Question

The gene SLC24A5 encodes an antiporter membrane protein that exchanges sodium for calcium (R. Ginger et al., JBC, 2007). This process has a role in the synthesis of the melanosomes that cause skin pigmentation. A mutation in this gene affecting a single amino acid occurs in humans. The homozygous mutant gene is found in 99% of humans with European origins. Both the wild type and mutant display codominance. A. Representing the wild-type form of the gene as +/+ and the mutant form of the gene as m/m for two homozygous parents, construct a Punnett square for this cross using the first grid below. Annotate your representation to identify the phenotypes with high (H), intermediate (I), and low (L) melanosome production. Use the second grid to represent an F2 generation from the offspring of the first cross. Use annotation to show the phenotype. B. Draw sister chromatids at anaphase II for both parents in the F1 generation and annotate your drawing to identify each genotype of the gametes using the cells of the Punnett square. C. Explain which of Mendel’s laws is violated by codominance. D. Suppose that these data were available to evaluate the claim that the wild-type and mutant forms of SLC24A5 are codominant: $\begin{array}{|c|c|}\hline \text { Phenotype } & {\text { Observed }} & {\text { Expected }} \\ \hline \mathrm{H} & {1206} \\ \hline \mathrm{H} & {1206} \\ \hline 1 & {2238} \\ \hline \mathrm{L} & {1124} \\ \hline\end{array}$ Complete the table. Explain the values expected in terms of the genotype of the offspring. E. Using a c2 statistic at the 95% confidence level, evaluate the claim that the wild-type and mutant forms of SCLO24A5 are codominant. The definition of the statistic $X_{c}^{2}=\sum \frac{\left(O_{i}-E_{i}\right)^{2}}{E_{i}}$ where X is the chi-square test statistic, c is the significant level of the test (we will use 0.05), O is the observed value for variable i, and E is the expected value for variable i. The Chi-square statistic table is provided in the AP Biology Exam.

Name: Problem 39, Chapter 12: Biology for AP Courses
Uploaded: 2020-11-10T19:44:08-08:00
Duration: 11 min 31 s
Channel: Bryan Lynn

   The gene SLC24A5 encodes an antiporter membrane protein that exchanges sodium for calcium (R. Ginger et al., JBC, 2007). This process has a role in the synthesis of the melanosomes that cause skin pigmentation. A mutation in this gene affecting a single amino acid occurs in humans. The homozygous mutant gene is found in 99% of humans with European origins. Both the wild type and mutant display codominance.

A. Representing the wild-type form of the gene as +/+ and the mutant form of the gene as m/m for two homozygous parents, construct a Punnett square for this cross using the first grid below. Annotate your representation to identify the phenotypes with high (H), intermediate (I), and low (L) melanosome production. Use the second grid to represent an F2 generation from the offspring of the first cross. Use annotation to show the phenotype.
B. Draw sister chromatids at anaphase II for both parents in the F1 generation and annotate your drawing to identify each genotype of the gametes using the cells of the Punnett square.
C. Explain which of Mendel’s laws is violated by codominance.
D. Suppose that these data were available to evaluate the claim that the wild-type and mutant forms of SLC24A5 are codominant:
$\begin{array}{|c|c|}\hline \text { Phenotype } & {\text { Observed }} & {\text { Expected }} \\ \hline \mathrm{H} & {1206} \\ \hline \mathrm{H} & {1206} \\ \hline 1 & {2238} \\ \hline \mathrm{L} & {1124} \\ \hline\end{array}$
Complete the table. Explain the values expected in terms of the genotype of the offspring.
E. Using a c2 statistic at the 95% confidence level, evaluate the claim that the wild-type and mutant forms of SCLO24A5 are codominant. The definition of the statistic $X_{c}^{2}=\sum \frac{\left(O_{i}-E_{i}\right)^{2}}{E_{i}}$ where X is the chi-square test statistic, c is the significant level of the test (we will use 0.05), O is the observed value for variable i, and E is the expected value for variable i. The Chi-square statistic table is provided in the AP Biology Exam.

Biology for AP Courses

Julianne Zedalis,…

Chapter 12, Problem 39

Instant Answer

Solved by Expert Bryan Lynn

11/10/2020

Step 1

We can construct a Punnett square for this cross as follows: | | + | + | |---|---|---| | m | +m| +m| | m | +m| +m| All offspring from this cross will have the genotype +m, which will result in an intermediate level of melanosome production due to codominance. Show more…

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The gene SLC24A5 encodes an antiporter membrane protein that exchanges sodium for calcium (R. Ginger et al., JBC, 2007). This process has a role in the synthesis of the melanosomes that cause skin pigmentation. A mutation in this gene affecting a single amino acid occurs in humans. The homozygous mutant gene is found in 99% of humans with European origins. Both the wild type and mutant display codominance. A. Representing the wild-type form of the gene as +/+ and the mutant form of the gene as m/m for two homozygous parents, construct a Punnett square for this cross using the first grid below. Annotate your representation to identify the phenotypes with high (H), intermediate (I), and low (L) melanosome production. Use the second grid to represent an F2 generation from the offspring of the first cross. Use annotation to show the phenotype. B. Draw sister chromatids at anaphase II for both parents in the F1 generation and annotate your drawing to identify each genotype of the gametes using the cells of the Punnett square. C. Explain which of Mendel’s laws is violated by codominance. D. Suppose that these data were available to evaluate the claim that the wild-type and mutant forms of SLC24A5 are codominant: $\begin{array}{|c|c|}\hline \text { Phenotype } & {\text { Observed }} & {\text { Expected }} \\ \hline \mathrm{H} & {1206} \\ \hline \mathrm{H} & {1206} \\ \hline 1 & {2238} \\ \hline \mathrm{L} & {1124} \\ \hline\end{array}$ Complete the table. Explain the values expected in terms of the genotype of the offspring. E. Using a c2 statistic at the 95% confidence level, evaluate the claim that the wild-type and mutant forms of SCLO24A5 are codominant. The definition of the statistic $X_{c}^{2}=\sum \frac{\left(O_{i}-E_{i}\right)^{2}}{E_{i}}$ where X is the chi-square test statistic, c is the significant level of the test (we will use 0.05), O is the observed value for variable i, and E is the expected value for variable i. The Chi-square statistic table is provided in the AP Biology Exam.

