Co-Dominance/Multiple Alleles

Till now we have seen that the F1 generation resembles either of the two parents and not both. However, in co-dominance, the F1 generation resembles both parents. The human blood group is an excellent example of co-dominance. The ABO blood groups are controlled by a single gene I, but it has multiple alleles (IA, IB, and i). RBC with allele 'i' does not possess  any sugar polymer on its surface.

The plasma membrane of the RBCs contains sugar polymers, and the kind of sugars is controlled by the gene. Since humans are diploid, each person contains any two alleles. For example, an offspring will possess a blood group A when both of his/her parents possess IA allele. 

Following are the possible blood type combinations:

These are the possible combinations of blood types. 

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PEDIGREE ANALYSIS: PRACTICE PROBLEMS

https://youtu.be/4nfdm6hT-r8

Q1. Identify the inheritance pattern in the following picture.

The above pedigree is Autosomal recessive. 

• Recessive because there is skipping of generation.
• Autosomal recessive because, it does not fulfill X-linked recessive condition.

For X-linked recessive, following conditions must be fulfilled:

1. More boys affected than girls
2. Female does not affect female
3. Affected mom has to have affected sons as well as dad

Q2. Identify the mode of inheritance.

The above pedigree is X-linked dominant because of the following reasons:

1. Dominant because there is no skipping of generation.
2. X-linked dominant condition must follow DDD rule (the first D represents dominant condition; the second D represents affected D; the third D represents daughters). It means, in a dominant condition, if dad is affected, all his daughters will be affected too.
3. The above pedigree follows DDD rule.




Q3. Identify the mode of inheritance.

Autosomal dominant

No DDD rule followed in the above pedigree

Q4. Identify the pattern of inheritance.

Autosomal recessive

No X-linked recessive condition followed.

Q5. Identify the pattern of inheritance.

X-linked recessive

Followed X-linked recessive condition.
It is recessive because two unaffected parents produced affected offspring -- a characteristic of recessive trait. 


Q6. Find the pattern of inheritance in the following pedigree and write the genotype of III 7

 Q7. Find the pattern of inheritance and the genotype of the parents

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PEDIGREE ANALYSIS

There are four different types of inheritance that are considered in pedigree analysis. These are mentioned below:

• Autosomal Dominant

(Examples: Polycystic kidney disease, Huntington's disease, hereditary spherocytosis)

• Autosomal Recessive

(Examples: Sickle cell anemia,cystic fibrosis, Tay-Sachs disease)

• X-linked Dominant

(Example: Rett syndrome, Vitamin-D resistant rickets, Alport syndrome)

• X-linked Recessive

(Example: hemophilia, color blindness)


Pedigree Symbols:

Tips to solving Pedigree Problems:

Autosomal dominant

- when father transfers disease to his son -- it is always an autosomal trait. 
- affected parents can have unaffected children.

- Pedigrees of autosomal dominant disorders show affected males and females in each generation and also show affected men and women transmitting the condition to equal proportions of their sons and daughters.

Autosomal recessive:

- there is skipping of generation 
- unaffected parents can gave affected children 

X-linked dominant:

- the disorder never tranfers from father to son
- it follows DDD rule (D= dominant, D= diseased dad, D= diseased daughters). It means, if father is affected, all daughters will be affected.

X-linked recessive: 

- males are more affected than females
- disease tends to tranfer from mother to son and father to daughter
- disease never tranfers from father to son


If both parents are shown healthy -- it is often a recessive trait 

When one or both the parents are shown as diseased -- it is often a dominant trait 

Now solve the following problem:

-- Since this is the first problem, we would check for all the 4 traits:

1. Autosomal dominant
2. Autosomal recessive
3. X-linked dominant
4. X-linked recessive

A. Let's check for the autosomal dominant trait. 

In the picture shown above, there are 4 generations. We are considering whether the mode is autosomal dominant or not. We choose 'A' as dominant allele and 'a' as recessive allele. We are only concerned with autosomes and not with sex chromosomes.

Generation I: Mom is diseases and dad is healthy. So the allele for mom could be 'Aa' or 'AA', but we consider 'Aa'. For dad, we take recessive alleles 'aa'.

Generation II: The offspring is a healthy daughter. So, her genotype would be 'aa' -- one 'a' comes from mom and one 'a' comes from dad. She marries a healthy man.

Generation III: there are two sons and one daughter. However, one son is diseased and rest others are healthy. Since one son is diseased, his genotype should be 'Aa', which is not possible because his parents possess only 'aa' genotype. That is why, it is not an autosomal dominant trait. 

B. Let's check for the autosomal recessive trait. 

In the above picture, we are checking whether its an autosomal recessive trait or not.

Generation I: Since mom is diseased, so her recessive genotype would be 'aa' and the genotype of dad would be 'Aa'. (For a recessive trait to appear, it has to be homozygous recessive).

Generation II: The offspring is a daughter and she is healthy and therefore, her genotype would be 'Aa' ('A' comes from father and 'a' comes from the mother). She marries a healthy man.

Generation III: there are two sons and one daughter. However, one son is diseased and rest others are healthy. The genotypes of all the offspring in the third generation can be satisfied from their mom and dad. For example, since mom and dad for generation III has a genotype of 'Aa' so, the genotypes of offspring could be 'Aa' and 'aa'.

Generation IV: All the genotypes are possible from their parental cross. And therefore, it is an autosomal recessive trait. 

