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How is the probability of Rh type inheritance calculated?

How is the probability of Rh type inheritance calculated?


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I found somewhere on the Internet that, if parents are both Rh-positive, their children have 93.75% chance of being RH-positive and 6.25% chance of being RH-negative.

My question is: How is this calculated?


There are four possible parental genotype combinations (R = Rh+ allele, r = Rh- allele):

m f RR RR Rr RR RR Rr Rr Rr

Only the last can produce Rh- offspring. The probabilty that theRr Rrcombination leads to arrgenotype in the F1 generation is $1 over 4$. Since there are four combinations, the probability that the Rh+ parents areRh Rhis also $1over4$. Thus the probability of the offspring to be Rh- is $${1over 4} cdot {1 over 4} = {1 over 16} = 0.0625$$ Thus, the probability of Rh+ in the F1 generation is $$1 - {1 over 16} = {15 over 16} = 0.9375$$

Note: This is the rationale behind the values you state in your question. I am pretty sure that these four combinations are not equally distributed in a population. You would have to need to know the allele frequencies.


How is the probability of Rh type inheritance calculated? - Biology

PROBABILITY/STATISTICS IN INHERITANCE

We know that when two people who are both heterozygous for a simple Mendelian autosomal gene alpha have a child, the probability that the child will show the dominant phenotype is 3/4. Let's ask a somewhat more complex question. If this couple has a total of four children, what is the probability that 3 of the 4 will show the dominant phenotype? To answer this, we will first derive the appropriate formula and then use it to calculate the numerical answer. The same formula allows us to understand the expected statistical distribution of the various possible phenotype patterns in four-child ( or any other size) families in a large population.


1. Review: How can some simple overall probabilities be calculated by combining the multiplication and addition "rules" we covered earlier?

Let's start with a very simple case: asking about gender probabilities in families of three children.

What is the probability that all three children in a family will be the same gender?
P(all female)= 1/2 x 1/2 x 1/2 = 1/8
P(all male ) = 1/2 x 1/2 x 1/2 = 1/8
P(all one gender) = P(all female) + P(all male) = 1/8 + 1/8 = 1/4

What is the probability that a three-child family is two girls and one boy?
Each possible birth order has P=1/8. That is, P(G,G,B)=P(G,B,G)=P(B,G,G)=1/8.
So, P(2G,1B)= 3/8 and P(1G,2B)= 3/8.

This allows us to write the overall gender probability distribution for families of three children as follows:
1/8 will be three girls
3/8 will be two girls and one boy
3/8 will be one girl and two boys
1/8 will be three boys
Adding it all up, we have 1/8 + 3/8 + 3/8 + 1/8 = 1 (100%)


2. How can we understand and use "Pascal's triangle" and the "general rule for repeated trials of events with constant probabilities", formula 1 (page 161 in textbook)?

Consider the numerators of the fractions in the above three-child family gender equation:
1 , 3 , 3 , 1. These numbers are the coefficients in the expansion of the term (p + q) cubed. In general, the coefficients of any such binomial expansion < the term (p+q) raised to any power>give the "number of ways" that something can happen.

Figure 4.21, "Pascal's triangle", shows these coefficients for the expansion of (p + q) raised to any power up to 10. The numbers in any row can be used just as described above. For example, assume that over the next two decades you have 6 kids. There are 64 possible gender birth orders, with 20 of these resulting in you having three girls and three boys.

The terms p and q are the individual probabilities for a specific outcome from a single "event". For "gender" calculations, the probabilities p and q are equal, both = 1/2 (the equal probabilities of male and female births).

For "dominant : recessive phenotype" type of calculations, p and q will usually not be equal. For a simple Mendelian inheritance from two heterozygotic parents, p will = 3/4 (if AA and Aa give dominant phenotype) and q will = 1/4 (aa gives recessive phenotype).

Generalizing this, we get to the formula on page 161 in your text that is "the general rule for repeated trials of events with constant probabilities". The term (n!/s!t!) is the number of possible ways (orders) of getting a certain net outcome ( "a total of n with s of one and t of the other"). This number can either be calculated or taken directly from Pascal's triangle.


3. Sample problem: You and your mate are both heterozygous for some simple Mendelian gene alpha (i.e., each of you has genotype Aa, and both of you show the dominant phenotype) on chromosome #1. Over the next decade, you proceed to have four children. What is the probability that 3 of your children will show the dominant phenotype and one will show the recessive phenotype? What are the probabilities of the other possible outcomes?

