fertilization to predict whether baby dragons with wings and without wings would be equally likely to have big horns. What was your prediction? Use the class results shown above to complete the following table. Fraction that have big horns Baby dragons with wings Baby dragons without wings Results Do your results match your prediction?

Transcript

00:01 Not yet to answer this question, let's talk about inheritance.

00:03 Apparently in this case, there is no background information, but in this case, the homo -sigous -dominant for w or the heterocybin for w is going to call for wings, and the homo -sigrisisive is for? no wins.

00:18 No wings.

00:20 And for the h -g -in -bio has the homocylus dominant and the heterocylus are going to call for, for in this case, big horns, and the homo -sacigizisive is for.

00:35 Small horns.

00:39 Now, it says, compile the data for all the baby dragons produced by all the students in the following chart.

00:47 Well, as i told you, there is no background information here, so i'm going to fill this as much as i can.

00:53 You have the following chart.

00:56 You have this chart, okay? so you have a cross between a heterocygose, a model that is heterocygote for both genes, and a father that is homozygib excessive for both genes.

01:05 So if we perform that cross, you're going to have.

01:07 Well, you first have to find the gametes, and you're going to find the gamut by mixing each of these w .ailil with each of this h alleles, in the same here.

01:16 So in this pattern, you have four possible gametes that are this, this, this, this, and this.

01:21 Okay, again, by mixing each of these double you adiles with each of these h alleles.

01:26 And in this case, for the father, there is only one different possibility that is this gamut.

01:31 Okay, so you're going to cross this with this, then this with this, and so on.

01:35 So in this case, the genotype for each of these offspring here are going to be.

01:42 In this case, you have the homocygos recessive with the heterozygos.

01:52 In this case, you have the homocygotech with the heterocygose.

01:56 In this case, you have the heterocyboceph with the homocybozygos excessive.

02:00 And in this case, you have the heterozygoyles with the heterocytes.

02:05 So you have the genotypes for each of these babies.

02:10 Now it's a number of babies with this genotype.

02:12 So if we suppose that there are four babies, okay, then you just have to multiply each of these fractions by four.

02:21 Because in this case, for this genotype, okay, for this genotype, you find it, this is homozygonquessence for both age.

02:32 Okay, so i'm sorry to get a mistake here.

02:35 You got this.

02:36 Now, for this specific genotype, you find it only once.

02:40 Okay, so you have that.

02:41 Out of four possibilities, only one is homozygotesis for both genes.

02:46 So you have one out of four that is one quarter.

02:49 And if you have four baby dragons, then one quarter multiplied by four is equal to one.

02:54 So you have one.

02:55 Again, we are supposing that you have four baby dragons, okay? but anyway, let's suppose we have eight.

03:02 Just to make another example.

03:05 If you have eight, and the probability for getting this genotype here is one quarter, one out of four.

03:11 You have one quarter multiplied by 8 and this is equal to 2.

03:16 So practically 2 is the number of babies with this genotype here.

03:21 Now it says phenotype, wings or no wings.

03:25 And this is for no wings.

03:26 So you have no wings and also small horns.

03:33 So you have small horns.

03:35 In this case, for this genotype, you only find it once because you don't find this genotype here or here or here.

03:45 So this is also 1 quarter multiplied by 8 is equal to 2.

03:49 And this genotype goes for, again, no wings and big horns.

03:55 Now, this genotype, you also find it once, so you have 1 quarter multiplied by 8, this is equal to.

04:01 And the phenotype is going to be for wings, for the presence of wings, and also for the legislative age that is small horns...

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