Introduction to Punnett Squares

Question 1 (1.2)

A Punnett Square is a method that was originally invented by an early 20th century English geneticist called Reginald Punnett. A Punnett Square is used to predict the chances of offspring inheriting their parents' genetic traits. In order to interpret Punnett Square diagrams, there are certain genetic terms that are involved. Provide your detailed explanations for the following terms:

Answer: To provide the detailed explanation for the terminology related to the Punnet Square diagram to define the genetic trait, here the diagram of monohybrid and dihybrid crosses are elucidated:

Monohybrid Cross

Parent: TT (tall)[homozygous] × tt (dwarf) [homozygous]

F1 generation: Tt (tall) hybrid

F2 generation:

It is the cross between two individuals having the genotype which is homozygous in nature or the genotypes which is completely dominant or recessive in nature that leads to a complete reverse phenotypic traits in the offspring. Here for example the inheritance pattern of plant height when tall pea plant was crossed with dwarf plant and other traits are ignored then the phenotypic ratio of the tall: dwarf plant in the F2 generation is 3:1 and the genotypic ratio is 1:2:1, homozygous tall (1): heterozygous (2): homozygous dwarf (1). Here the dominant trait of height was tall and the recessive trait was small. In case of F1 generation all the plants were tall hybrid, the chances were 100% and the offspring from the F1 generation when cross they will produce the ratio in F2 generation. (Jorde, et al, 2015)

Dihybrid cross

When two pairs of allele are present then the inheritance pattern is bit different and that can be illustrated with dihybrid cross. In case of the cross among the tall and purple flower (TTPP) tree with short and white flower (ttpp) tree produce the offspring in the F1 generation are TtPp, all tall and purple flower, but in the F2 generation the phenotypic ratio is 9:3:3:1 [Tall and purple: tall and white: short and purple: short and white] and the genotype will be for 9 tall and purple [1:2:2:4], 3 tall and white [1:2], 3 short and purple [1:2] and 1 short and white. (Jorde, et al, 2015)

Parent: TTPP [Tall, purple flower] × ttpp [short, white flower]

F1 generation: TtPp [All tall and purple flower]

F2 generation:

gene and allele: Gene is termed as the function unit of DNA which contains a specific trait that will pass through the generation and allele is termed as the different variation of that specific gene type. Gene can have more versions and individual human have two allele or version for each gene. (Ettorre, 2013)

homozygote and heterozygote: The homozygotes are those individuals who have the same allele for a particular gene in the locus, whereas heterozygotes are for those individuals whose gene contains the different allele in pair. The heterozygotes are sometimes advantageous as for some cases the single allele is causing the disease and the another copy remains good, then it will be advantage for that individual and this is termed as heterozygote advantage. (Ettorre, 2013)

dominant and recessive: The two versions of the gene, that is the alleles can be two types, either dominant or recessive. Dominant allele imparts the effect on the phenotype of the offspring when only one of its copy is present (heterozygous). For example the brown eye trait is the dominant one so if the offspring get one copy of this gene then the eye colour will be brown. For recessive allele the both copies have to be present to impart the effect of the trait. For example blue eye colour is recessive one, so both the copies have to be present in the offspring to get blue coloured eye. (Ettorre, 2013)

genotype and phenotype: these are the two fundamental terms in the law of inheritance and in genetics. Genotype is the expression of the allele that carried in the particular organism and it is inherited from the biological parents to its offspring. The genotype cannot be observed but need to determine by genetic testing. For example in case of someone with albinism, they must have Tyr gene mutated version in their genotype. In contrast phenotype is the observable characteristic of the organisms gene makeup and is combined with the effect of environmental factors as well. The example of phenotypes are eye colour, hair colour and height etc. (Ettorre, 2013)

Question 2a (1.1, 1.2, 1.3 dominant/recessive/monohybrid crosses)

The recessive allele for red hair is shown as ‘b’ and the dominant allele for black hair is shown as ‘B’. A woman who is homozygous for black hair marries a man who is homozygous recessive for red hair.

Using a Punnett Square, predict the hair colour outcome of their children. What’s the percentage outcome of them having a child with red hair or black hair? Explain your reasons.

