Principles Of Inheritance And Variation

To determine the number of heterozygous red-flowered plants in a population under Hardy-Weinberg equilibrium, we consider the allelic frequencies for the red and white flower color phenotypes.Let represent the frequency of the allele for the red color, and represent the frequency of the allele for the white color. Since the population is in Hardy-Weinberg equilibrium, the sum of these frequencies is 1: Given that 490 out of 1000 plants have white flowers (a recessive phenotype), we can deduce the frequency of the white allele, , from the expression for the homozygous recessive genotype: Thus, the frequency of the white allele is: From this, we calculate the frequency of the red allele as: The squared frequency of the homozygous dominant genotype (red) is: The frequency of the heterozygous genotype (red) is given by: The number of heterozygous red-flowered plants can then be calculated by multiplying this frequency by the total population: Therefore, there are 420 heterozygous red-flowered plants in the population.

From the Punnett square, it is clear that all embryos will have the Ww genotype. However, as the seed coat colour is determined solely by the maternal genes, the seed coat colour will be brown.

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As one parent is homozygous dominant QQ while the other is homozygous recessive qq , progeny will all be heterozygous Qq. Hence no amount of progeny has QQ alleles.

Let the allelic frequency of the normal gene be denoted by p and that of the IAT gene by q. In a Hardy-Weinberg Equilibrium, sum of allelic frequencies is 1, i.e.$ \mathbf{= 0 . 3} =0.7$.

In an island with a population of 10,000, four individuals have sickle cell anemia, a recessive autosomal disease. To determine how many individuals are heterozygous for the disease allele, we apply the Hardy-Weinberg equilibrium concepts.The frequency of alleles is denoted as follows:Let be the frequency of the normal (dominant) allele.Let be the frequency of the recessive allele.According to the Hardy-Weinberg principle, the sum of allele frequencies must equal 1: Moreover, the distribution of genotypes can be expressed as: Since sickle cell anemia is a recessive condition, affected individuals are homozygous recessive ( ).Calculating : Then, calculate : Next, determine the frequency of heterozygous individuals, given by : Thus, the number of heterozygous individuals is: Therefore, there are 392 individuals on the island expected to be heterozygous for the disease allele.

The genetic disorder described in the pedigree chart is recessive because it skips a generation. It is classified as autosomal since it affects both males and females equally. Since individual III-2 is affected, this indicates she must be homozygous for the recessive disorder, meaning she received the recessive allele from each parent. As a result, individual II-3 must be heterozygous, carrying one normal allele and one recessive allele.

The disease is autosomal as there is no father-to-son, i.e. male to male transfer. It is dominant as it is continuous across generations, not skipping any.

Soluton:Each F1 plant isPpQq, but thegamete proportions are not 1:1:1:1, soP and Q are linked.Cross 1 (♀ R1 × ♂ R2)Gametes produced:35%pQ35%Pq15%PQ15%pqThemost frequent gametes (parental types)arepQ and Pq, so the F1 has genes inrepulsion (trans) phase:Thus the true-breeding parents are:PPQQppqqSo,R3 = PPQQR4 = ppqqAnswer: ✔Option A

Given factsHoney bee males (drones) arehaploid (n).Honey bee females (workers and queens) arediploid (2n).This system is calledhaplodiploidy.Option analysisOption A:Honey bee males cannot have daughters but can have sons.Males are haploid and arise fromunfertilized eggs.Malesproduce sperm, but sperm is only used when afemale (queen)fertilizes her eggs.A male bee's sperm can fertilize eggs → producingfemale offspring (daughters).However, because malesdon’t lay eggs themselves, theycannot directly have either sons or daughters.Also,male bees never produce gametes that produce sons, becausesons come from unfertilized eggs laid by females, not from males.✅ Therefore, the statement“Honey bee males cannot have daughters but can have sons”isincorrect— in fact, males haveneither sons nor daughters.Option B:Honey bee males are produced from unfertilized eggs and females are produced from fertilized eggs.✅ This iscorrect— that’s the basis of haplodiploidy.Option C:A honey bee male does not have a father but has a grandfather.✅ True:A male comes from anunfertilized egg→ thusno father.The queen (his mother) is diploid, and she herself came from afertilized egg, so her sire (the male who fertilized her egg) is thegrandfatherof that male bee.Option D:Honey bee males form gametes by mitosis and females form gametes by meiosis.✅ True:Males are haploid, so to produce haploid sperm, they can only divide bymitosis—no further reduction division needed.Females are diploid; to produce haploid eggs, they usemeiosis.✅Final Answer:Theincorrect statementisOption A.