Principles of Inheritance and Variation

Polygenic inheritance refers to the inheritance of a trait controlled by two or more genes. When human disorders are determined by mutation in the single gene then they are transmitted to the offspring as per Mendelian principle. Polygenic trait shows non-Mendelian inheritance pattern.

As in the $\mathrm{F}_1$ generation the carrier female and non-affected (normal, not carrier) had affected male child that means the genetic disorder is sex-linked recessive.

The consanguineous mating between female ( $X^c X$ ) and male ( $X^c Y$ )

Out of 4 child only one is carrier i.e. $\frac{1}{4}$.

A classical dihybrid cross performed by Mendel involves.

A cross which was made between a pure round yellow seeded pea plant (RRYY) with wrinkled green seeded plant (rryy). Yellow colour is dominant over green and round seed shape over wrinkled seed shape.

Phenotypic ratio in F2 generation

Let's analyze both statements to determine their correctness with respect to the concept of polygenic inheritance.

Statement I: When many alleles of a single gene govern a character, it is called polygenic inheritance.

This statement is incorrect. Polygenic inheritance involves multiple genes (each with two or more alleles) contributing to a single phenotype, rather than many alleles of a single gene. So, this statement misrepresents the definition of polygenic inheritance.

Statement II: In Polygenic inheritance, the effect of each allele is additive.

This statement is correct. In polygenic inheritance, each allele of the different genes that contribute to a trait typically has a small, additive effect on the phenotype. For example, characteristics like human height and skin color are determined by the cumulative effect of multiple genes.

Based on the above analysis:

Option A

Statement I is true but Statement II is false

Option B

Statement I is false but Statement II is true

Option C

Both Statement I and Statement II are true

Option D

Both Statement I and Statement II are false

The correct answer is Option B: Statement I is false but Statement II is true.

To match List-I with List-II correctly, let's analyze each type of inheritance and its corresponding example:

Incomplete dominance: This is when the heterozygote displays a phenotype that is intermediate between the two homozygous phenotypes. The classic example of this is the flower color in Antirrhinum (snapdragon). In this case, the heterozygous flowers have a color that is a blend of both parental colors, which is different from either homozygous parent. Therefore, Incomplete dominance matches with "Flower colour in Antirrhinum".

Co-dominance: This refers to a situation where both alleles in a heterozygote are fully expressed, resulting in a phenotype that displays both traits equally. A common example of this is blood groups in humans. In the case of the ABO blood group system, both A and B alleles are expressed in individuals with AB blood type. Therefore, Co-dominance matches with "Blood groups in human".

Pleiotropy: This occurs when one gene influences multiple traits. Phenylketonuria (PKU) is an example where a single genetic defect in the phenylalanine hydroxylase gene results in multiple symptoms in different organs. Therefore, Pleiotropy matches with "Phenylketonuria".

Polygenic inheritance: This refers to situations where multiple genes influence a single trait. An example of this is skin color in humans, where multiple genes contribute to the phenotype. Therefore, Polygenic inheritance matches with "Skin colour in human".

Based on this analysis, the correct matching would be:

(A) Incomplete dominance - (II) Flower colour in Antirrhinum

(B) Co-dominance - (I) Blood groups in human

(C) Pleiotropy - (IV) Phenylketonuria

(D) Polygenic inheritance - (III) Skin colour in human

Therefore, the correct option is:

Option B: A-II, B-I, C-IV, D-III

As per the type of given progeny the genotype of the parent will be TTRr.

According to the above cross progeny will be Tall round and Tall wrinkle.

Aneuploidy is a chromosomal disorder characterized by an abnormal number of chromosomes, which is not an exact multiple of the haploid set. The primary causes of aneuploidy are related to the addition (gain) or deletion (loss) of chromosomes or chromosomal segments.

Let's examine the options given:

A. Substitution - This involves replacing one segment of a chromosome with another, which generally does not alter the chromosome number.

B. Addition - This involves the gain of an extra chromosome or chromosomal segment, which directly causes aneuploidy.

C. Deletion - This involves the loss of a chromosome or chromosomal segment, which also directly causes aneuploidy.

D. Translocation - This involves the rearrangement of parts between nonhomologous chromosomes and typically does not change the overall chromosome number.