Answer: The woman is homozygous with black hair (BB), which is dominant trait, marries the man with homozygous recessive trait of red hair (bb), the resulting offspring will show the chance of 100% of black hair and no red hair offspring. The cross is showing in the punnet square format.

Parent: BB (Black hair)×bb (red hair)

F1: Bb (hybrid) black hair

All the offspring shows the hair colour black and the percentage of such offspring is 100% and there is no chance of the hair colour red as the parents are homozygous so they have same pair of the allele in their chromosome and it is BB for black hair which is dominant and bb for red hair which is recessive, so in the phenotype of the offspring the dominant allele will be expressed as the genotype of the offspring’s are Bb. To get the expression of dominant trait only one copy need to be present in the offspring, which has been found from the monohybrid cross. (Bulik-Sullivan,et al, 2015)

Question 2b (1.1, 1.2, 1.3 co-dominant/monohybrid crosses)

Explain the term codominance (1.1).

Answer: Codominance is the factor when both the copies of the allele are present in the individual but not dominating each other in terms of their trait and both the traits are appeared. For example A and B alleles for blood type coexist in the genotype and get AB blood group.

Some chickens can have black feathers, white feathers or a ‘checkered’ mixture of black and white feathers. The gene for the colour of their feathers is determined by a co-dominant genetic inheritance pattern. The allele for black feathers is B and the allele for white feathers is W. (1.1)

What must the genotype for black chickens be?

Answer: The genotype of the black chicken is BB

What must the genotype for white chickens be?

Answer: The genotype for white chickens are WW

What must the genotype for checkered chickens be?

Answer: The genotype for checkered chickens are BW.

If two ‘checkered’ chickens were crossed, what is the percentage probability of their chicks being black feathered, white feathered or ‘checkered’ feathered? Use a Punnett Square to support your answer and to give the resulting genotypes as well. (1.1, 1.2, 1.3)

Answer: When these checkered chickens were crossed they will give the chicks in the percentage of black feathered (25%), white feathered (25%) and checkered feather (50%). This can be observed from the Punnett square. (Bulik-Sullivan,et al, 2015; Vizmanos, et al, 2020 )

Parent: BW × BW (Checkered)

Offspring:

The resulting genotype is 1:2:1 for black feathered: 2 checkered feathered: white feathered.

Question 3 (1.1, 1.2, 1.3 dominant/recessive/dihybrid crosses)

In humans, the allele for being able to roll the tongue is dominant and is shown as ‘T’. It is dominant to the allele for not being able to roll the tongue, which is shown as ‘t’. Also, the allele for having unattached earlobes, shown as ‘U’ is dominant to the recessive allele for having attached earlobes, which is shown as ‘u’.

In the following couples, what would the ratio and percentage of the phenotypes be that can be predicted in their offspring? Produce a Punnett Square for each of the following couples and use the FOIL method to show how you arrive at your predicted phenotype outcomes. NB: Be very careful that your computer doesn’t alter your choice of upper or lower case letters in your Punnett Square. You may do these Punnett Squares by hand and photograph or scan them into your assignment if you’d rather.

Female partner: is heterozygous for being able to roll her tongue and homozygous recessive for having attached earlobes.

Male partner: is homozygous recessive for not being able to roll his tongue and is homozygous recessive for having attached earlobes.

Answer: Genotype of female partner is Ttuu and male partner is ttuu.

By using the FOIL method the gametes are Tu, Tu, tu and tu, now exclude the repetation gametes are 2 Tu and 2 tu for female. For male four are tu types.

The percentage of offspring having the rolling of tongue and unattached earlobes are 50% and the offspring having attached earlobe and not able to roll the tongue is 50%. (Vizmanos, et al, 2020)

Female partner: is homozygous recessive and cannot roll her tongue and is heterozygous for having unattached ear lobes.

Male partner: is homozygous recessive and cannot roll his tongue and is also homozygous recessive for having attached earlobes

Answer: The genotype of female partner is ttUu and male partner is ttuu. The FOIL method is applied to generate the female gametes are tU, tu, tU and tu and male gametes are four tu. The offspring ratio are:

The percentage of offspring are 50% of not able to roll tongue and unattached earlobes (tUtu) and 50% of not able to roll the tongue and attached earlobes (tutu). (Vizmanos, et al, 2020)

Question 4 (1.2, 1.3, 2.1 monohybrid crosses)

Explain, in detail, how genetically the sex of a baby is determined. Support your answer with a Punnett Square to show the predicted percentage outcomes for a male and female baby.

