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The mother of a family with 10 children has blood type \(\mathrm{Rh}^{+}\). She also has a very rare condition (elliptocytosis, phenotype \(\mathrm{E}\) ) that causes red blood cells to be oval rather than round in shape but that produces no adverse clinical effects. The father is \(\mathrm{Rh}^{-}\) (lacks the \(\mathrm{Rh}^{+}\) antigen) and has normal red blood cells (phenotype e). The children are \(1 \mathrm{Rh}^{+} \mathrm{e}, 4 \mathrm{Rh}^{+} \mathrm{E},\) and \(5 \mathrm{Rh}^{-}\) e. Information is available on the mother's parents, who are \(\mathrm{Rh}^{+} \mathrm{E}\) and \(\mathrm{Rh}^{-}\) e. One of the 10 children (who is \(\mathrm{Rh}^{+} \mathrm{E}\) ) marries someone who is \(\mathrm{Rh}^{+} \mathrm{e}\), and they have an \(\mathrm{Rh}^{+} \mathrm{E}\) child. a. Draw the pedigree of this whole family. b. Is the pedigree in agreement with the hypothesis that the \(R h^{+}\) allele is dominant and \(R h^{-}\) is recessive? c. What is the mechanism of transmission of elliptocytosis? d. Could the genes governing the \(\mathrm{E}\) and \(\mathrm{Rh}\) phenotypes be on the same chromosome? If so, estimate the map distance between them, and comment on your result.

Short Answer

Expert verified
The pedigree confirms Rh dominance and elliptocytosis as dominant. Genes likely unlinked.

Step by step solution

01

Analyze the Blood Type Inheritance

The mother's phenotype is \(Rh^{+} E\), while the father's phenotype is \(Rh^{-} e\). According to the given information, the dominant Rh factor is \(Rh^{+}\), so the mother's genotype must be \(Rh^{+} Rh^{-}\) and for elliptocytosis, it must be E (dominant inheritance). The father's genotype is \(Rh^{-}Rh^{-}\) and e. Therefore, children should inherit one allele from each parent.
02

Construct the Pedigree

For 10 children, the reported distribution is: 1 \(Rh^{+} e\), 4 \(Rh^{+} E\), and 5 \(Rh^{-} e\). Based on this, draw a pedigree chart including both parents and all children, indicating each child’s phenotype. Include the grandparents \(Rh^{+} E\) and \(Rh^{-} e\). Note the marriage and child of the \(Rh^{+} E\) child.
03

Evaluate Dominance Hypothesis

Examine whether the children's phenotypes are consistent with the hypothesis that \(Rh^{+}\) is dominant over \(Rh^{-}\). Since children show a mix consistent with one parent's dominant \(Rh^{+}\) and the other's \(Rh^{-}\), this supports \(Rh^{+}\) being dominant.
04

Analyze Elliptocytosis Inheritance

Elliptocytosis appears in the mother and most children. Its range of expression implies it is dominant. Hence, it follows the autosomal dominant pattern, as it appears when just one dominant E is present.
05

Examine Potential Gene Linkage

Check gene placement by phase: \(Rh\) and E. Relatedness is examined through recombination frequency. With 5 out of 10 not recombined (showing discordant transmission \(Rh^{-}e\)), consider 50% recombinant indicating independent assortment. Therefore, genes are unlinked, possibly on different chromosomes as per Mendel's law.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pedigree Analysis
Pedigree analysis is a graphical representation of a family tree that displays how genetic traits are passed down through generations. By marking individuals with their phenotypes or genotypes, we can trace the inheritance patterns of specific traits, such as the Rh blood group and elliptocytosis. In this exercise, the pedigree for a family shows the distribution of Rh phenotypes and elliptocytosis, using the symbols for Rh-positive (Rh+) and Rh-negative (Rh-) individuals, along with those with and without elliptocytosis (E or e).

By examining the pedigree, we determine how these traits are inherited among the children. This helps to reveal the underlying genetic principles like dominance or recessivity. For instance, when constructing the pedigree, we can see how the Rh factor and elliptocytosis are inherited from the parents to the children. Once constructed, this visualization can assist in analyzing the inheritance mechanism and in confirming hypotheses regarding dominance of particular alleles.
Rh Blood Group System
The Rh blood group system is crucial in determining blood types and is an excellent example of dominant and recessive allele interaction. The system is primarily concerned with the presence or absence of the Rh antigen on the surface of red blood cells, where Rh+ means the presence of antigen and Rh- its absence.

In this family, the mother's Rh+ status suggests she carries at least one Rh+ allele, with the Rh+ allele being dominant over Rh-. Thus, her genotype is likely Rh+Rh-, highlighting that individuals can express Rh+ phenotype even with only one positive allele. The father, who is Rh-, must have an Rh-Rh- genotype, passing only Rh- alleles to the offspring. This compatible mix of genotypes leads to the observation that the Rh+ allele indeed expresses dominance over the Rh- allele, as evidenced by the variation in the children's Rh status.
Elliptocytosis Inheritance
Elliptocytosis is a hereditary red blood cell condition characterized by oval-shaped cells instead of the typical round shape. While it doesn't generally adversely affect health, it shows a dominant inheritance pattern. This means that only one copy of the dominant allele (E) is necessary to show the trait.