E. Inversion - This involves a part of the chromosome being reversed end to end, which does not usually cause a change in the chromosome number.

Based on the explanation above, the most appropriate causes of aneuploidy from the options provided would be B (Addition) and C (Deletion).

Therefore, the correct choice is:

Option D

B and C only

The correct answer is Option D: B and E. Here's why:

Let's break down the blood group system and genotypes:

• Blood Groups: The ABO blood group system is determined by three alleles: $I^A$, $I^B$, and $i$.
• Genotypes: The combination of these alleles determines the blood group. Here are the possible genotypes and their corresponding blood groups:
• $I^A I^A$ or $I^A i$ = Blood group A
• $I^B I^B$ or $I^B i$ = Blood group B
• $I^A I^B$ = Blood group AB
• $i i$ = Blood group O
• Rh Factor: The Rh factor is another blood group system. It's either positive (+) or negative (-). A person inherits one Rh allele from each parent.

Now, let's analyze the given information:

• Mother: $A+$ - Possible genotypes: $I^A I^A$ or $I^A i$ and Rh positive (can be $D D$ or $D d$)
• Father: $B+$ - Possible genotypes: $I^B I^B$ or $I^B i$ and Rh positive (can be $D D$ or $D d$)
• Child: $A+$ - Possible genotypes: $I^A I^A$ or $I^A i$ and Rh positive (can be $D D$ or $D d$)

Now let's see why options B and E are correct:

Option B:

$I^A I^A\left|I^B i\right| I^A i$

• Mother: $I^A I^A$ (Blood group A) and Rh positive (can be $D D$ or $D d$)
• Father: $I^B i$ (Blood group B) and Rh positive (can be $D D$ or $D d$)
• Child: $I^A i$ (Blood group A) and Rh positive (can be $D D$ or $D d$)

Option E:

$I^A i\left|I^B i\right| I^A i$

• Mother: $I^A i$ (Blood group A) and Rh positive (can be $D D$ or $D d$)
• Father: $I^B i$ (Blood group B) and Rh positive (can be $D D$ or $D d$)
• Child: $I^A i$ (Blood group A) and Rh positive (can be $D D$ or $D d$)

Both Option B and E are valid possibilities because they fulfill the given blood group information for the parents and child.

The other options are incorrect because they either don't match the child's blood group or don't allow for the possibility of the father's blood group.

The correct answer is:

Mendel's Law of Dominance states that in a heterozygote, the dominant allele will mask the expression of the recessive allele. Let's analyze the options:

• A. Out of one pair of factors one is dominant and the other is recessive: This is the core principle of Mendel's Law of Dominance.


• C. Factors occur in pairs in normal diploid plants: This is a fundamental concept in genetics, as diploid organisms have two copies of each chromosome, thus two copies of each gene (factors).


• D. The discrete unit controlling a particular character is called factor: Mendel used the term "factor" to describe what we now know as genes.


• E. The expression of only one of the parental characters is found in a monohybrid cross: This is a direct consequence of the law of dominance, where the dominant trait masks the expression of the recessive trait in the F1 generation of a monohybrid cross.

• B. Alleles do not show any expression and both the characters appear as such in F2 generation: This statement is incorrect. While the recessive allele is not expressed in the F1 generation, it reappears in the F2 generation in a 3:1 ratio (dominant:recessive). This is due to the Law of Segregation, not the Law of Dominance.

Pink colour flower in snapdragon have genotype $\mathrm{R r}$

Red flowered snapdragon have genotype RR when they both are crossed

So the progeny that we get are red and pink flowered plants only.

To determine the genotype of a black seed plant that could either be homozygous dominant ($\mathrm{BB}$) or heterozygous ($\mathrm{Bb}$), you need to perform a test cross. A test cross involves crossing the individual in question with a homozygous recessive individual. In this scenario, that would be a plant with white seed color, or genotype $\mathrm{bb}$.