Answer: In case of an offspring the allele is inherited in each copies from their parent. The variation in the sex chromosome is observed in case of male as they have X and Y chromosome, so the nature of the offspring depends on the presence of XX chromosome which signifies female and XY chromosome which signifies male. So the individual’s sex is always determined by the presence of the Y chromosome. Thus in case of human the aneuploidy like 47,XXY and 47,XYY are determined as male and 45, X and 47, XXX are female. The sex-determining- region (SRY gene) is situated in the Y chromosome, thus the male who looks like male but have two X chromosome and one Y chromosome are considered to be male as the SRY gene is present in the Y chromosome. The diploid germ cell in the female after meiosis will give the gametes X and X in the egg and the diploid germ cell in male will give X and Y in the sperm, when they get fertilized the diploid offspring are produced either have XX or XY and the ratio is 1:1, so the chances of having male or female baby in 50% and always equal. (Bachtrog, et al, 2014)

The Punnett square is showing here to determine the sex of a baby by genetic inheritance:

Question 5a (1.1 & 1.4 autosomal dominant)

Give your explanation of how an autosomal dominant disease is inherited and the particular features of this form of inheritance pattern. (1.4)

From the family tree diagram below, what is the genotype of the children that are predicted to inherit Huntington’s Disease? (1.1)

Answer (i): Autosomal dominance is the inheritance pattern of some diseases, which is passed through the next generation in the autosome not by the sex chromosomal inheritance and dominant means only one copy is efficient for the disease to pass into the offspring and to get expressed. Huntington’s disease is the most common form of the autosomal disorder which is dominant in nature. In case of the autosomal disorders the rate of affected offspring for both male and female are same as this is not related with the sex chromosome. Autosomal inherited disease pattern do not skip the generation if it is a dominant one because if one parent is affected with the disease and if they are heterozygous then also the disease will be expressed and there will be definite chance of 50% to get inherit the disease gene. The main key feature of this autosomal and dominant inheritance pattern is thus no chance to skip the disease and the determination is easier, if the offspring get affected by the disease then one of the parent must have one copy of the faulty gene. By following the pedigree chart the dominant autosomal inheritance pattern can be determined. (McKusick, 2014)

(iv) answer: From the family tree diagram the genotype of the children having the chance to inherit Huntington’s disease are Hh (heterozygous) as the H is the dominant allele present in the female partner and h is not affected recessive allele present in the male partner. So the chance of Hh (affected heterozygous) and hh (not affected heterozygous) are same. (McKusick, 2014)

Question 5b (1.1 & 1.4 autosomal recessive)

From the pedigree diagram below, what is the genotype of the third generation child marked as ‘X’? (1.1)

Explain why the child marked as ‘X’ on the pedigree diagram below has inherited the cystic fibrosis disease whilst her parents that are carriers in the second generation are not affected. (1.4)

Answer (i): From the pedegree diagram the genotype of third generation child with X marking is cc (recessive homozygous), as the disease Cystic Fibrosis is autosomal recessive so both the copies have to be affected. The dominant trait is C and recessive here c, so the affectec child must have the genotype of cc. (McKusick, 2014)

(ii): This disease Cyctic fibrosis is an autosomal recessive one, so the disease allele is present in the autosome not in the sex chromosome and as it is a recessive disorder so both the copies of the gene, I mean both allele have to be faulty, then only the child get affected and the disease will be expressed in the next generation. If only one copy of the gene is present in the offspring then they are termed as carrier, that means they will carry the disease but will not be affected by it, only if they get married with partner who is also carrier then only there is 25% chance that the offspring will have the disease. This is what exactly happened with the case of X child in the third generation, whose grandmother was the carrier of having genotype Cc (heterozygous) but grandfather was not the carrier also, there children become carrier with the genotype Cc, who is the mother of the X child. The father of this child is also carrier whose geneotype is Cc, so there will be chance of 25% than the child will be affected, either male or female child as the disease is autosomal, here the girl X get affected as during the cross the genotype of one offsring will be cc (homozygous recessive). This kind of inheritance pattern generally skipps one generation and expressed in the next generation which is in contrast to the inheritence pattern of autosomal dominant pattern. (McKusick, 2014)