C. Let's check for the X-linked dominant trait. 

D. Let's check for the X-linked recessive trait. 

Let's discuss another question on Pedigree. Look at the following pedigree:

In the above pedigree, we have to find any one out of the following four given situations:

a) Autosomal Dominant

b) Autosomal recessive

c) X-linked dominant

d) X-linked recessive

If you see the pedigree, there are three generations; however, there is no skipping of disease. It means it is a DOMINANT condition. So, we can discard the recessive conditions. 

Let's check first for X-linked dominant condition. The trick is DDD rule -- the first D represents dominant; the second D represents diseased dad and the third D represents disease daughther. In other words, if the situation is dominant, so a disease dad will transfer disease to all his daughters. This condition is not shown above, it means, it's an AUTOSOMAL DOMINANT trait.

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DISCUSSION QUSTIONS ON LINKAGE AND RECOMBINATION

Q1. An individual with cd genes was crossed with wild type + +. On test crossing F1, the progeny was + c 105, + d 115, cd 880, and + + 900. Distance between cd genes is:

a. 11 map units
b. 5.5 map units
c. 44 map units
d. 88 map units

Solution: The simplest way to solve this problem is to find the recombination frequency percentage, which is equal to the map distance.

Q2. A plant of genotype C/C;d/d is crossed to c/c;D/D and an F1 test crossed to c/c;d/d. If the genes are linked and 40 map units apart, the percentage of c/c;d/d recombinants will be?

A. 10%
B. 20%
C. 30%
D. 40%


Q3. The maximum recombination frequency between two genes is:

A. 100%
B. 80%
C. 50%
D. 10%

Q4. In Drosophila, the two genes w and sn are X-linked and 25 map units apart.  A female fly of genotype w+ sn+/w sn is crossed to a male from a wild-type line.  What percentage of male progeny will be w+ sn?

A. 0
B. 12.5%
C. 25%
D. 37.5%

Q5. Recombination frequencies ___

A. are the same for all genes.
B. arise from specific genetic exchange.
C. are the same for cis and trans heterozygotes.
D. decrease with distance

Q6. In crossing over

A. Genetic exchange occurs before chromosome replication
B. The probability of its occurrence decreases with increasing distance between the genes exchanged
C. Occurs between two loci very close together
D. The reciprocal exchange between homologous chromosomes is random

Q7. Which of the following is suitable for experiment on linkage?

A. aaBB x aaBB
B. AABB x aabb
C. AaBb x AaBb
D. AAbb x AaBB

Q8. AB genes are linked. What is the genotype of the progeny in a cross between AB/ab and ab/ab?

A. AABB and aabb
B. AaBb and aabb
C. AAbb and aaBB
D. AaBb and AaBb

Q9. Three genes a b and c show crossing over of 20% between a and b, 28% between b and c and 8% between a and c. Sequence of genes will be:

A. b a c
B. a b c
C. a c b
D. b c a

Q10. Distance between the genes a, b, c and d in map units is a-d = 35, b-c = 1, a-b = 6, c-d = 1.5 and a-c = 5. Find out the sequence of genes?
A. a d c b
B. a c d b
C. a b c d
D. a c b d
E. a d b c


Answer key:

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LECTURE 4: LINKAGE AND RECOMBINATION

Thomas Hunt Morgan and his colleagues discovered the variation that arises due to sexual reproduction.  Morgan worked on Drosophila.

Why Drosophila to study linkage and recombination?

• Small genome size with only 4 pairs of chromosomes
• Genome size = 1.2 x 10^8 bp (approx 14,000 genes)
• Short life cycle (2 weeks)
• Controlled cross possible
• Many offspring from one cross
• Rich source of mutation

There are some simple notations that need to be considered while studying linkage and recombination:

• Letters describe mutant phenotype, for example (sw) for short wing
• Plus sign (+) is used to denote wild-type allele (sw+)
• Wild type alleles are more common in nature
• Lower case letter is used to show recessive wild type sw+ allele
• Upper case is used to denote mutant allele dominant to wild type allele.

Unlinked versus Linked Genetic Inheritance:

A cross has been shown above between a wild type drosophila (female) and a short winged male. Since wild type is dominant, the F1 generation shows a wild - type phenotype. 

-In Mendelian inheritance, we discussed about test cross, the monohybrid test cross ratio is 1:1 and the dihybrid test cross ratio is 1 : 1 : 1 : 1

-Test cross is also used to find linked genes.

- In the following cross, we have considered that the genes are unlinked

-Let's see the expected and observed value of the following test cross.

-- According to Mendel's test cross experiment, the expected ratio should be 250 : 250 : 250 : 250, however, the observed ratio was quite different (415 : 410 : 90 : 85). So the observed data does not fit the expected data and this means that the genes are not unlinked. 

=> Now take another situation where we consider that the genes are linked. 

-The above picture shows the types of gametes being formed from b/sw+ b/sw

-The figure above shows the test cross, however, genes are linked.

- The figure above shows that the genes are totally linked (means closed enough that no crossing over is possible) and therefore no recombination is done. Therefore, we can only expect parental genotypes.

The above picture shows that if F1 is crossed with a recessive parent (test cross), the ratio would differ depending on whether alleles are unlinked, partially linked or fully linked.

The above picture shows the expected and observed number of progeny formed after a test cross. The picture also shows how to calculate the recombination frequency. The recombinant frequency calculated is the distance between two genes. 

Well, the distance between two genes have already been given, which is 20cM. It means that the recombinant frequency is 20%. Since there are two recombinants so each will have 10% frequency. And each parent will have (100-20 =80/2 = 40%). 

So, AB/ab = 40%

Ab/ab = 10%

aB/ab = 10%

ab/ab = 40%

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