A test cross is used because it can reveal whether the black seed plant carries the recessive $\mathrm{b}$ allele. When crossed with a homozygous recessive ($\mathrm{bb}$) plant:

• If the black seed plant is homozygous dominant ($\mathrm{BB}$), all offspring will inherit one $\mathrm{B}$ allele from the black seed parent and one $\mathrm{b}$ allele from the white seed parent, resulting in all offspring having black seeds ($\mathrm{Bb}$).
• If the black seed plant is heterozygous ($\mathrm{Bb}$), there is a 50% chance that an offspring will inherit the $\mathrm{B}$ allele and a 50% chance of inheriting the $\mathrm{b}$ allele from the black seed parent. This results in approximately half of the offspring having black seeds ($\mathrm{Bb}$) and half having white seeds ($\mathrm{bb}$).

By observing the seed colors of the offspring, you can determine whether the black seed plant was homozygous dominant or heterozygous. If any white seeds appear among the offspring, the black seed plant must be heterozygous ($\mathrm{Bb}$). If no white seeds appear, the black seed plant is likely homozygous dominant ($\mathrm{BB}$).

Therefore, the correct option for crossing to determine the genotype of the black seed plant is:
Option B $\mathrm{bb}$

Let's analyze each term in List-I with the definitions presented in List-II:

• A. Two or more alternative forms of a gene: These are called alleles. Thus, A matches with III 'Allele'.
• B. Cross of F$_1$ progeny with homozygous recessive parent: This specific type of cross is known as a test cross, often used to determine the genotype of an individual having dominant phenotype. Therefore, B matches with IV 'Test cross'.
• C. Cross of F$_1$ progeny with any of the parents: This is defined as a back cross, which can be used for multiple purposes, including the testing of the parental genes in the offspring. So, C matches with I 'Back cross'.
• D. Number of chromosome sets in plant: This describes how many sets of chromosomes are present, i.e., the level of ploidy of the organism. Therefore, D matches with II 'Ploidy'.

From the analysis:

• A matches with III
• B matches with IV
• C matches with I
• D matches with II

Accordingly, the correct answer that matches all the descriptions is:

The ABO blood group is determined by alleles I^A, I^B, and i, which are responsible for producing A, B, and O blood types, respectively. The I^A and I^B alleles are codominant, meaning both can be expressed if both are present, whereas the i allele is recessive.

Given the blood types in the problem statement:

• Father's blood group: $\mathrm{B}^{+}$ - Possible genotypes: I^BI^B or I^Bi
• Mother's blood group: $\mathrm{A}^{+}$ - Possible genotypes: I^AI^A or I^Ai
• Child's blood group: $\mathrm{O}^{+}$ - Possible genotype: ii

To find the correct pairing:

The child has an O blood type, meaning their genotype must be ii, indicating they inherited an i allele from each parent. Therefore, both parents must have at least one i allele.

Given this information:

• Father's Genotype: It must include i since the child inherited i. Thus, the father's genotype could be I^Bi.
• Mother's Genotype: It must also include i for the same reason, so the mother's genotype could be I^Ai.
• Child's Genotype: It is confirmed as ii.

Matching this analysis with the provided options:

• Option A: I^Bi/I^Ai/ii - This matches the reasoning provided.
• Option B: I^BI^B/I^AI^A/ii - This is incorrect because it suggests that neither parent has a recessive i allele to pass to the child, which is necessary for the child’s blood type O.
• Option C: iI^B/iI^A/I^AI^B - This is incorrect and does not match the required genotypes for the child to inherit ii.
• Option D: I^Ai/I^Bi/I^Ai - This includes a typo in the children's genotype and misalignment with blood types.
• Option E: iI^B/iI^A/I^AI^B - This is incorrect, mismatches the inheritance mechanism.

So the correct pairing based on the genotypes and blood groups mentioned, along with genetic laws, would be Option "A", i.e., I^Bi/I^Ai/ii.

So the answer is: Option A

Option A : (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Explanation :

(A) Monohybrid Cross : Monohybrid crosses are those that involve one pair of contrasting traits. In monohybrid crosses, the phenotypic ratio of the F2 generation is typically 3:1 (III).

