<< Chapter < Page Chapter >> Page >

Law of segregation

Observing that true-breeding pea plants with contrasting traits gave rise to F 1 generations that all expressed the dominant trait and F 2 generations that expressed the dominant and recessive traits in a 3:1 ratio, Mendel proposed the law of segregation    . This law states that paired unit factors (genes) must segregate equally into gametes such that offspring have an equal likelihood of inheriting either factor. For the F 2 generation of a monohybrid cross, the following three possible combinations of genotypes result: homozygous dominant, heterozygous, or homozygous recessive. Because heterozygotes could arise from two different pathways (receiving one dominant and one recessive allele from either parent), and because heterozygotes and homozygous dominant individuals are phenotypically identical, the law supports Mendel’s observed 3:1 phenotypic ratio. The equal segregation of alleles is the reason we can apply the Punnett square to accurately predict the offspring of parents with known genotypes. The physical basis of Mendel’s law of segregation is the first division of meiosis in which the homologous chromosomes with their different versions of each gene are segregated into daughter nuclei. This process was not understood by the scientific community during Mendel’s lifetime ( [link] ).

Homologous pairs of chromosomes line up at the metaphase plate during metaphase I of meiosis. The homologous chromosomes with their different versions of each gene are segregated into daughter nuclei.
The first division in meiosis is shown.

Test cross

Beyond predicting the offspring of a cross between known homozygous or heterozygous parents, Mendel also developed a way to determine whether an organism that expressed a dominant trait was a heterozygote or a homozygote. Called the test cross    , this technique is still used by plant and animal breeders. In a test cross, the dominant-expressing organism is crossed with an organism that is homozygous recessive for the same characteristic. If the dominant-expressing organism is a homozygote, then all F 1 offspring will be heterozygotes expressing the dominant trait ( [link] ). Alternatively, if the dominant-expressing organism is a heterozygote, the F 1 offspring will exhibit a 1:1 ratio of heterozygotes and recessive homozygotes ( [link] ). The test cross further validates Mendel’s postulate that pairs of unit factors segregate equally.

In a test cross, a parent with a dominant phenotype but unknown genotype is crossed with a recessive parent. If the parent with the unknown phenotype is homozygous dominant, all the resulting offspring will have at least one dominant allele. If the parent with the unknown phenotype is heterozygous, 50 percent of the offspring will inherit a recessive allele from both parents and will have the recessive phenotype.
A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.

Art connection

This illustration shows a monohybrid cross. In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy. Each parent produces one kind of gamete, resulting in an F_{1} generation with a dominant yellow phenotype and the genotype Yy. Self-pollination of the F_{1} generation results in an F_{2} generation with a 3 to 1 ratio of yellow to green peas. One out of three of the yellow pea plants has a dominant genotype of YY, and 2 out of 3 has the heterozygous genotype Yy. The homozygous recessive plant has the green phenotype and the genotype yy.
This Punnett square shows the cross between plants with yellow seeds and green seeds. The cross between the true-breeding P plants produces F 1 heterozygotes that can be self-fertilized. The self-cross of the F 1 generation can be analyzed with a Punnett square to predict the genotypes of the F 2 generation. Given an inheritance pattern of dominant–recessive, the genotypic and phenotypic ratios can then be determined.

In pea plants, round peas ( R ) are dominant to wrinkled peas ( r ). You do a test cross between a pea plant with wrinkled peas (genotype rr ) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the parent plant is homozygous dominant or heterozygous?

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
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, Concepts of biology for the university of georgia. OpenStax CNX. Aug 09, 2013 Download for free at http://legacy.cnx.org/content/col11520/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Concepts of biology for the university of georgia' conversation and receive update notifications?

Ask