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Part 2: Heredity and 
Mendelian Genetics

Part 2: Heredity and 
Mendelian Genetics. The Inheritance of Two Traits. Shhhh! Peas pay attention. Mendel’s Second Experiment: A Dihybrid Cross. Mendel’s second major 
experiment involved the crossing 
of two pea plants that differed in 
 TWO traits.

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Part 2: Heredity and 
Mendelian Genetics

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  1. Part 2: Heredity and 
Mendelian Genetics

  2. The Inheritance of Two Traits Shhhh! Peas pay attention

  3. Mendel’s Second Experiment: A Dihybrid Cross Mendel’s second major 
experiment involved the crossing 
of two pea plants that differed in 
TWO traits.

  4. Mendel wanted to discover if 
the inheritance of one trait is 
influenced by the inheritance of 
another trait. (i.e. Did pea colourinfluence pea shape?)

  5. Once again Mendel crossed two 
plants that were pure breeding. This time he 
observed two different 
traits.

  6. The Determination of Gametes 
for a Dihybrid Cross. When generating the possible 
gametes for a dihybrid cross, one must 
predict the entire complement of 
possible gametes. If the F1 generation 
are hybrids, then each parent can each 
produce four possible gametes.

  7. The possible gametic alleles: Each hybrid F1 individual has the 
genotype: RrYy The possible gametic alleles are: RY, Ry, rY, ry

  8. Nine different genotypes and four 
different phenotypes result from a 
dihybrid cross of F1 plants.

  9. Mendel discovered that all of the 
F1 generation were hybrid for 
both traits (round and yellow). Mendel then crossed the F1 
generation and discovered that the F2 generation had a phenotypic ratio of 9:3:3:1. The F2 generations of other dihybrid crosses also showed this ratio.

  10. Of the 551 plants in Mendel’s F2 generation, he observed the 
following traits: 320 round yellow 101 wrinkled yellow 104 round green 26 wrinkled green

  11. Mendel’s Law of Independent 
Assortment The inheritance of alleles for one 
trait does not influence the inheritance of alleles for another trait. This means that the combination of 
alleles in the offspring may not occur 
in either parent. (i.e. they sort 
independently.)

  12. Probability Genotypic and phenotypic ratios are determined by the probability of inheriting a certain trait. The probability of an event is the likelihood that the event will occur. Probability can be 
expressed by the following formula:

  13. In humans, free earlobes are 
controlled by the dominant allele E, 
and attached earlobes by the recessive 
allele e. The widow’s peak hairline is 
regulated by the dominant allele H, 
while the straight hairline is controlled 
by the recessive allele h

  14. Sample Problem 1 What are the probabilities of obtaining F1 offspring 
with the following characteristics if one parents is 
homozygous dominant for both traits and the other is 
heterozygous dominant for both ? • widow’s peak and free earlobes • straight hairline and free earlobes • widow’s peak and attached earlobes • straight hairline and attached earlobes Check your answers by completing a Punnet Square!

  15. Sample Problem 2 What is the probability that a child from the mating of the EeHh × EeHh parents would be a male with a widow’s peak and have attached earlobes? Write your probability as a percentage!

  16. Mono and Dihybrid Cross Assignment Coming Soon!

  17. Beyond Mendel’s Laws Although his research and 
examination of patterns of 
inheritance in the pea plant was 
revolutionary in genetics; Mendel did 
not have a complete understanding of 
the patterns of inheritance.

  18. There are three other mechanisms of inheritance we 
will discuss in this course. Incomplete Dominance Co-dominance Multiple Alleles

  19. Incomplete Dominance Not all traits are completely 
dominant or recessive as Mendel 
suggested. Some traits are incompletely 
dominant, that is, there can be an 
intermediate expression of a 
particular trait when the genotype is 
heterozygous.

  20. The snapdragon flower is 
incompletely dominant for flower 
colour. Two red alleles are required for a 
red flower i.e. RR Two white alleles are required for a 
white flower i.e. R’R’ A heterozygous set of alleles is 
required for a pink flower i.e. RR’ The R’ allele is called “R prime”

  21. Incomplete Dominance in the 
Snapdragon Flower

  22. Sample 1. Determine the F1 phenotypic ratio of a cross 
between a pink and a white snapdragon.

  23. Co-dominant Inheritance In some cases both alleles for a 
trait may be dominant. Such alleles are said to be co-
dominant because both alleles are 
expressed in a heterozygous 
individual.

  24. Shorthorn cattle have the 
mechanism of co-dominant 
inheritance for their coat colour. The expression of both alleles occurs and 
there is no blending of traits. The homozygous red coat alleles are:  CRCR (called “C superscript R) The homozygous white coat alleles are:  CWCW The heterozygous roan coat alleles are:  CRCW

  25. Sample 1. Find the F1 phenotypes of a cross between a red cow and a 
roan bull.

  26. Multiple Allelic Inheritance For some traits more than two 
alleles control the expression in a 
species Although a single individual cannot 
have more than two alleles for each 
trait, different individuals can have 
different pairs of alleles.

  27. Human blood types have 
multiple allele inheritance.

  28. Sex Linkage - X and Y Some traits are inherited from genes on the sex chromosomes This is known as sex-linked (most often X-linked) inheritance. Colour blindness, hemophilia, and baldness are sex-linked traits.

  29. A male with hemophilia mates with a woman with no hemophiliac gene. What is the probability of producing sons or daughters who have hemophilia?

  30. Pedigree Charts Pedigree analysis is useful when the traits of 
many generations of offspring have been 
recorded. A pedigree chart can be used to trace the 
passing of an allele from parents to 
offspring. A pedigree chart contains a number of 
symbols.

  31. 1. Indicate whether each family member is homozygous or 
heterozygous for shortsightedness, or homozygous for normal 
vision. 2. If couple 4 and 5 in row II had another child, what genotype might 
the child have? (Hint: What genotype is possible but not shown in 
the chart?) Would the child have normal vision or be shortsighted?

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