(B) Dihybrid Cross : Dihybrid crosses are those that involve two pairs of contrasting traits. In dihybrid crosses, the phenotypic ratio of the F2 generation is typically 9:3:3:1 (IV).

(C) Incomplete Dominance : Incomplete dominance is when neither allele for a specific trait is dominant over the other. As a result, the phenotype of the heterozygote is somewhere in between the two homozygotes. The phenotypic ratio in such a case is typically 1:2:1 (II).

(D) Test Cross : A test cross involves crossing an individual exhibiting the dominant phenotype (but unknown genotype) with a homozygous recessive individual. The phenotypic ratio can be 1:1 (I) if the dominant individual was a heterozygote. If the dominant individual was a homozygote, all offspring will exhibit the dominant phenotype.

The correct answer is Option B : Sickle cell anemia.

Sickle cell anemia is caused by a single nucleotide mutation in the beta-globin gene, which is a part of hemoglobin. This mutation causes the amino acid glutamic acid to be replaced by valine at the sixth position of the beta-globin chain. The resulting hemoglobin, called hemoglobin S, can deform red blood cells into a sickle shape, especially under low oxygen conditions. This change can cause various complications including pain, anemia, and increased risk of infection.

The other disorders listed - Thalassemia, Haemophilia, and Phenylketonuria - are also genetic disorders, but they are caused by different mutations.

Parent generation (P) :

• One parent is homozygous for yellow seeds (dominant trait, represented by 'YY') and red flowers (dominant trait, represented by 'RR'). Therefore, this parent's genotype is YYRR.
• The other parent is homozygous for green seeds (recessive trait, represented by 'yy') and white flowers (recessive trait, represented by 'rr'). Therefore, this parent's genotype is yyrr.

First filial generation (F₁) :

• When these two parents cross, all of the offspring in the F₁ generation will inherit one allele from each parent for both traits, leading to a genotype of YyRr. This genotype results in a phenotype of yellow seeds and pink flowers (as red is incompletely dominant over white, resulting in pink when both are present).

Second filial generation (F₂) :

• Now, when these F₁ generation plants self-fertilize, we can get a range of genotypes. This is due to the law of independent assortment which states that the alleles for yellow/green seeds and the alleles for red/pink/white flowers will sort independently of each other into the gametes.

For each trait, the F₁ parent can produce four types of gametes (YR, Yr, yR, yr). The Punnett square method can be used to find out the different combinations of these gametes. However, many of these combinations will lead to the same phenotypes due to the dominance of certain traits.

If we only look at the different phenotypes (physical appearances), we get :

1. Yellow seeds and red flowers
2. Yellow seeds and pink flowers
3. Yellow seeds and white flowers
4. Green seeds and red flowers
5. Green seeds and pink flowers
6. Green seeds and white flowers

These are the 6 different phenotypes expected in the F₂ generation.

In the case of the grasshopper, which has an XO/XX system :

(A) It is not an example of female heterogamety. It is an example of male heterogamety because males have two different types of gametes (XO and OO), while females have only one type (XX).

(B) This statement is correct. Males produce two different types of gametes, either with (X) or without (O) an X chromosome.

(C) This statement is incorrect because the total number of chromosomes is not the same in both males and females. Females have XX, and males have XO.

(D) This statement is correct. All eggs bear an additional X chromosome besides the autosomes.

Therefore, the correct answer is Option A : (B) and (D) only.

Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and ‘mapped’ their position on the chromosome.

Sutton and Boveri proposed chromosomal theory of inheritance.

Henking discovered X-chromosome.

Thomas Hunt Morgan proved chromosomal theory of inheritance and proposed the concept of linkage.

When a single gene affects multiple phenotypic expression, the gene is called pleiotropic gene and the phenomenon is called pleiotropism.

Option B :

B and E only

Explanation :

B. Such an individual has overall masculine development. However, the feminine development is also expressed. People with Klinefelter syndrome are male (XY), but they often have certain physical characteristics that may be typically associated with female development, such as wider hips, less body hair, and sometimes breast tissue development.