In the given family context, the mother has elliptocytosis and passes it to four of her children with the genotype Rh+E. The prevalence of elliptocytosis in multiple children reinforces its autosomal dominant transmission, where even if a child inherits only one dominant E allele, the phenotype is expressed. An autosomal dominant trait requires just one affected parent to possess the dominant allele and it can manifest in approximately half of the offspring if one parent exhibits the trait.
Gene Linkage and Recombination
Gene linkage refers to the tendency of genes that are located close to each other on a chromosome to be inherited together. When genes are not linked, they follow Mendel's law of independent assortment, which means they segregate independently into gametes. Gene recombination occurs due to the exchange of genetic material during meiosis, and it's used to measure the distance between genes on a chromosome.

In our family scenario, we have elucidated potential linkage between Rh and elliptocytosis by analyzing recombinant offspring. With offspring distribution showing 5 out of 10 with Rh- e (neither Rh+ nor E from the mother), a 50% recombination rate implies no linkage between these two traits. In genetics, linkage is discernible at recombination rates less than 50%. Thus, this full recombination supports that Rh and elliptocytosis genes are located on separate chromosomes or are far apart, corroborating independent assortment as per Mendel's laws.

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Most popular questions from this chapter

In the tiny model plant Arabidopsis, the recessive allele hyg confers seed resistance to the drug hygromycin, and her, a recessive allele of a different gene, confers seed resistance to herbicide. A plant that was homozygous hyg/hyg. her/her was crossed with wild type, and the \(\mathrm{F}_{1}\) was selfed. Seeds resulting from the \(\mathrm{F}_{1}\) self were placed on petri dishes containing hygromycin and herbicide. a. If the two genes are unlinked, what percentage of seeds are expected to grow? b. In fact, 13 percent of the seeds grew. Does this percentage support the hypothesis of no linkage? Explain. If not, calculate the number of map units between the loci. c. Under your hypothesis, if the \(\mathrm{F}_{1}\) is testcrossed, what proportion of seeds will grow on the medium containing hygromycin and herbicide?

In the fungus Neurospora, a strain that is auxotrophic for thiamine (mutant allele \(t\) ) was crossed with a strain that is auxotrophic for methionine (mutant allele \(m\) ). Linear asci were isolated and classified into the following groups. $$\begin{array}{lcccccc}\text { Spore pair } & {}{} {\text { Ascus types }} \\\ \hline 1 \text { and } 2 & t+ & t+ & t+ & t+ & t m & t m \\ 3 \text { and } 4 & t+ & t m & +m & ++ & t m & ++ \\ 5 \text { and } 6 & +m & ++ & t+ & t m & ++ & t+ \\ 7 \text { and } 8 & +m & +m & +m & +m & ++ & +m \\ \hline \text { Number } & 260 & 76 & 4 & 54 & 1 & 5 \\ \hline\end{array}$$ a. Determine the linkage relations of these two genes to their centromere(s) and to each other. Specify distances in map units. b. Draw a diagram to show the origin of the ascus type with only one single representative (second from right

A plant of genotype $$\begin{array}{cc}A & B \\\\\hline \hline a & b\end{array}$$ is testcrossed with $$\begin{array}{cc}a & b \\\\\hline \hline a & b\end{array}$$ If the two loci are \(10 \mathrm{m}\). \(u\). apart, what proportion of progeny will be \(A B / a b ?\)

The \(A\) locus and the \(D\) locus are so tightly linked that no recombination is ever observed between them. If \(A d / A d\) is crossed with \(a D / a D\) and the \(F_{1}\) is intercrossed, what phenotypes will be seen in the \(\mathrm{F}_{2}\) and in what proportions?

In a tetrad analysis, the linkage arrangement of the \(p\) and \(q\) loci is as follows: Assume that in region i, there is no crossover in 88 percent of meioses and there is a single crossover in 12 percent of meioses; in region ii, there is no crossover in 80 percent of meioses and there is a single crossover in 20 percent of meioses; and there is no interference (in other words, the situation in one region does not affect what is going on in the other region) What proportions of tetrads will be of the following types? (a) \(\mathrm{M}_{\mathrm{f}} \mathrm{M}_{\mathrm{I}}, \mathrm{PD}\) \((\mathbf{b}) \mathrm{M}_{1} \mathrm{M}_{\mathrm{l}}, \mathrm{NPD}\) \((\mathbf{c}) M_{1} M_{11}, T ;\) (d) \(\mathrm{M}_{\mathrm{D}} \mathrm{M}_{\mathrm{l}}, \mathrm{T}\) \((\mathbf{e}) M_{1} M_{\Perp}, P D\) \((\mathbf{f}) M_{11} M_{11}, N P D ;\) (g) \(\mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{U}}\), T. (Note: Here the M pattern written first is the one that pertains to the \(p\) locus.) Hint: The easiest way to do this problem is to start by calculating the frequencies of asci with crossovers in both regions, region i, region ii, and neither region. Then determine what \(\mathrm{M}_{1}\) and \(\mathrm{M}_{\mathrm{II}}\) patterns result.

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