If we were looking at thousands of such families, we know that the overall ratio of dominant to recessive phenotypes in the children would average out to 3:1, as shown by a simple Punnett square. But for one couple having four children, what's the probability, P(3D,1r)?

To calculate P(3D,1r), we use formula 1 for the case n=4, s=3, t=1, p=3/4, q=1/4.
P(3D,1r)= 4!/3! x (3/4)cubed x (1/4) = 4 x (27/64) x 1/4 = .42 ( 42% )

By also calculating the other four possibilities, we can construct a graph that shows the statistical distribution you would expect to see in a large population.

Problem S-4: "Heterozygous parents have three children".

Modify the sample problem above to do the calculations for you and your mate (both Aa) having THREE children. Do the calculations for the probabilities that all three, two, one, or none of the three children will show the dominant phenotype from gene alpha. Construct the graph and compare the result with the "four children" graph done in class.


Scientific journal articles for further reading

Hoekstra C, Zhao ZZ, Lambalk CB, Willemsen G, Martin NG, Boomsma DI, Montgomery GW. Dizygotic twinning. Hum Reprod Update. 2008 Jan-Feb14(1):37-47. Epub 2007 Nov 16. Review. PubMed: 18024802.

Machin G. Familial monozygotic twinning: a report of seven pedigrees. Am J Med Genet C Semin Med Genet. 2009 May 15151C(2):152-4. doi: 10.1002/ajmg.c.30211. PubMed: 19363801.

Mbarek H, Steinberg S, Nyholt DR, Gordon SD, Miller MB, McRae AF, Hottenga JJ, Day FR, Willemsen G, de Geus EJ, Davies GE, Martin HC, Penninx BW, Jansen R, McAloney K, Vink JM, Kaprio J, Plomin R, Spector TD, Magnusson PK, Reversade B, Harris RA, Aagaard K, Kristjansson RP, Olafsson I, Eyjolfsson GI, Sigurdardottir O, Iacono WG, Lambalk CB, Montgomery GW, McGue M, Ong KK, Perry JR, Martin NG, Stefánsson H, Stefánsson K, Boomsma DI. Identification of Common Genetic Variants Influencing Spontaneous Dizygotic Twinning and Female Fertility. Am J Hum Genet. 2016 May 598(5):898-908. doi: 10.1016/j.ajhg.2016.03.008. Epub 2016 Apr 28. Pubmed: 27132594.

Painter JN, Willemsen G, Nyholt D, Hoekstra C, Duffy DL, Henders AK, Wallace L, Healey S, Cannon-Albright LA, Skolnick M, Martin NG, Boomsma DI, Montgomery GW. A genome wide linkage scan for dizygotic twinning in 525 families of mothers of dizygotic twins. Hum Reprod. 2010 Jun25(6):1569-80. doi: 10.1093/humrep/deq084. Epub 2010 Apr 8. PubMed: 20378614. Free full-text available from PubMed Central: PMC2912534.


Part 4: Pedigree Analysis

We will trace the inheritance pattern of the autosomal recessive trait albinism through four generations. The legend is as follows:

In the pedigree chart below determine the genotypes of each individual. Use a Punnet Square analysis to help you. Remember that the genotype of affected individuals is nn. If you cannot determine both gene pairs of a normal individual, indicate the genotype as N_. Put the genotype next to each symbol.


Blood transfer table

Because of the existence of blood types, not every person&aposs blood can be transferred to a person in need. We even created a separate blood donor calculator to help you avoid confusion when donating or receiving blood!

The rules of blood transfer are:

A person with an AB blood type can receive blood from everyone else, but can only donate blood for other individuals with AB type.

A person with an A or B blood type can receive blood from everyone except AB, and can donate blood to other individuals with the same type. They can also donate blood to people with type AB.

A person with a 0 blood type can receive blood only from individuals with the same blood type. However, they can donate blood to individuals with all blood types.

A person with an Rh+ blood type can donate only to individuals with Rh+ blood type, but can receive blood from both Rh+ and Rh-.

A person with an Rh- blood type can donate to individuals with both Rh+ and Rh- blood types, but can receive blood only from Rh-.