Question 6 (1.1 codominance, 2.2)

Answer: The colour of the far coat for cat is X chromosomal inheritance and the black colour Xᴮ is the black coat colour and the XG is the ginger colour, when the male cat with ginger cat mate with female black cat the inheritance pattern shows the codominance. So the genotype of the offspring will be 100% tortoise coat colouring female offspring (XᴮXᴳ)and the male offspring shows 100% of black coat colouring (XᴮY). As the inheritance is by the X chromosome so the male get the X chromosome from the mother cat so in all the cases they are black in colour. (Elston, et al, 2012)

Question 7 (1.4, 2.2)

Draw a family tree showing this couple’s probable outcomes for their male and female offspring. Identify on the family tree the genotypes and percentage possibility of the offspring being carriers of haemophilia, being haemophiliac, or being free of the condition.

Answer: The mother is the carrier of haemophilia (XXh) and the father is not having haemophilia (XY), so the inheritance pattern in the offspring will be: XX, XXh, XY, XhY, so 25% chance of girl child with no haemophilia, 25% girl as the carrier, 25% boy shows normal trait and 25% chance for boy having haemophilia. (Oldenburg, et al, 2010)

Explain what an ‘x-linked recessive’ inheritance pattern entails, and provide examples from your family tree in part a) to support your points. (1.4 & 2.2)

Answer: The X-linked recessive inheritance pattern means the disease that are associated with the X-chromosome will be transmitted through inheritance in the offspring. As the disorder is recessive in nature so both the allele have to be faulty for the offspring to get affected. If one of the X chromosome in the female offspring shows the faulty allele then if another X chromosome is not faulty, then the girl child will be a carrier one, who will carry the gene but not express it and to become affected the girl child must have two faulty gene in the two X chromosomes, as the inheritance is recessive in nature. For the boy child the scenario is different , they cannot be carries because is one of their X chromosome get the allele of haemophilia, then they will have the disease as the other chromosome is Y. So the chance of affected child is more for male than female. In the family tree also it is observed that 50% of boy child is affected and 50% non-affected, there is no carrier option, but in case of girl child 50% are carrier and 50% of them are normal. (Oldenburg, et al, 2010)

Question 8 (3.1)

There are two main causes of heritable variation; continuous and discontinuous. Explain in detail these two causes. Include in your explanation an account of recombination and mutation.

Answer: Heritable variation is also known as the variance that is the additional variable trait give rise during the inheritance of a certain trait. This variation is reflected as a small difference between two persons and they can be continuous and discontinuous as well. The discontinuous variation means when the individuals descent into the separate class or in terms of categories which is based on those features that are not measurable across the range which is complete. That means those individual belongs to the wide categories either have the trait or they do not have it. For example an individual human must have one blood group or another but not anything in between. These discrete data analysis of discontinuous variation can be calculated by chi-square test. The environmental factor is less in case of such variation. In contrast continuous variation is generated by the combined effect of a number of gene also termed as polygenic inheritance. In case of such variation type there will be a complete range of data starting from one end and ends to other end. They are significantly affected with the environmental factors. For example milk production by a cow not only depend and varies due to its genetic make-up but also affected by environment. All the data from this variation can be plotted as histogram and when fitted they give normal bell shaped distribution. (1000 Genomes Project Consortium,2015)

The cause of genetic variation are mainly gene mutation and the recombination. The mutations that occur into the human genetic make up are in the chromosome but most common form of mutation is in the gene and single gene mutation, this will alter the sequence of the gene and will give rise to different gene or codon. That will either produce faulty protein or may lead to inherited diseases. For example in case of sickle cell anaemia base substitution will cause the production of faulty amino acid that ultimately leads to disease. The recombination of homologous chromosome during cross over is another cause of genetic variation, which will produce recombinant chromosome. (1000 Genomes Project Consortium,2015)

Question 9 (3.2)

References

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