E. Such individuals are sterile. Often, individuals with Klinefelter syndrome produce little to no sperm and are therefore usually infertile. However, there are cases where fertility treatments can help some men with Klinefelter syndrome to father children.

For the other options :

A. This disorder was first described by Langdon Down (1866). This is incorrect. Klinefelter's syndrome was first described by Dr. Harry Klinefelter in the 1940s, not by Langdon Down.

C. The affected individual is short statured. This is incorrect. In fact, individuals with Klinefelter's syndrome are often taller than average.

D. Physical, psychomotor and mental development is retarded. This is also incorrect. While individuals with Klinefelter syndrome may have some learning difficulties or delays, particularly with language and speech, it is not accurate or appropriate to say that their physical, psychomotor, and mental development is "retarded". They may face some challenges, but with support they can lead healthy, productive lives.

The symbol representing mating between relatives (consanguineous mating) in human pedigree analysis is

Sickle cell anaemia is autosomal recessive disorder, whereas, haemophilia is sex linked recessive disorder.

Both sickle cell anaemia and haemophilia are the genetic disorders related to blood.

Crossing over and exchange of genetic material between homologous chromosomes occurs during pachytene stage of meiosis. The enzyme involved in this process is recombinase.

Sutton and Boveri proposed chromosomal theory of inheritance. Thomas Morgan experimentally verified the chromosomal theory of inheritance. Gregor Mendel proposed laws of inheritance.

(G) Green pod colour is dominant over yellow pod colour (g)

(I) Inflated pod is dominant over constricted pod (i)

Expected percentage of F₂ progeny with yellow – inflated pod in this cross is

${1 \over 4} \times {3 \over 4} \times 100 = 18.75\% $

The symptoms described in the question are characteristic of Down syndrome, a genetic disorder caused by the presence of an extra copy of chromosome 21. This results in a trisomy of chromosome 21.

Option A : Trisomy of chromosome 21 - Individuals with Down syndrome typically exhibit a range of physical and developmental features, such as a small round head, furrowed tongue, partially open mouth, and a broad palm with a characteristic single palmar crease. They also tend to have delayed physical, psychomotor, and mental development. Karyotype analysis of an individual with Down syndrome will show three copies of chromosome 21 instead of the normal two copies.

The other options describe different chromosomal abnormalities:

Option B : 47 chromosomes with XXY sex chromosomes - This karyotype is characteristic of Klinefelter syndrome, a condition in which males have an extra X chromosome. Affected individuals have 47 chromosomes, and their symptoms are different from those described in the question.

Option C : 45 chromosomes with XO sex chromosomes - This karyotype is characteristic of Turner syndrome, a condition in which females have only one X chromosome. Affected individuals have 45 chromosomes, and their symptoms are different from those described in the question.

Option D : 47 chromosomes with XYY sex chromosomes - This karyotype is characteristic of XYY syndrome, a condition in which males have an extra Y chromosome. Affected individuals have 47 chromosomes, and their symptoms are different from those described in the question.

Hence (b) & (c) are correct which makes option (c) as correct option.

Grasshopper is an example of XO type of sex determination in which the males have only one X-chromosome besides the autosomes, whereas females have a pair of X-chromosomes.

Gregor J. Mendel, conducted hybridisation experiments on garden peas and selected 14 true breeding pea plant varieties (seven contrasting traits). Contrasting traits studied were smooth or wrinkled seeds, yellow or green seeds, inflated on constricted pods, green or yellow pods, tall or dwarf plants, violet or white flowers and axial or terminal flower positions.

Haemophilia is a X-linked recessive disorder. Thalassemia is an autosomal recessive disorder. Sickle cell anaemia is an autosomal recessive disorder.

Myotonic dystrophy is an autosomal dominant disorder i.e. it occurs due to the presence of autosomal linked dominant trait.

Closely located genes do not show independent assortment. Mendel's law of independent assortment holds good for those genes which are located on different chromosomes.

If mother of man is colour blind, then man will also be colour blind as colour blindness is a X-linked recessive trait and shows criss-cross inheritance.

1% recombination frequency = 1 centi Morgan

To place the genes on a linear chromosome, decreasing order of recombination frequency will be considered.