Normal Distribution

The normal distribution or Gaussian distribution is a continuous probability distribution that follows the function of:

where &mu is the mean and σ 2 is the variance. Note that standard deviation is typically denoted as σ. Also, in the special case where &mu = 0 and σ = 1, the distribution is referred to as a standard normal distribution. Above, along with the calculator, is a diagram of a typical normal distribution curve.

The normal distribution is often used to describe and approximate any variable that tends to cluster around the mean. For example, the heights of male students in a college, the leaf sizes on a tree, the scores of a test, etc. Use the "Normal Distribution" calculator above to determine the probability of an event with a normal distribution lying between two given values (i.e. P in the diagram above) for example, the probability of the height of a male student is between 5 and 6 feet in a college. Finding P as shown in the above diagram involves standardizing the two desired values to a z-score by subtracting the given mean and dividing by the standard deviation, as well as using a Z-table to find probabilities for Z. If for example it is desired to find the probability that a student at a university has a height between 60 inches and 72 inches tall given a mean of 68 inches tall with a standard deviation of 4 inches, 60 and 72 inches would be standardized as such:

Given &mu = 68 σ = 4
(60 - 68)/4 = -8/4 = -2
(72 - 68)/4 = 4/4 = 1

The graph above illustrates the area of interest in the normal distribution. In order to determine the probability represented by the shaded area of the graph, use the standard normal Z-table provided at the bottom of the page. Note that there are different types of standard normal Z-tables. The table below provides the probability that a statistic is between 0 and Z, where 0 is the mean in the standard normal distribution. There are also Z-tables that provide the probabilities left or right of Z, both of which can be used to calculate the desired probability by subtracting the relevant values.

For this example, to determine the probability of a value between 0 and 2, find 2 in the first column of the table, since this table by definition provides probabilities between the mean (which is 0 in the standard normal distribution) and the number of choice, in this case 2. Note that since the value in question is 2.0, the table is read by lining up the 2 row with the 0 column, and reading the value therein. If instead the value in question were 2.11, the 2.1 row would be matched with the 0.01 column and the value would be 0.48257. Also note that even though the actual value of interest is -2 on the graph, the table only provides positive values. Since the normal distribution is symmetrical, only the displacement is important, and a displacement of 0 to -2 or 0 to 2 is the same, and will have the same area under the curve. Thus, the probability of a value falling between 0 and 2 is 0.47725 , while a value between 0 and 1 has a probability of 0.34134. Since the desired area is between -2 and 1, the probabilities are added to yield 0.81859, or approximately 81.859%. Returning to the example, this means that there is an 81.859% chance in this case that a male student at the given university has a height between 60 and 72 inches.

The calculator also provides a table of confidence intervals for various confidence levels. Refer to the Sample Size Calculator for Proportions for a more detailed explanation of confidence intervals and levels. Briefly, a confidence interval is a way of estimating a population parameter that provides an interval of the parameter rather than a single value. A confidence interval is always qualified by a confidence level, usually expressed as a percentage such as 95%. It is an indicator of the reliability of the estimate.


Red cell antigens and antibody

Brian Castillo , . Amer Wahed , in Transfusion Medicine for Pathologists , 2018

Rh System

The Rh blood group system , ISBT number (004)/symbol (RH)/CD number (CD240D/CD240CE) is complex and contains many antigens that are highly immunogenic. It was first discovered in 1939 but confirmed in 1940 in a series of experiments performed by Landsteiner and Weiner in which they evaluated the immunologic response of rabbits injected with RBCs collected from rhesus monkey. It currently contains 52 antigens which include 14 polymorphic (D, C, E, e, F, Ce, G hr s , C G , RH26 (c-like), cE, hr B , and Rh41), 26 low prevalence (C w , C x , V ^ , E w , VS ^ , CE, D w , hr h , Go a , Rh32, Rh33, Rh35, Be a , Evans, Tar, Crawford, Riv, JAL, STEM, FPTT, BARC, JAHK, DAK ^ , LOCR, and CENR), and 12 high prevalence (Hr0, Hr, Rh29, H b , Rh39, Nou, Sec, Dav, MAR, CEST, CELO, and CEAG) antigens. Rh-associated glycoprotein (RhAG) is essential for expression of Rh antigens. The RhAG blood group system, ISBT number (030)/symbol (RhAG)/CD number (CD241) was elevated to its own blood group system in 2008. It currently contains two low prevalence (Ol a , RhAG4) and 2 high prevalence (Dclos, DSLK) antigens. The Rh antigenic determinants are encoded by the RHD (D) and RHCE (C, E, c, e) genes, while the RhAG antigenic determinants are encoded by the RhAG gene [1] .