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Key Concepts

Punnett Square

A Punnett square is a grid used in genetics to predict the genotypes and phenotypes of offspring from a particular cross by organizing all possible gamete combinations. It provides a visual representation of how alleles can combine based on parental genotype contributions, allowing for the calculation of expected ratios of phenotypes and genotypes in the offspring.

Codominance

Codominance is a genetic scenario where both alleles at a locus are fully expressed in the heterozygote, with neither allele being recessive. In codominance, the phenotype of the heterozygote is distinct, exhibiting traits from both alleles simultaneously. This pattern deviates from the classic dominant-recessive relationship and is important in understanding more complex inheritance patterns.

Meiosis and Sister Chromatids

Meiosis is the cellular process that produces gametes and involves two rounds of division. During anaphase II, sister chromatids separate and are distributed into different gametes. Understanding this process is crucial for genetics because it explains how alleles can be assorted into gametes and how genetic variability is generated through the separation of homologous chromosomes and sister chromatids.

Mendel’s Laws

Mendel’s Laws, including the law of segregation and the law of independent assortment, describe the fundamental principles of heredity. However, these laws rely on the assumption of complete dominance, which is violated in cases of codominance. Codominance challenges the conventional view of dominance by displaying both alleles in the phenotype, highlighting that the inheritance patterns can be more complex than Mendel’s original observations.

Chi-square Test in Genetics

The chi-square test is a statistical method used in genetics to compare observed data with expected data under a specific genetic model. It quantifies the discrepancy between the observed frequencies of phenotypes and those predicted by Mendelian ratios, helping researchers determine whether deviations are due to random chance or indicate a departure from the proposed model of inheritance.

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Transcript

00:01 So here we're talking about mendel's experiments and heredity.

00:05 And we are starting with having two parent genes.

00:11 So this is for skin color, a plus plus gene being producing melan, and then the mm being the mutant, which is not producing it.

00:26 So the mm is going to be white and european folks, and the plus plus would be darker skin tones.

00:34 And so part a, this problem wants us to do a couple of the ponent squares.

00:41 So the first generation is going to be to cross the two parents, right? so we would take m -m, this will be our first generation, and plus plus, and cross that.

00:58 And so we're going to get a plus m for every offspring.

01:07 And since this is codominence, both are expressed, so we end up getting an intermediate level of skin color.

01:17 Now when we move on to the second generation, we're going to be crossing the offspring from the first generation, and they were the same.

01:25 So we're just doing a plus m cross with a plus m.

01:31 And so when we do this cross, we get some variation.

01:34 We get plus plus.

01:36 We get 2 plus m and 2 mm.

01:43 And so this is going to give us, the plus plus is going to have a higher amount of pigmentation.

01:50 The plus m is being intermediate due to the co -dominance and then the mm being low because it's the double mutant here.

02:03 Okay, and so part b wants us to do a little drawing and so a sister chromatid right is when we have the two chromosomes connected there.

02:19 Now our parents from the f1 generation were homozygous, we would have an m, m, and then another sister chromatid from parent number two, which was plus plus.

02:33 And anaphae's two is the one where it splits up into the gametes, right? so we're going to end up with one gamete that has an m, and we'll have another one that's the same way.

02:52 It'll be a lot.

02:52 It'll then this one will also split up into two gametes, of course with the plus.

03:04 And so how these gametes relate is that they are what we're putting in the sort of column headings, you could call it in the row headings.

03:21 So these gametes are the information we're using to create our crosses.

03:32 Okay, so this little drawing here was part b and if we move on to part c, this is just a quick little question about mendel's laws.

03:44 And so which one is violated by co -dominance? so mendel's law of independent assortment basically states that one allele does not affect another.

04:04 So this is that if you're allel for height, maybe doesn't affect your allel for eye color, but but in the case of co -dominance, they are impacting each other.

04:31 Right? so if you have a red flower petal and a white flower pedal and they're co -dominant and they make some pink flower petals, then the white one is making the red one lighter or the red one is making the white one darker.

04:46 But they do affect each other.

04:49 So it's going to be this law of independent assortment, which is violated.

04:55 Okay.

04:55 Now we can move on to part d so this is a long problem here and we are given a grid so we have the phenotype of high intermediate and low and we are able to observe some quantities okay so we have 1206 for high 2238 for intermediate and 1124 for for low and we want to know what we would expect.

05:41 Well, if we go back up real quick to our punt in square, we can see that 25 % are going to have a high...

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