<< Chapter < Page Chapter >> Page >

Random segregation

This diagram shows the genetics experiment conducted by Mendel. The top panel shows the offspring from first generation cross and the bottom panel shows the offspring from the second generation cross.
In the formation of gametes, it is equally likely that either one of a pair alleles from one parent will be passed on to the offspring. This figure follows the possible combinations of alleles through two generations following a first-generation cross of homozygous dominant and homozygous recessive parents. The recessive phenotype, which is masked in the second generation, has a 1 in 4, or 25 percent, chance of reappearing in the third generation.

Mendel’s observation of pea plants also included many crosses that involved multiple traits, which prompted him to formulate the principle of independent assortment. The law states that the members of one pair of genes (alleles) from a parent will sort independently from other pairs of genes during the formation of gametes. Applied to pea plants, that means that the alleles associated with the different traits of the plant, such as color, height, or seed type, will sort independently of one another. This holds true except when two alleles happen to be located close to one other on the same chromosome. Independent assortment provides for a great degree of diversity in offspring.

Mendelian genetics represent the fundamentals of inheritance, but there are two important qualifiers to consider when applying Mendel’s findings to inheritance studies in humans. First, as we’ve already noted, not all genes are inherited in a dominant–recessive pattern. Although all diploid individuals have two alleles for every gene, allele pairs may interact to create several types of inheritance patterns, including incomplete dominance and codominance.

Secondly, Mendel performed his studies using thousands of pea plants. He was able to identify a 3:1 phenotypic ratio in second-generation offspring because his large sample size overcame the influence of variability resulting from chance. In contrast, no human couple has ever had thousands of children. If we know that a man and woman are both heterozygous for a recessive genetic disorder, we would predict that one in every four of their children would be affected by the disease. In real life, however, the influence of chance could change that ratio significantly. For example, if a man and a woman are both heterozygous for cystic fibrosis, a recessive genetic disorder that is expressed only when the individual has two defective alleles, we would expect one in four of their children to have cystic fibrosis. However, it is entirely possible for them to have seven children, none of whom is affected, or for them to have two children, both of whom are affected. For each individual child, the presence or absence of a single gene disorder depends on which alleles that child inherits from his or her parents.

Autosomal dominant inheritance

In the case of cystic fibrosis, the disorder is recessive to the normal phenotype. However, a genetic abnormality may be dominant to the normal phenotype. When the dominant allele is located on one of the 22 pairs of autosomes (non-sex chromosomes), we refer to its inheritance pattern as autosomal dominant    . An example of an autosomal dominant disorder is neurofibromatosis type I, a disease that induces tumor formation within the nervous system that leads to skin and skeletal deformities. Consider a couple in which one parent is heterozygous for this disorder (and who therefore has neurofibromatosis), Nn , and one parent is homozygous for the normal gene, nn . The heterozygous parent would have a 50 percent chance of passing the dominant allele for this disorder to his or her offspring, and the homozygous parent would always pass the normal allele. Therefore, four possible offspring genotypes are equally likely to occur: Nn , Nn , nn , and nn . That is, every child of this couple would have a 50 percent chance of inheriting neurofibromatosis. This inheritance pattern is shown in [link] , in a form called a Punnett square    , named after its creator, the British geneticist Reginald Punnett.

Questions & Answers

how do you get the 2/50
Abba Reply
number of sport play by 50 student construct discrete data
Aminu Reply
width of the frangebany leaves on how to write a introduction
Theresa Reply
Solve the mean of variance
Veronica Reply
Step 1: Find the mean. To find the mean, add up all the scores, then divide them by the number of scores. ... Step 2: Find each score's deviation from the mean. ... Step 3: Square each deviation from the mean. ... Step 4: Find the sum of squares. ... Step 5: Divide the sum of squares by n – 1 or N.
kenneth
what is error
Yakuba Reply
Is mistake done to something
Vutshila
Hy
anas
hy
What is the life teble
anas
hy
Jibrin
statistics is the analyzing of data
Tajudeen Reply
what is statics?
Zelalem Reply
how do you calculate mean
Gloria Reply
diveving the sum if all values
Shaynaynay
let A1,A2 and A3 events be independent,show that (A1)^c, (A2)^c and (A3)^c are independent?
Fisaye Reply
what is statistics
Akhisani Reply
data collected all over the world
Shaynaynay
construct a less than and more than table
Imad Reply
The sample of 16 students is taken. The average age in the sample was 22 years with astandard deviation of 6 years. Construct a 95% confidence interval for the age of the population.
Aschalew Reply
Bhartdarshan' is an internet-based travel agency wherein customer can see videos of the cities they plant to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400 a. what is the probability of getting more than 12,000 hits? b. what is the probability of getting fewer than 9,000 hits?
Akshay Reply
Bhartdarshan'is an internet-based travel agency wherein customer can see videos of the cities they plan to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400. a. What is the probability of getting more than 12,000 hits
Akshay
1
Bright
Sorry i want to learn more about this question
Bright
Someone help
Bright
a= 0.20233 b=0.3384
Sufiyan
a
Shaynaynay
How do I interpret level of significance?
Mohd Reply
It depends on your business problem or in Machine Learning you could use ROC- AUC cruve to decide the threshold value
Shivam
how skewness and kurtosis are used in statistics
Owen Reply
yes what is it
Taneeya
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Anatomy & Physiology. OpenStax CNX. Feb 04, 2016 Download for free at http://legacy.cnx.org/content/col11496/1.8
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Anatomy & Physiology' conversation and receive update notifications?

Ask