The RHD and RHCE genes are both located on chromosome 1p36.11 and each consists of 10 exons distributed over 69 kbp of gDNA in opposite orientation with their 3’ end facing each other. The opposite orientation and the presence of a “hairpin” formation allow homologous DNA segments in close proximity to engage in gene recombination. Collectively they encode two protein structures compromised of 416 AA that differ by approximately 32–35 AA depending on the antigenic expression of the RhCE haplotype. Both protein structures are multipass glycoproteins with each spanning the membrane 12 times with an endocellular N-terminal and C-terminal domains. Approximately, 30,000–32,000 copies of the RHD protein are found on the surface of RBCs. In total, there are 100,000–200,000 RhD and RhCE structures in combination on RBCs. The Rh proteins (RhD and RhCE) form a core complex with the RhAG glycoprotein. Similar to the Rh protein, it spans the membrane 12 times and has an endocellular N-terminal and C-terminal domains. There are approximately 100,000–200,000 copies per RBCs. Presence of the RhAG protein is essential for Rh-antigen expression. Additionally, Rh/RhAG complex interacts with band 3, GPA, GPB, LW, CD47, ankyrin, and protein 4.2. This interaction is essential for the erythrocyte membrane integrity, as noted by the presence of stomatocytes in the Rhnull phenotype.


Punnett Squares

The above listing can be more compactly demonstrated by using a Punnett square. This type of diagram is named after Reginald C. Punnett. Although it can be used for more complicated situations than the ones that we will consider, other methods are easier to use.

A Punnett square consists of a table listing all of the possible genotypes for offspring. This is dependent upon the genotypes of the parents being studied. The genotypes of these parents are typically denoted on the outside of the Punnett square. We determine the entry in each cell in the Punnett square by looking at the alleles in the row and column of that entry.

In what follows we will construct Punnett squares for all possible situations of a single trait.


Punnett Square Practice

7. Determining the probability of inheriting a combination can be accomplished using a punnett square. Consider two parents with the genotypes BbGg.

What color eyes do these parents have? __________________

8. A punnett square has been set up below, fill in the boxes to show the possible offspring of this cross.

9. According to the cross, how many offspring will have brown eyes?

How many will have green eyes?

How many will have blue eyes?


Recombination Frequency

To determine how close together on a gene two alleles are using reproductive data alone – that is, to solve gene mapping problems – scientists look at the difference between the predicted phenotypic ratios in a population of offspring and the actual ratios.

This is done by crossing a "dihybrid" parent with an offspring that shows both recessive traits. In the case of your alien linkage biology, this means crossing a purple-haired, round-headed alien (which, in the case of a dihybrid organism, has the genotype PpRr) with the least likely product of such a mating – a yellow-haired, flat-headed alien (pprr).

Suppose this yields the following data for 1,000 offspring:

  • Purple hair, round head: 404
  • Purple hair, flat head: 100
  • Yellow hair, round head: 98
  • Yellow hair, flat head: 398

The key to solving linkage mapping problems is to recognize that if two genes are not linked, then the offspring's genotypes and phenotypes should be produced in essentially equal amounts. However, as you can see from the results of the data, they are not produced in equal amounts.

For linked genes, the parental chromosomal configurations (aka PpRr, purple-haired round headed, and pprr, yellow-haired flat-headed) are over-represented in number, while the recombinant configurations (Pprr and ppRr) are far fewer.

This allows for the calculation of the recombination frequency, which is simply recombinant offspring divided by total offspring:

(100 + 98) ÷ (100 + 398 + 404 + 98) = 0.20

Geneticists calculate the corresponding percentage to assign the degree of genetic linkage, which has units of "centimorgans," or cM. In this case, the value is 0.20 times 100, or 20%.

The lower the recombination frequency, the more closely the genes are physically linked.

Think about it like this: The closer genes are to each other on a chromosome, the closer they are physically. This physical closeness makes the likelihood of recombination (and thus separation of the linked genes) much lower since they're so close. While recombination frequency isn't a direct measure of this physical closeness, it does give us an estimate of that closeness.

To sum up: You estimate that two genes with a large recombination frequency are likely to be farther apart, while those with a small recombination frequency are likely to be closer together.


Watch the video: ABO blood group Rh factor and probability (December 2022).