BIO1011 Lecture Notes - Lecture 9: Quantitative Trait Locus, Y Chromosome, Haemophilia

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5 Jul 2018

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LECTURE 9 & 10 PATTERN OF INHERITANCE I & II

Mendel’s Law of Segregation

* If the blending model of inheritance were

correct, the F1 hybrids from a cross between

purple -flowered and white-flowered pear plants

would have pale purple flowers, a trait

intermediate between those of the P generation.

*All the F1 offspring had flowers just as purple

as the purple-flowered parents.

*What happen to white flowers?

*If it were lost, then the F1 plants could produce

only purple-flowered offspring in the F2

generation.

*But when Mendel allowed the F2 self-pollinate,

the white-flower trait reappeared in the F2

generation.

* Mendel reason that the heritable factor for

white flowers did not disappear in the F1 plants

but was somehow hidden, or masked, when

purple-flower factor was present.

* In Mendel’s terminology, purple flower colour is

a dominant trait, and white flower colour is a

recessive trait.

* Reappearance of white-flowered plants in the

F2 generation was evidence that the heritable

factor causing white flowers had not been

diluted or destroyed by coexisting with the

purple-flower factor in the F1 hybrids.

Mendel’s Model

A. First, alternative versions of genes account for variations in inherited characters.

A. Alleles: Alternative versions of a gene.

B. Second, for each character, an organism inherits two copies (that is, two alleles) of a

gene, one from each parent.

C. Third, if the two alleles at a locus differ, then one, the dominant alleles, determines

the organisms’ appearance; the other, the recessive allele, has no noticeable effect

on the organism’s appearance.

D. Fourth, the law of segregation, states that the two the two allele for a heritable

character segregate (separate from each other) during gamete formation and end up in

different gametes.

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Monohybrid Cross

✤A cross involving a single gene.

✤To ﬁnd the chance of the next baby having albanism, we construct a

punnett square.

Dihybrid Cross

Genotype : AA : Aa : aa = 1:2:1

Phenotype : 3 Pigmentation : 1 Albino

✦Phenotypic ratio 9:3:3:1 is

typical of a dihybrid cross in

the F2 generation where the

F1 are double

heterozygotes.

✦P 1 = Pure Breed (original

gene, no cross)

✦F 1 = First generation

9 : 3 : 3 :1

IS THE TYPICAL OF A DIHYBRID

CROSS WHEN BOTH PARENTS

ARE DOUBLE

HETEROZYGOUS.

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Mende“’sPrincip“eofIndependentAssort”ent

✤Alleles of one gene on one chromosome controlling one trait assort

independently of alleles of another gene on another chromosome

controlling a different trait.

✤LINKED GENES = ASSORT DEPENDENTLY

✤NON - LINK GENES = ASSORT INDEPENDENTLY

TESTCROSS

✤Invo“veoneparentwhois

ho”ozygousrecessive(aa/aabb)

withanotherparentwhoshowsthe

do”inanttrait.

✤LawofIndependentAssort”ent:

Twoor”oregenesassort

independent“y-thatis,eachpairof

a““e“essegregatesindependent“yof

eachotherpairofa““e“esduring

ga”etesfor”ation.

✤Purposeofthistest:

1.Identifythegenotypesofthe

do”inanttraitaseitherho”ozygousor

heterozygous.

2.Estab“ishifgenesare“inkedina

dihybridcross.

DETECTINGLINKAGE

✤Observingthe4c“assesofoffspringyou

canidentifyifthey’re“inkedornot:

1.Ifthefourc“assesofoffspringare

notequa“inphenotypicproportion,the

genesare“inked

2.Ifthephenotypicproportionofthe

fourc“assesofoffspringisequa“,thegenes

arenot“inked.

REMEMBER:LINKEDGENESAREFOUND

ONTHESAMECHROMOSOME

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Related Questions

NOTE: There are two parts to this problem (a & b), with each part having their own questions.

a) One dark purple and one white flowered plants were crossed to each other. A total of 20 seeds were harvested and planted. The following phenotypes were observed in the F1 generation:

5 dark purple, 9 light purple and 6 white

1. Calculate the phenotypic ratio

2. What is the most likely pattern of inheritance in this exercise?

3. How would you represent the genotype of the dark purple, light purple and white flowered plants?

b) Two light purple flowered plants were crossed with each other. What is the probability that a group of six offspring from this cross will be pomposed of two white, 1 light purple, and three dark purple-flowered plants? Clearly show work for the equation, the math, and the probability.

Equation:

Math:

Probability:

azuresheep826

1. A true breeding tall pea line is crossed with atrue breeding short line. The F1 plants are all short. The F1 arecrossed with themselves to produce an F2. If the difference (tallvs. short) is due to a single locus, what proportion of the F2individuals do you expect to be tall?

2. Imagine that a researcher crossed the peas in theexample above, and counted 100 F2. 86 were short. 14 were tall. Doa chisquare test to see whether the results conform to a singlelocus cross with the allele coding for short plants dominant to theallele coding for tall plants.

3. A clematis from a true-breeding line with whiteflowers and short corollas (line A) is crossed to a clematis from atrue-breeding line with deep red flowers and long corollas (lineB). The resulting offspring, the F1, are all pink-flowered, withcorollas of medium length. An individual from the F1 iscrossed to an individual from line A. The resulting F2 offspringare as follows: 25% white flowers and short corollas, 25% whiteflowers with medium corollas, 25% pink flowers with short corollas,25% pink flowers with medium corollas.
What is the most likely genetic mechanism determining flower colorand corolla length? If a plant from the F1 listed above were selfed(crossed with itself), what proportion of the F2 offspring wouldyou expect to be white flowered with short corollas?

4. A man and a woman walk into a genetic counselorâsoffice. They have just had a child diagnosed with situs inversus, acomplete transposition of the internal organs from left to right,and vice versus. This condition usually produces no obvioussymptoms. It is caused by an autosomal recessive allele. What isthe chance that their next child will have the same condition? Whatmight this suggest about the parents?
Another man and woman walk into the same office. The father is anachondroplasic dwarf. The mother is of normal stature. (Look upachondroplasia, it is a lethal dominant). What is the chance thattheir first child will be a dwarf?

Please show work! (:

carminebeaver663

BIO1011 Lecture Notes - Lecture 9: Quantitative Trait Locus, Y Chromosome, Haemophilia

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